A Solution Manual For
ADVANCED ENGINEERING MATHEMATICS.
ERWIN KREYSZIG, HERBERT KREYSZIG,
EDWARD J. NORMINTON. 10th edition. John
Wiley USA. 2011
Nasser M. Abbasi
November 13, 2025
Compiled on November 13, 2025 at 7:40pm
Contents
1 Lookup tables for all problems in current book
5
2 Book Solved Problems
17
3
CONTENTS
4
Chapter
1
Lookup tables for all problems in
current book
Local contents
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page
174 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3.
Extended Power Series Method: Frobenius Method page 186 . . . . . . . . .
Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.4.
Bessels Equation page 195 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200 . . . . . . . . . . . . . . . . . . .
Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201 . . .
Chapter 6. Laplace Transforms. Problem set 6.2, page 216 . . . . . . . . . .
Chapter 6. Laplace Transforms. Problem set 6.3, page 224 . . . . . . . . . .
Chapter 6. Laplace Transforms. Problem set 6.4, page 230 . . . . . . . . . .
5
6
7
8
9
10
11
13
15
chapter 1 . lookup tables for all problems in current book
1.1
6
Chapter 5. Series Solutions of ODEs. Special
Functions. Problem set 5.1. page 174
Table 1.1: Lookup table for all problems in current section
ID
problem
8566
6
ODE
(x + 1) y 0 = y
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
7
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
Using series expansion around x = 0.
8567
7
y 0 = −2xy
Using series expansion around x = 0.
8568
8
y 0 x − 3y = k
Using series expansion around x = 0.
8569
9
y 00 + y = 0
Using series expansion around x = 0.
8570
10
xy − y 0 + y 00 = 0
Using series expansion around x = 0.
8571
11
y 00 − y 0 + x2 y = 0
Using series expansion around x = 0.
8572
12
−x2 + 1 y 00 − 2y 0 x + 2y = 0
Using series expansion around x = 0.
8573
13
y 00 + x2 + 1 y = 0
Using series expansion around x = 0.
8574
14
y 00 − 4y 0 x + 4x2 − 2 y = 0
Using series expansion around x = 0.
8575
16
y 0 + 4y = 1
5
y(0) =
4
Using series expansion around x = 0.
8576
17
y 00 + 3y 0 x + 2y = 0
y(0) = 1
y 0 (0) = 1
Using series expansion around x = 0.
Continued on next page
chapter 1 . lookup tables for all problems in current book
7
Table 1.1 Lookup table
Continued from previous page
ID
problem
ODE
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
−x2 + 1 y 00 − 2y 0 x + 30y = 0
8577
y(0) = 0
15
y 0 (0) =
8
18
Using series expansion around x = 0.
8578
(x − 2) y 0 = xy
y(0) = 4
19
Using series expansion around x = 0.
1.2
Chapter 5. Series Solutions of ODEs. Special
Functions. Problem set 5.3. Extended Power
Series Method: Frobenius Method page 186
Table 1.2: Lookup table for all problems in current section
ID
problem
8579
2
ODE
(x − 2)2 y 00 + (2 + x) y 0 − y = 0
Using series expansion around x = 0.
8580
3
y 00 x + 2y 0 + xy = 0
Using series expansion around x = 0.
8581
4
y 00 x + y = 0
Using series expansion around x = 0.
8582
5
y 00 x + (2x + 1) y 0 + (x + 1) y = 0
Using series expansion around x = 0.
8583
6
y 00 x + 2x3 y 0 + x2 − 2 y = 0
Using series expansion around x = 0.
8584
7
y 00 + (x − 1) y = 0
Using series expansion around x = 0.
8585
8
y 00 x + y 0 + xy = 0
Using series expansion around x = 0.
8586
9
2x(x − 1) y 00 − (x + 1) y 0 + y = 0
Using series expansion around x = 0.
Continued on next page
chapter 1 . lookup tables for all problems in current book
8
Table 1.2 Lookup table
Continued from previous page
ID
problem
8587
10
ODE
y 00 x + 2y 0 + 4xy = 0
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
Using series expansion around x = 0.
8588
11
y 00 x + (−2x + 2) y 0 + (x − 2) y = 0
Using series expansion around x = 0.
8589
12
x2 y 00 + 6y 0 x + 4x2 + 6 y = 0
Using series expansion around x = 0.
8590
13
y 00 x + (1 − 2x) y 0 + (x − 1) y = 0
Using series expansion around x = 0.
8591
15
2(1 − x) xy 00 − (1 + 6x) y 0 − 2y = 0
Using series expansion around x = 0.
00
8592
16
(1 − x) xy +
1
+ 2x y 0 − 2y = 0
2
Using series expansion around x = 0.
8593
17
4y 00 x + y 0 + 8y = 0
Using series expansion around x = 0.
8594
18
4 t2 − 3t + 2 y 00 − 2y 0 + y = 0
Using series expansion around t = 0.
8595
19
2 t2 − 5t + 6 y 00 + (2t − 3) y 0 − 8y = 0
Using series expansion around t = 0.
8596
20
3t(1 + t) y 00 + y 0 t − y = 0
Using series expansion around t = 0.
1.3
Chapter 5. Series Solutions of ODEs. Special
Functions. Problem set 5.4. Bessels Equation page
195
chapter 1 . lookup tables for all problems in current book
9
Table 1.3: Lookup table for all problems in current section
ID
problem
ODE
4
2
x y +yx+ x −
y=0
49
2 00
8597
2
0
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
Using series expansion around x = 0.
8598
y 00 x + y 0 +
3
y
=0
4
Using series expansion around x = 0.
1
−2x
y + e
−
y=0
9
00
8599
4
Using series expansion around x = 0.
8600
x + 34 y
xy +
=0
4
2 00
6
Using series expansion around x = 0.
8601
x2 y 00 + y 0 x +
7
(x2 − 1) y
=0
4
Using series expansion around x = 0.
8602
(2x + 1)2 y 00 + 2(2x + 1) y 0 + 16(x + 1) xy = 0
8
Using series expansion around x = 0.
1.4
Chapter 5. Series Solutions of ODEs. Special
Functions. Problem set 5.5. Bessel Functions Y(x).
General Solution page 200
Table 1.4: Lookup table for all problems in current section
ID
problem
8603
1
ODE
x2 y 00 + y 0 x + x2 − 6 y = 0
Solved? Maple
Mma
Sympy
3
3
3
7
3
3
3
3
3
3
3
3
3
3
3
3
Using series expansion around x = 0.
8604
2
y 00 x + 5y 0 + xy = 0
Using series expansion around x = 0.
8605
3
9x2 y 00 + 9y 0 x + 36x4 − 16 y = 0
Using series expansion around x = 0.
8606
4
y 00 + xy = 0
Using series expansion around x = 0.
Continued on next page
chapter 1 . lookup tables for all problems in current book
10
Table 1.4 Lookup table
Continued from previous page
ID
problem
8607
5
ODE
4y 00 x + 4y 0 + y = 0
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
7
Using series expansion around x = 0.
8608
y 00 x + y 0 + 36y = 0
6
Using series expansion around x = 0.
8609
y 00 + k 2 x2 y = 0
7
Using series expansion around x = 0.
8610
y 00 + k 2 x4 y = 0
8
Using series expansion around x = 0.
8611
y 00 x − 5y 0 + xy = 0
9
Using series expansion around x = 0.
1.5
Chapter 5. Series Solutions of ODEs. REVIEW
QUESTIONS. page 201
Table 1.5: Lookup table for all problems in current section
ID
problem
8612
11
ODE
4y + y 00 = 0
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
7
3
3
3
7
3
3
3
7
3
3
3
3
Using series expansion around x = 0.
8613
12
y 00 x + (1 − 2x) y 0 + (x − 1) y = 0
Using series expansion around x = 0.
8614
13
(x − 1)2 y 00 − (x − 1) y 0 − 35y = 0
Using series expansion around x = 0.
8615
14
16(x + 1)2 y 00 + 3y = 0
Using series expansion around x = 0.
8616
15
x2 y 00 + y 0 x + x2 − 5 y = 0
Using series expansion around x = 0.
8617
16
x2 y 00 + 2x3 y 0 + x2 − 2 y = 0
Using series expansion around x = 0.
Continued on next page
chapter 1 . lookup tables for all problems in current book
11
Table 1.5 Lookup table
Continued from previous page
ID
problem
8618
17
ODE
y − (x + 1) y 0 + y 00 x = 0
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
Using series expansion around x = 0.
8619
y 00 x + 3y 0 + 4x3 y = 0
18
Using series expansion around x = 0.
8620
y 00 +
19
y
=0
4x
Using series expansion around x = 0.
8621
y 00 x + y 0 − xy = 0
20
Using series expansion around x = 0.
1.6
Chapter 6. Laplace Transforms. Problem set 6.2,
page 216
Table 1.6: Lookup table for all problems in current section
ID
problem
ODE
y0 +
8622
1
26y
97 sin (2t)
=
5
5
y(0) = 0
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
Using Laplace transform method.
8623
2
y 0 + 2y = 0
3
y(0) =
2
Using Laplace transform method.
8624
3
y 00 − y 0 − 6y = 0
y(0) = 11
y 0 (0) = 28
Using Laplace transform method.
8625
4
y 00 + 9y = 10 e−t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
Continued on next page
chapter 1 . lookup tables for all problems in current book
12
Table 1.6 Lookup table
Continued from previous page
ID
problem
ODE
y
=0
4
y(0) = 12
y 0 (0) = 0
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
y 00 −
8626
5
Using Laplace transform method.
8627
6
y 00 − 6y 0 + 5y = 29 cos (2t)
16
y(0) =
5
31
y 0 (0) =
5
Using Laplace transform method.
8628
7
y 00 + 7y 0 + 12y = 21 e3t
7
y(0) =
2
0
y (0) = −10
Using Laplace transform method.
8629
8
y 00 − 4y 0 + 4y = 0
81
y(0) =
10
39
y 0 (0) =
10
Using Laplace transform method.
8630
9
y 00 − 4y 0 + 3y = 6t − 8
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
y
t2
=
25
50
y(0) = −25
y 0 (0) = 0
y 00 +
8631
10
Using Laplace transform method.
9y
= 9t3 + 64
4
y(0) = 1
63
y 0 (0) =
2
y 00 + 3y 0 +
8632
11
Using Laplace transform method.
Continued on next page
chapter 1 . lookup tables for all problems in current book
13
Table 1.6 Lookup table
Continued from previous page
ID
problem
8633
ODE
y 00 − 2y 0 − 3y = 0
y(4) = −3
y 0 (4) = −17
12
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
Using Laplace transform method.
8634
y 0 − 6y = 0
y(−1) = 4
13
Using Laplace transform method.
8635
y 00 + 2y 0 + 5y = 50t − 100
y(2) = −4
y 0 (2) = 14
14
Using Laplace transform method.
8636
y 00 + 3y 0 − 4y = 6 e2t−3
3
y
=4
2
0 3
y
=5
2
15
Using Laplace transform method.
1.7
Chapter 6. Laplace Transforms. Problem set 6.3,
page 224
Table 1.7: Lookup table for all problems in current section
ID
8637
problem
18
ODE
9y 00 − 6y 0 + y = 0
y(0) = 3
y 0 (0) = 1
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
Using Laplace transform method.
8638
19
y 00 + 6y 0 + 8y = e−3t − e−5t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
8639
20
y 00 + 10y 0 + 24y = 144t2
19
y(0) =
12
y 0 (0) = −5
Using Laplace transform method.
Continued on next page
chapter 1 . lookup tables for all problems in current book
14
Table 1.7 Lookup table
Continued from previous page
ID
problem
ODE
00
y + 9y =
8640
21
8 sin (t) 0 < t < π
0
π<t
y(0) = 0
y 0 (0) = 4
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
7
3
3
3
7
3
3
3
7
3
3
3
3
3
3
3
7
3
3
3
3
Using Laplace transform method.
00
0
y + 3y + 2y =
8641
22
4t 0 < t < 1
8
1<t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
00
0
y + y − 2y =
8642
23
3 sin (t) − cos (t) 0 < t < 2π
3 sin (2t) − cos (2t)
2π < t
y(0) = 1
y 0 (0) = 0
Using Laplace transform method.
00
0
y + 3y + 2y =
8643
24
1 0<t<1
0
1<t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
00
y +y =
8644
25
t 0<t<1
0
1<t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
00
0
y + 2y + 5y =
8645
26
10 sin (t) 0 < t < 2π
0
2π < t
y(π) = 1
y 0 (π) = 2 e−π − 2
Using Laplace transform method.
00
y + 4y =
8646
27
8t2 0 < t < 5
0
5<t
y(1) = 1 + cos (2)
y 0 (1) = 4 − 2 sin (2)
Using Laplace transform method.
chapter 1 . lookup tables for all problems in current book
1.8
15
Chapter 6. Laplace Transforms. Problem set 6.4,
page 230
Table 1.8: Lookup table for all problems in current section
ID
8647
problem
ODE
y 00 + 4y = δ(t − π)
y(0) = 8
y 0 (0) = 0
3
Solved? Maple
Mma
Sympy
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
7
3
Using Laplace transform method.
8648
y 00 + 16y = 4δ(t − 3π)
y(0) = 2
y 0 (0) = 0
4
Using Laplace transform method.
8649
y 00 + y = δ(t − π) − δ(t − 2π)
y(0) = 0
y 0 (0) = 1
5
Using Laplace transform method.
8650
y 00 + 4y 0 + 5y = δ(t − 1)
y(0) = 0
y 0 (0) = 3
6
Using Laplace transform method.
00
0
−t
4y + 24y + 37y = 17 e
8651
7
1
+δ t−
2
y(0) = 1
y 0 (0) = 1
Using Laplace transform method.
8652
8
y 00 + 3y 0 + 2y = 10 sin (t) + 10δ(t − 1)
y(0) = 1
y 0 (0) = −1
Using Laplace transform method.
8653
9
y 00 + 4y 0 + 5y = (1 − Heaviside (t − 10)) et − e10 δ(t − 10)
y(0) = 0
y 0 (0) = 1
Using Laplace transform method.
Continued on next page
chapter 1 . lookup tables for all problems in current book
16
Table 1.8 Lookup table
Continued from previous page
ID
8654
problem
10
ODE
π
y 00 + 5y 0 + 6y = δ t −
+ cos (t) Heaviside (t − π)
2
y(0) = 0
y 0 (0) = 0
Solved? Maple
Mma
Sympy
3
3
7
3
3
3
3
3
3
3
3
3
Using Laplace transform method.
8655
11
y 00 + 5y 0 + 6y = Heaviside (t − 1) + δ(t − 2)
y(0) = 0
y 0 (0) = 1
Using Laplace transform method.
8656
12
y 00 + 2y 0 + 5y = 25t − 100δ(t − π)
y(0) = −2
y 0 (0) = 5
Using Laplace transform method.
Chapter
2
Book Solved Problems
Local contents
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page
174 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3.
Extended Power Series Method: Frobenius Method page 186 . . . . . . . . .
Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.4.
Bessels Equation page 195 . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200 . . . . . . . . . . . . . . . . . . .
Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201 . . .
Chapter 6. Laplace Transforms. Problem set 6.2, page 216 . . . . . . . . . .
Chapter 6. Laplace Transforms. Problem set 6.3, page 224 . . . . . . . . . .
Chapter 6. Laplace Transforms. Problem set 6.4, page 230 . . . . . . . . . .
17
18
100
255
305
374
454
508
547
chapter 2 . book solved problems
2.1
18
Chapter 5. Series Solutions of ODEs. Special
Functions. Problem set 5.1. page 174
Local contents
2.1.1 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.2 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.3 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.4 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.5 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.6 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.7 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.8 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.9 Problem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.10 Problem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.11 Problem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.12 Problem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1.13 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
25
31
34
40
46
53
60
66
73
79
86
94
chapter 2 . book solved problems
2.1.1
19
Problem 6
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
24
24
Internal problem ID [8566]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 6
Date solved : Tuesday, November 11, 2025 at 12:03:41 PM
CAS classification : [_separable]
(x + 1) y 0 = y
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving first order ode. Let
y 0 = f (x, y)
Where f (x, y) is analytic at expansion point x0 . We can always shift to x0 = 0 if x0 is not
zero. So from now we assume x0 = 0 . Assume also that y(x0 ) = y0 . Using Taylor series
(x − x0 )2 00
(x − x0 )3 00
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2 df
x3 d 2 f
= y0 + xf +
+
+ ···
2 dx x0 ,y0 3! dx2 x0 ,y0
∞
X
xn+1 dn f
= y0 +
(n + 1) ! dxn x0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
But
df
∂f
∂f
=
+
f
dx
∂x ∂y
d2 f
d df
=
dx2
dx dx
∂ df
∂ df
=
+
f
∂x dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
=
+
f
∂x dx2
∂y dx2
..
.
(1)
(2)
(3)
And so on. Hence if we name F0 = f (x, y) then the above can be written as
F0 = f (x, y)
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
=
Fn−1 +
F0
∂x
∂y
(4)
(5)
chapter 2 . book solved problems
20
For example, for n = 1 we see that
d
(F0 )
dx
∂
∂F0
=
F0 +
F0
∂x
∂y
∂f
∂f
=
+
f
∂x ∂y
F1 =
Which is (1). And when n = 2
d
(F1 )
dx
∂
∂F1
=
F1 +
F0
∂x
∂y
∂ ∂f
∂f
∂ ∂f
∂f
=
+
f +
+
f f
∂x ∂x ∂y
∂y ∂x ∂y
∂ df
∂ df
=
+
f
∂x dx
∂y dx
F2 =
Which is (2) and so on. Therefore (4,5) can be used from now on along with
y(x) = y0 +
∞
X
xn+1
Fn |x0 ,y0
(n
+
1)
!
n=0
Hence
y
x+1
dF0
F1 =
dx
∂F0 ∂F0
=
+
F0
∂x
∂y
=0
dF1
F2 =
dx
∂F1 ∂F1
=
+
F1
∂x
∂y
=0
dF2
F3 =
dx
∂F2 ∂F2
=
+
F2
∂x
∂y
=0
dF3
F4 =
dx
∂F3 ∂F3
=
+
F3
∂x
∂y
=0
F0 =
And so on. Evaluating all the above at initial conditions x(0) = 0 and y(0) = y(0) gives
F0 = y(0)
F1 = 0
F2 = 0
F3 = 0
F4 = 0
Substituting all the above in (6) and simplifying gives the solution as
y = (x + 1) y(0) + O x6
(6)
chapter 2 . book solved problems
21
Since x = 0 is also an ordinary point, then standard power series can also be used. Writing
the ODE as
y 0 + q(x)y = p(x)
y
y0 −
=0
x+1
Where
q(x) = −
1
x+1
p(x) = 0
Next, the type of the expansion point x = 0 is determined. This point can be an ordinary
point, a regular singular point (also called removable singularity), or irregular singular point
(also called non-removable singularity or essential singularity). When x = 0 is an ordinary
point, then the standard power series is used. If the point is a regular singular point, Frobenius
series is used instead. Irregular singular point requires more advanced methods (asymptotic
methods) and is not supported now. Hopefully this will be added in the future. x = 0 is
called an ordinary point q(x) has a Taylor series expansion around the point x = 0. x = 0
is called a regular singular point if q(x) is not not analytic at x = 0 but xq(x) has Taylor
series expansion. And finally, x = 0 is an irregular singular point if the point is not ordinary
and not regular singular. This is the most complicated case. Now the expansion point x = 0
is checked to see if it is an ordinary point or not. Let the solution be represented as power
series of the form
∞
X
y=
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
Substituting the above back into the ode gives
(x + 1)
∞
X
!
nan x
n−1
=
n=1
∞
X
an x n
(1)
n=0
Which simplifies to
∞
X
!
nan x
n
+
n=1
∞
X
!
nan x
n−1
+
n=1
∞
X
(−an xn ) = 0
(2)
n =0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
nan xn−1 =
n =1
∞
X
(n + 1) an+1 xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
n=1
!
nan xn
+
∞
X
n=0
!
(n + 1) an+1 xn
+
∞
X
(−an xn ) = 0
(3)
n =0
For 1 ≤ n, the recurrence equation is
nan + (n + 1) an+1 − an = 0
(4)
chapter 2 . book solved problems
22
Solving for an+1 , gives
an+1 = −
an (n − 1)
n+1
(5)
For n = 1 the recurrence equation gives
2a2 = 0
Which after substituting the earlier terms found becomes
a2 = 0
For n = 2 the recurrence equation gives
a2 + 3a3 = 0
Which after substituting the earlier terms found becomes
a3 = 0
For n = 3 the recurrence equation gives
2a3 + 4a4 = 0
Which after substituting the earlier terms found becomes
a4 = 0
For n = 4 the recurrence equation gives
3a4 + 5a5 = 0
Which after substituting the earlier terms found becomes
a5 = 0
For n = 5 the recurrence equation gives
4a5 + 6a6 = 0
Which after substituting the earlier terms found becomes
a6 = 0
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
y = a0 x + a0 + . . .
Collecting terms, the solution becomes
y = (x + 1) a0 + O x6
(3)
chapter 2 . book solved problems
23
Figure 2.1: Slope field (x + 1) y 0 = y
3 Maple. Time used: 0.004 (sec). Leaf size: 10
Order:=6;
ode:=(x+1)*diff(y(x),x) = y(x);
dsolve(ode,y(x),type='series',x=0);
y = y(0) (1 + x)
Maple trace
Methods for first order ODEs:
--- Trying classification methods --trying a quadrature
trying 1st order linear
<- 1st order linear successful
Maple step by step
•
•
•
Let’s solve
d
(1 + x) dx
y(x) = y(x)
Highest derivative means the order of the ODE is 1
d
y(x)
dx
Solve for the highest derivative
d
y(x) = y(x)
dx
1+x
Separate variables
d
y(x)
dx
1
= 1+x
Integrate both sides with respect to x
R dxd y(x)
R 1
dx
=
dx + C 1
y(x)
1+x
Evaluate integral
ln (y(x)) = ln (1 + x) + C 1
Solve for y(x)
y(x) = eC 1 (1 + x)
Redefine the integration constant(s)
y(x) = C 1 (1 + x)
y(x)
•
•
•
•
chapter 2 . book solved problems
24
3 Mathematica. Time used: 0.001 (sec). Leaf size: 9
ode=(1+x)*D[y[x],x]==y[x];
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1 (x + 1)
3 Sympy. Time used: 0.143 (sec). Leaf size: 10
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((x + 1)*Derivative(y(x), x) - y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="1st_power_series",x0=0,n=6)
y(x) = C1 + C1 x + O x6
chapter 2 . book solved problems
2.1.2
25
Problem 7
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
30
30
Internal problem ID [8567]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 7
Date solved : Tuesday, November 11, 2025 at 12:03:41 PM
CAS classification : [_separable]
y 0 = −2yx
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving first order ode. Let
y 0 = f (x, y)
Where f (x, y) is analytic at expansion point x0 . We can always shift to x0 = 0 if x0 is not
zero. So from now we assume x0 = 0 . Assume also that y(x0 ) = y0 . Using Taylor series
(x − x0 )2 00
(x − x0 )3 00
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2 df
x3 d 2 f
= y0 + xf +
+
+ ···
2 dx x0 ,y0 3! dx2 x0 ,y0
∞
X
xn+1 dn f
= y0 +
(n + 1) ! dxn x0 ,y0
n=0
0
But
df
∂f
∂f
=
+
f
dx
∂x ∂y
d2 f
d df
=
dx2
dx dx
∂ df
∂ df
=
+
f
∂x dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
=
+
f
∂x dx2
∂y dx2
..
.
(1)
(2)
(3)
And so on. Hence if we name F0 = f (x, y) then the above can be written as
F0 = f (x, y)
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
=
Fn−1 +
F0
∂x
∂y
(4)
(5)
chapter 2 . book solved problems
26
For example, for n = 1 we see that
d
(F0 )
dx
∂
∂F0
=
F0 +
F0
∂x
∂y
∂f
∂f
=
+
f
∂x ∂y
F1 =
Which is (1). And when n = 2
d
(F1 )
dx
∂
∂F1
=
F1 +
F0
∂x
∂y
∂ ∂f
∂f
∂ ∂f
∂f
=
+
f +
+
f f
∂x ∂x ∂y
∂y ∂x ∂y
∂ df
∂ df
=
+
f
∂x dx
∂y dx
F2 =
Which is (2) and so on. Therefore (4,5) can be used from now on along with
∞
X
xn+1
y(x) = y0 +
Fn |x0 ,y0
(n + 1) !
n=0
Hence
F0 = −2yx
dF0
F1 =
dx
∂F0 ∂F0
=
+
F0
∂x
∂y
= 4x2 − 2 y
dF1
F2 =
dx
∂F1 ∂F1
=
+
F1
∂x
∂y
= −8x3 + 12x y
dF2
F3 =
dx
∂F2 ∂F2
=
+
F2
∂x
∂y
= 4y 4x4 − 12x2 + 3
dF3
F4 =
dx
∂F3 ∂F3
=
+
F3
∂x
∂y
15
4
2
= −32y x − 5x +
x
4
And so on. Evaluating all the above at initial conditions x(0) = 0 and y(0) = y(0) gives
F0 = 0
F1 = −2y(0)
F2 = 0
F3 = 12y(0)
F4 = 0
(6)
chapter 2 . book solved problems
27
Substituting all the above in (6) and simplifying gives the solution as
1 4
2
y = 1 − x + x y(0) + O x6
2
Since x = 0 is also an ordinary point, then standard power series can also be used. Writing
the ODE as
y 0 + q(x)y = p(x)
y 0 + 2yx = 0
Where
q(x) = 2x
p(x) = 0
Next, the type of the expansion point x = 0 is determined. This point can be an ordinary
point, a regular singular point (also called removable singularity), or irregular singular point
(also called non-removable singularity or essential singularity). When x = 0 is an ordinary
point, then the standard power series is used. If the point is a regular singular point, Frobenius
series is used instead. Irregular singular point requires more advanced methods (asymptotic
methods) and is not supported now. Hopefully this will be added in the future. x = 0 is
called an ordinary point q(x) has a Taylor series expansion around the point x = 0. x = 0
is called a regular singular point if q(x) is not not analytic at x = 0 but xq(x) has Taylor
series expansion. And finally, x = 0 is an irregular singular point if the point is not ordinary
and not regular singular. This is the most complicated case. Now the expansion point x = 0
is checked to see if it is an ordinary point or not. Let the solution be represented as power
series of the form
∞
X
y=
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
Substituting the above back into the ode gives
∞
X
nan xn−1 = −2
n=1
∞
X
!
an x n
x
(1)
n=0
Which simplifies to
∞
X
!
nan x
n−1
+
n=1
∞
X
!
2an x
1+n
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
nan x
n−1
=
n =1
∞
X
n =0
∞
X
(1 + n) a1+n xn
n=0
2an x1+n =
∞
X
2an−1 xn
n=1
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
!
!
∞
∞
X
X
(1 + n) a1+n xn +
2an−1 xn = 0
(3)
n=0
n=1
chapter 2 . book solved problems
28
For 1 ≤ n, the recurrence equation is
(1 + n) a1+n + 2an−1 = 0
(4)
Solving for a1+n , gives
a1+n = −
2an−1
1+n
For n = 1 the recurrence equation gives
2a2 + 2a0 = 0
Which after substituting the earlier terms found becomes
a2 = −a0
For n = 2 the recurrence equation gives
3a3 + 2a1 = 0
Which after substituting the earlier terms found becomes
a3 = 0
For n = 3 the recurrence equation gives
4a4 + 2a2 = 0
Which after substituting the earlier terms found becomes
a4 =
a0
2
For n = 4 the recurrence equation gives
5a5 + 2a3 = 0
Which after substituting the earlier terms found becomes
a5 = 0
For n = 5 the recurrence equation gives
6a6 + 2a4 = 0
Which after substituting the earlier terms found becomes
a6 = −
a0
6
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
(5)
chapter 2 . book solved problems
29
Substituting the values for an found above, the solution becomes
1
y = a0 − a0 x 2 + a0 x 4 + . . .
2
Collecting terms, the solution becomes
y=
1 4
1 − x + x a0 + O x 6
2
2
(3)
Figure 2.2: Slope field y 0 = −2yx
3 Maple. Time used: 0.003 (sec). Leaf size: 24
Order:=6;
ode:=diff(y(x),x) = -2*x*y(x);
dsolve(ode,y(x),type='series',x=0);
1 4
2
y = 1 − x + x y(0) + O x6
2
Maple trace
Methods for first order ODEs:
--- Trying classification methods --trying a quadrature
trying 1st order linear
<- 1st order linear successful
Maple step by step
•
•
•
Let’s solve
d
y(x) = −2xy(x)
dx
Highest derivative means the order of the ODE is 1
d
y(x)
dx
Solve for the highest derivative
d
y(x) = −2xy(x)
dx
Separate variables
d
y(x)
dx
y(x)
= −2x
chapter 2 . book solved problems
•
•
•
•
30
Integrate both sides with respect to x
R dxd y(x)
R
dx = −2xdx + C 1
y(x)
Evaluate integral
ln (y(x)) = −x2 + C 1
Solve for y(x)
2
y(x) = e−x +C 1
Redefine the integration constant(s)
2
y(x) = C 1 e−x
3 Mathematica. Time used: 0.001 (sec). Leaf size: 20
ode=D[y[x],x]==-2*x*y[x];
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
4
x
2
y(x) → c1
−x +1
2
3 Sympy. Time used: 0.140 (sec). Leaf size: 19
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(2*x*y(x) + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="1st_power_series",x0=0,n=6)
y(x) = C1 − C1 x2 +
C 1 x4
+ O x6
2
chapter 2 . book solved problems
2.1.3
31
Problem 8
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
33
33
Internal problem ID [8568]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 8
Date solved : Tuesday, November 11, 2025 at 12:03:41 PM
CAS classification : [_separable]
y 0 x − 3y = k
Using series expansion around x = 0.
Since this is an inhomogeneous, then let the solution be
y = yh + yp
Where yh is the solution to the homogeneous ode y 0 x − 3y = 0,and yp is a particular solution
to the inhomogeneous ode. First, we solve for yh Let the homogeneous solution be represented
as Frobenius power series of the form
y=
∞
X
an xn+r
n=0
Then
y0 =
∞
X
(n + r) an xn+r−1
n=0
Substituting the above back into the ode gives
∞
X
!
(n + r) an x
n+r−1
x−3
n=0
∞
X
!
an x
n+r
=0
(1)
n=0
Which simplifies to
∞
X
!
an x
n+r
(n + r)
+
n=0
∞
X
−3an xn+r = 0
(2A)
n =0
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives Substituting all the above in Eq (2A) gives the
following equation where now all powers of x are the same and equal to n + r.
∞
X
n=0
!
an x
n+r
(n + r)
+
∞
X
−3an xn+r = 0
n =0
The indicial equation is obtained from n = 0. From Eq (2) this gives
an xn+r (n + r) − 3an xn+r = 0
(2B)
chapter 2 . book solved problems
32
When n = 0 the above becomes
a0 xr r − 3a0 xr = 0
The corresponding balance equation is found by replacing r by m and a by c to avoid confusing
terms between particular solution and the homogeneous solution. Hence the balance equation
is
(xm m − 3xm ) c0 = k
This equation will used later to find the particular solution.
Since a0 6= 0 then the indicial equation becomes
(r − 3) xr = 0
Since the above is true for all x then the indicial equation simplifies to
r−3=0
Solving for r gives the root of the indicial equation as
r=3
From the above we see that there is no recurrence relation since there is only one summation
term. Therefore all an terms are zero except for a0 . Hence
yh = a0 x3 + O x6
Now the particular solution is found Solving the balance equation (xm m − 3xm ) c0 = k for
[c0 , m] gives
c0 = −
k
3
m=0
Hence the particular solution is
y p = c0 x m
k
=−
3
The solution is
y = yh + yp
y = c1 x 3 + O x 6
−
k
3
3 Maple. Time used: 0.025 (sec). Leaf size: 24
Order:=6;
ode:=diff(y(x),x)*x-3*y(x) = k;
dsolve(ode,y(x),type='series',x=0);
3
y = c1 x 1 + O x
Maple trace
6
k
6
+ − +O x
3
chapter 2 . book solved problems
33
Methods for first order ODEs:
--- Trying classification methods --trying a quadrature
trying 1st order linear
<- 1st order linear successful
Maple step by step
•
•
•
Let’s solve
d
x dx
y(x) − 3y(x) = k
Highest derivative means the order of the ODE is 1
d
y(x)
dx
Solve for the highest derivative
d
y(x) = 3y(x)+k
dx
x
Separate variables
d
y(x)
dx
= x1
Integrate both sides with respect to x
R dxd y(x)
R
dx = x1 dx + C 1
3y(x)+k
Evaluate integral
ln(3y(x)+k)
= ln (x) + C 1
3
Solve for y(x)
3C 1 3
y(x) = e 3 x − k3
Redefine the integration constant(s)
y(x) = C 1 x3 − k3
3y(x)+k
•
•
•
•
3 Mathematica. Time used: 0.005 (sec). Leaf size: 15
ode=x*D[y[x],x]-3*y[x]==k;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
k
y(x) → − + c1 x3
3
7 Sympy
from sympy import *
x = symbols("x")
k = symbols("k")
y = Function("y")
ode = Eq(-k + x*Derivative(y(x), x) - 3*y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="1st_power_series",x0=0,n=6)
ValueError : ODE -k + x*Derivative(y(x), x) - 3*y(x) does not match hint 1
st_power_series
chapter 2 . book solved problems
2.1.4
34
Problem 9
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
39
39
Internal problem ID [8569]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 9
Date solved : Tuesday, November 11, 2025 at 12:03:42 PM
CAS classification : [[_2nd_order, _missing_x]]
y 00 + y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.1)
(2.2)
(2)
(3)
chapter 2 . book solved problems
35
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = −y
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −y 0
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
=y
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= y0
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= −y
(7)
chapter 2 . book solved problems
36
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = −y(0)
F1 = −y 0 (0)
F2 = y(0)
F3 = y 0 (0)
F4 = −y(0)
Substituting all the above in (7) and simplifying gives the solution as
1 2
1 4
1 3
1 5 0
y = 1 − x + x y(0) + x − x +
x y (0) + O x6
2
24
6
120
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
∞
X
n(n − 1) an xn−2 +
n=2
∞
X
!
an x n
=0
(1)
=0
(2)
n=0
Which simplifies to
∞
X
!
n(n − 1) an xn−2
+
n=2
∞
X
!
an x n
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
(n + 2) an+2 (n + 1) x
n=0
n
+
∞
X
!
an x
n
=0
(3)
n=0
For 0 ≤ n, the recurrence equation is
(n + 2) an+2 (n + 1) + an = 0
(4)
Solving for an+2 , gives
an+2 = −
an
(n + 2) (n + 1)
(5)
chapter 2 . book solved problems
37
For n = 0 the recurrence equation gives
2a2 + a0 = 0
Which after substituting the earlier terms found becomes
a2 = −
a0
2
For n = 1 the recurrence equation gives
6a3 + a1 = 0
Which after substituting the earlier terms found becomes
a3 = −
a1
6
For n = 2 the recurrence equation gives
12a4 + a2 = 0
Which after substituting the earlier terms found becomes
a0
24
a4 =
For n = 3 the recurrence equation gives
20a5 + a3 = 0
Which after substituting the earlier terms found becomes
a5 =
a1
120
For n = 4 the recurrence equation gives
30a6 + a4 = 0
Which after substituting the earlier terms found becomes
a6 = −
a0
720
For n = 5 the recurrence equation gives
42a7 + a5 = 0
Which after substituting the earlier terms found becomes
a7 = −
a1
5040
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
chapter 2 . book solved problems
38
Substituting the values for an found above, the solution becomes
1
1
1
1
y = a0 + a1 x − a0 x 2 − a1 x 3 + a0 x 4 +
a1 x 5 + . . .
2
6
24
120
Collecting terms, the solution becomes
1 2
1 4
1 3
1 5
y = 1 − x + x a0 + x − x +
x a1 + O x 6
2
24
6
120
(3)
At x = 0 the solution above becomes
1 2
1 4
1 3
1 5 0
y = 1 − x + x y(0) + x − x +
x y (0) + O x6
2
24
6
120
Figure 2.3: Slope field y 00 + y = 0
3 Maple. Time used: 0.008 (sec). Leaf size: 39
Order:=6;
ode:=y(x)+diff(diff(y(x),x),x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2
1 4
1 3
1 5 0
y = 1 − x + x y(0) + x − x +
x y (0) + O x6
2
24
6
120
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful
Maple step by step
•
Let’s solve
d d
y(x) + y(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
chapter 2 . book solved problems
•
•
Characteristic polynomial of ODE
r2 + 1 = 0
Use quadratic
formula to solve for r
√ 0±
•
•
•
•
•
39
−4
r=
2
Roots of the characteristic polynomial
r = (−I, I)
1st solution of the ODE
y1 (x) = cos (x)
2nd solution of the ODE
y2 (x) = sin (x)
General solution of the ODE
y(x) = C 1 y1 (x) + C 2 y2 (x)
Substitute in solutions
y(x) = C 1 cos (x) + C 2 sin (x)
3 Mathematica. Time used: 0.001 (sec). Leaf size: 42
ode=D[y[x],{x,2}]+y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
4
x5
x3
x
x2
−
+ x + c1
−
+1
120
6
24
2
3 Sympy. Time used: 0.164 (sec). Leaf size: 29
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(y(x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
4
x
x2
x2
y(x) = C2
−
+ 1 + C1 x 1 −
+ O x6
24
2
6
chapter 2 . book solved problems
2.1.5
40
Problem 10
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
45
45
Internal problem ID [8570]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 10
Date solved : Tuesday, November 11, 2025 at 12:03:42 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 − y 0 + yx = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.3)
(2.4)
(2)
(3)
chapter 2 . book solved problems
41
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = y 0 − yx
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= (1 − x) y 0 − (x + 1) y
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= (−2x − 1) y 0 + y x2 − x − 1
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= x2 − 3x − 4 y 0 + y 2x2 + 3x − 1
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= 3x2 + 2x − 8 y 0 − y x3 − 3x2 − 8x − 3
(7)
chapter 2 . book solved problems
42
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = y 0 (0)
F1 = y 0 (0) − y(0)
F2 = −y 0 (0) − y(0)
F3 = −4y 0 (0) − y(0)
F4 = −8y 0 (0) + 3y(0)
Substituting all the above in (7) and simplifying gives the solution as
y=
1 3
1 4
1 5
1 2 1 3
1 4
1 5 0
1− x − x −
x y(0) + x + x + x − x − x y (0) + O x6
6
24
120
2
6
24
30
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
n(n − 1) an xn−2 −
nan xn−1 +
n=2
n=1
∞
X
!
an x n x = 0
(1)
n=0
Which simplifies to
∞
X
∞
X
!
n(n − 1) an xn−2
+
n=2
−nan x
n−1
n =1
+
∞
X
!
x1+n an
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (1 + n) xn
n=0
∞
X
−nan x
n−1
=
n =1
∞
X
(−(1 + n) a1+n xn )
n=0
∞
X
x1+n an =
n =0
∞
X
an−1 xn
n=1
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
n=0
!
(n + 2) an+2 (1 + n) xn
+
∞
X
n =0
(−(1 + n) a1+n xn ) +
∞
X
n=1
!
an−1 xn
=0
(3)
chapter 2 . book solved problems
43
n = 0 gives
2a2 − a1 = 0
a2 =
a1
2
For 1 ≤ n, the recurrence equation is
(n + 2) an+2 (1 + n) − (1 + n) a1+n + an−1 = 0
(4)
Solving for an+2 , gives
an+2 =
=
na1+n + a1+n − an−1
(n + 2) (1 + n)
a1+n
an−1
−
n + 2 (n + 2) (1 + n)
For n = 1 the recurrence equation gives
6a3 − 2a2 + a0 = 0
Which after substituting the earlier terms found becomes
a3 =
a1 a0
−
6
6
For n = 2 the recurrence equation gives
12a4 − 3a3 + a1 = 0
Which after substituting the earlier terms found becomes
a4 = −
a1
a0
−
24 24
For n = 3 the recurrence equation gives
20a5 − 4a4 + a2 = 0
Which after substituting the earlier terms found becomes
a5 = −
a1
a0
−
30 120
For n = 4 the recurrence equation gives
30a6 − 5a5 + a3 = 0
Which after substituting the earlier terms found becomes
a6 = −
a1
a0
+
90 240
For n = 5 the recurrence equation gives
42a7 − 6a6 + a4 = 0
(5)
chapter 2 . book solved problems
44
Which after substituting the earlier terms found becomes
a7 = −
a1
a0
+
1680 630
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
y = a0 + a1 x +
a1 x 2 a1 a0 3 a1
a0 4 a1
a0 5
+
−
x + − −
x + − −
x + ...
2
6
6
24 24
30 120
Collecting terms, the solution becomes
1 3
1 4
1 5
1 2 1 3
1 4
1 5
y = 1− x − x −
x a0 + x + x + x − x − x a1 + O x6 (3)
6
24
120
2
6
24
30
At x = 0 the solution above becomes
y=
1 3
1 4
1 5
1 2 1 3
1 4
1 5 0
1− x − x −
x y(0) + x + x + x − x − x y (0) + O x6
6
24
120
2
6
24
30
3 Maple. Time used: 0.007 (sec). Leaf size: 54
Order:=6;
ode:=x*y(x)-diff(y(x),x)+diff(diff(y(x),x),x) = 0;
dsolve(ode,y(x),type='series',x=0);
y=
1 3
1 4
1 5
1 2 1 3
1 4
1 5 0
1− x − x −
x y(0) + x + x + x − x − x y (0) + O x6
6
24
120
2
6
24
30
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
Let’s solve
d d
d
y(x) − dx
y(x) + xy(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
chapter 2 . book solved problems
45
d d
y(x)
dx dx
•
Assume series solution for y(x)
∞
P
y(x) =
ak x k
Rewrite ODE with series expansions
Convert x · y(x) to series expansion
∞
P
x · y(x) =
ak xk+1
k=0
◦
k=0
◦
Shift index using k− >k − 1
∞
P
x · y(x) =
ak−1 xk
◦
d
Convert dx
y(x) to series expansion
∞
P
d
y(x) =
ak k xk−1
dx
◦
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + 1) xk
dx
◦
d d
Convert dx
y(x) to series expansion
dx
∞
P
d d
y(x) =
ak k(k − 1) xk−2
dx dx
k=1
k=1
k=0
k=2
◦
Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=0
Rewrite ODE
∞with series expansions
P
k
2a2 − a1 +
(ak+2 (k + 2) (k + 1) − ak+1 (k + 1) + ak−1 ) x = 0
k=1
•
•
•
•
Each term must be 0
2a2 − a1 = 0
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 − ak+1 k + ak−1 − ak+1 = 0
Shift index using k− >k + 1
(k + 1)2 + 3k + 5 ak+3 − ak+2 (k + 1) + ak − ak+2 = 0
Recursion
relation that defines the series solution tothe ODE
∞
P
−ak +2ak+2
y(x) =
ak xk , ak+3 = kak+2
, 2a2 − a1 = 0
k2 +5k+6
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 63
ode=D[y[x],{x,2}]-D[y[x],x]+x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
5
x5
x4 x3
x
x4 x3 x2
y(x) → c1 −
−
−
+ 1 + c2 − −
+
+
+x
120 24
6
30 24
6
2
3 Sympy. Time used: 0.204 (sec). Leaf size: 37
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*y(x) - Derivative(y(x), x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
4
3
x
x3
x
x2 x
y(x) = C2 − −
+ 1 + C1 x − +
+ + 1 + O x6
24
6
24
6
2
chapter 2 . book solved problems
2.1.6
46
Problem 11
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
51
52
Internal problem ID [8571]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 11
Date solved : Tuesday, November 11, 2025 at 12:03:42 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 − y 0 + yx2 = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.6)
(2.7)
(2)
(3)
chapter 2 . book solved problems
47
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
Therefore (6) can be used from now on along with
∞
X
xn+2
y(x) = y0 + xy00 +
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = y 0 − yx2
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −x2 + 1 y 0 − x(x + 2) y
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= −2x2 − 4x + 1 y 0 + y(x + 1) x3 − x2 − 2
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
0
1 2
4
2
4
3
= x − 3x − 10x − 5 y + 2y x + 4x − x − x − 1
2
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= 3x4 + 12x3 − 4x2 − 18x − 17 y 0 − y x6 − 3x4 − 18x3 − 29x2 + 2x + 2
(4)
(5)
(6)
(7)
chapter 2 . book solved problems
48
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = y 0 (0)
F1 = y 0 (0)
F2 = y 0 (0) − 2y(0)
F3 = −5y 0 (0) − 2y(0)
F4 = −17y 0 (0) − 2y(0)
Substituting all the above in (7) and simplifying gives the solution as
y=
1 4
1 5
1 2 1 3
1 4
1 5 0
1 − x − x y(0) + x + x + x + x − x y (0) + O x6
12
60
2
6
24
24
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
n(n − 1) an xn−2 −
nan xn−1 +
n=2
n=1
∞
X
!
an x n x 2 = 0
(1)
n=0
Which simplifies to
∞
X
∞
X
!
n(n − 1) an xn−2
+
n=2
−nan x
n−1
n =1
+
∞
X
!
xn+2 an
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
∞
X
−nan x
n−1
=
n =1
∞
X
(−(n + 1) an+1 xn )
n=0
∞
X
xn+2 an =
n =0
∞
X
an−2 xn
n=2
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
n=0
!
(n + 2) an+2 (n + 1) xn
+
∞
X
n =0
(−(n + 1) an+1 xn ) +
∞
X
n=2
!
an−2 xn
=0
(3)
chapter 2 . book solved problems
49
n = 0 gives
2a2 − a1 = 0
a2 =
a1
2
n = 1 gives
6a3 − 2a2 = 0
Which after substituting earlier equations, simplifies to
6a3 − a1 = 0
Or
a3 =
a1
6
For 2 ≤ n, the recurrence equation is
(n + 2) an+2 (n + 1) − (n + 1) an+1 + an−2 = 0
(4)
Solving for an+2 , gives
an+2 =
nan+1 − an−2 + an+1
(n + 2) (n + 1)
=−
an−2
an+1
+
(n + 2) (n + 1) n + 2
For n = 2 the recurrence equation gives
12a4 − 3a3 + a0 = 0
Which after substituting the earlier terms found becomes
a4 =
a1
a0
−
24 12
For n = 3 the recurrence equation gives
20a5 − 4a4 + a1 = 0
Which after substituting the earlier terms found becomes
a5 = −
a1
a0
−
24 60
For n = 4 the recurrence equation gives
30a6 − 5a5 + a2 = 0
Which after substituting the earlier terms found becomes
a6 = −
17a1
a0
−
720
360
For n = 5 the recurrence equation gives
42a7 − 6a6 + a3 = 0
(5)
chapter 2 . book solved problems
50
Which after substituting the earlier terms found becomes
a7 = −
37a1
a0
−
5040 2520
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
y = a0 + a1 x +
a1 x 2 a1 x 3 a1
a0 4 a1
a0 5
+
+
−
x + − −
x + ...
2
6
24 12
24 60
Collecting terms, the solution becomes
y=
1 4
1 5
1 2 1 3
1 4
1 5
1 − x − x a0 + x + x + x + x − x a1 + O x 6
12
60
2
6
24
24
(3)
At x = 0 the solution above becomes
y=
1 4
1 5
1 2 1 3
1 4
1 5 0
1 − x − x y(0) + x + x + x + x − x y (0) + O x6
12
60
2
6
24
24
3 Maple. Time used: 0.006 (sec). Leaf size: 49
Order:=6;
ode:=diff(diff(y(x),x),x)-diff(y(x),x)+x^2*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
y=
1 4
1 5
1 2 1 3
1 4
1 5 0
1 − x − x y(0) + x + x + x + x − x y (0) + O x6
12
60
2
6
24
24
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Kummer
-> hyper3: Equivalence to 1F1 under a power @ Moebius
<- hyper3 successful: received ODE is equivalent to the 1F1 ODE
<- Kummer successful
<- special function solution successful
chapter 2 . book solved problems
51
Maple step by step
Let’s solve
d d
d
y(x) − dx
y(x) + x2 y(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Assume series solution for y(x)
∞
P
y(x) =
ak x k
•
•
k=0
◦
Rewrite ODE with series expansions
Convert x2 · y(x) to series expansion
∞
P
x2 · y(x) =
ak xk+2
k=0
◦
Shift index using k− >k − 2
∞
P
x2 · y(x) =
ak−2 xk
◦
d
Convert dx
y(x) to series expansion
∞
P
d
y(x) =
ak k xk−1
dx
◦
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + 1) xk
dx
◦
d d
Convert dx
y(x) to series expansion
dx
∞
P
d d
y(x) =
ak k(k − 1) xk−2
dx dx
◦
Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=2
k=1
k=0
k=2
k=0
Rewrite ODE with series expansions
∞
P
k
2a2 − a1 + (6a3 − 2a2 ) x +
(ak+2 (k + 2) (k + 1) − ak+1 (k + 1) + ak−2 ) x = 0
k=2
•
•
•
•
•
The coefficients of each power of x must be 0
[2a2 − a1 = 0, 6a3 − 2a2 = 0]
Solve for the dependent coefficient(s)
a2 = a21 , a3 = a61
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 − ak+1 k + ak−2 − ak+1 = 0
Shift index using k− >k + 2
(k + 2)2 + 3k + 8 ak+4 − ak+3 (k + 2) + ak − ak+3 = 0
Recursion
relation that defines the series solution to the ODE
∞
P
−ak +3ak+3
y(x) =
ak xk , ak+4 = kak+3
, a2 = a21 , a3 = a61
k2 +7k+12
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 56
ode=D[y[x],{x,2}]-D[y[x],x]+x^2*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
5
x5 x4
x
x 4 x3 x2
− −
+ 1 + c2 − +
+
+
+x
60 12
24 24
6
2
chapter 2 . book solved problems
3 Sympy. Time used: 0.250 (sec). Leaf size: 49
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x**2*y(x) - Derivative(y(x), x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
5
4
x4 r(3) x5 r(3)
x
x4
x
x
y(x) =
+
+ C2 − −
+ 1 + C 1 x − + + 1 + O x6
4
20
60 12
20 2
52
chapter 2 . book solved problems
2.1.7
53
Problem 12
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
59
59
Internal problem ID [8572]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 12
Date solved : Tuesday, November 11, 2025 at 12:03:43 PM
CAS classification : [_Gegenbauer]
−x2 + 1 y 00 − 2y 0 x + 2y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
dx2
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.9)
(2.10)
(2)
(3)
chapter 2 . book solved problems
54
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
55
To find y(x) series solution around x = 0. Hence
2(y 0 x − y)
x2 − 1
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
8(y 0 x − y) x
=
(x2 − 1)2
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
8(y 0 x − y) (5x2 + 1)
=−
(x2 − 1)3
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
240 x2 + 35 x(y 0 x − y)
=
(x2 − 1)4
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
48(y 0 x − y) (35x4 + 42x2 + 3)
=−
(x2 − 1)5
F0 = −
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = −2y(0)
F1 = 0
F2 = −8y(0)
F3 = 0
F4 = −144y(0)
Substituting all the above in (7) and simplifying gives the solution as
1 4
2
y = 1 − x − x y(0) + xy 0 (0) + O x6
3
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
−x2 + 1 y 00 − 2y 0 x + 2y = 0
Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n=2
n(n − 1) an xn−2
chapter 2 . book solved problems
56
Substituting the above back into the ode gives
!
!
!
∞
∞
∞
X
X
X
−x2 + 1
n(n − 1) an xn−2 − 2
nan xn−1 x + 2
an x n = 0
(1)
n=2
n=1
n=0
Which simplifies to
∞
X
(−xn an n(n − 1)) +
n =2
∞
X
!
n(n − 1) an xn−2
+
n=2
∞
X
(−2nan xn ) +
n =1
∞
X
!
2an xn
= 0 (2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
(−xn an n(n−1))+
n =2
∞
X
!
(n+2) an+2 (n+1) xn +
n=0
∞
X
(−2nan xn )+
n =1
∞
X
!
2an xn
= 0 (3)
n=0
n = 0 gives
2a2 + 2a0 = 0
a2 = −a0
For 2 ≤ n, the recurrence equation is
−nan (n − 1) + (n + 2) an+2 (n + 1) − 2nan + 2an = 0
(4)
Solving for an+2 , gives
an+2 =
(n − 1) an
n+1
For n = 2 the recurrence equation gives
−4a2 + 12a4 = 0
Which after substituting the earlier terms found becomes
a4 = −
a0
3
For n = 3 the recurrence equation gives
−10a3 + 20a5 = 0
Which after substituting the earlier terms found becomes
a5 = 0
(5)
chapter 2 . book solved problems
57
For n = 4 the recurrence equation gives
−18a4 + 30a6 = 0
Which after substituting the earlier terms found becomes
a6 = −
a0
5
For n = 5 the recurrence equation gives
−28a5 + 42a7 = 0
Which after substituting the earlier terms found becomes
a7 = 0
And so on. Therefore the solution is
∞
X
y=
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
1
y = a0 + a1 x − a0 x 2 − a0 x 4 + . . .
3
Collecting terms, the solution becomes
1 4
2
y = 1 − x − x a0 + a1 x + O x 6
3
(3)
At x = 0 the solution above becomes
1 4
2
y = 1 − x − x y(0) + xy 0 (0) + O x6
3
3 Maple. Time used: 0.007 (sec). Leaf size: 28
Order:=6;
ode:=(-x^2+1)*diff(diff(y(x),x),x)-2*diff(y(x),x)*x+2*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 4
2
y = 1 − x − x y(0) + xy 0 (0) + O x6
3
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful
chapter 2 . book solved problems
58
Maple step by step
Let’s solve
d d
d
(−x2 + 1) dx
y(x) − 2x dx
y(x) + 2y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
2x dx
y(x)
2y(x)
d d
y(x)
=
−
2
dx dx
x −1
x2 −1
•
Group terms with
y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
◦
◦
2x
d
y(x)
dx
x2 −1
− 2y(x)
=0
x2 −1
Check to see if x0 is a regular singular point
Define functions
P2 (x) = x22x−1 , P3 (x) = − x22−1
(x + 1) · P2 (x) is analytic at x = −1
d d
y(x) +
dx dx
((x + 1) · P2 (x))
◦
=1
x=−1
(x + 1)2 · P3 (x) is analytic at x = −1
(x + 1)2 · P3 (x)
=0
x=−1
◦
•
•
•
x = −1is a regular singular point
Check to see if x0 is a regular singular point
x0 = −1
Multiply by denominators
d d
d
(x2 − 1) dx
y(x)
+
2x
y(x)
− 2y(x) = 0
dx
dx
Change variables using x = u − 1 so that the regular singular point is at u = 0
d d
d
(u2 − 2u) du
y(u)
+
(2u
−
2)
y(u)
− 2y(u) = 0
du
du
Assume series solution for y(u)
∞
P
y(u) =
ak uk+r
k=0
◦
Rewrite ODE with series expansions
d
Convert um · du
y(u) to series expansion for m = 0..1
∞
P
d
um · du
y(u) =
ak (k + r) uk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
um · du
y(u) =
ak+1−m (k + 1 − m + r) uk+r
k=−1+m
d d
Convert um · du
y(u) to series expansion for m = 1..2
du
∞
P
d d
um · du
y(u)
=
ak (k + r) (k + r − 1) uk+r−2+m
du
k=0
◦
Shift index using k− >k + 2 − m
∞
P
d d
um · du
y(u)
=
ak+2−m (k + 2 − m + r) (k + 1 − m + r) uk+r
du
k=−2+m
Rewrite ODE with
∞series expansions
k+r
P
2
2 −1+r
−2a0 r u
+
−2ak+1 (k + 1 + r) + ak (k + r + 2) (k + r − 1) u
=0
k=0
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
−2r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term in the series must be 0, giving the recursion relation
−2ak+1 (k + 1)2 + ak (k + 2) (k − 1) = 0
Recursion relation that defines series solution to ODE
ak+1 = ak (k+2)(k−1)
2(k+1)2
Recursion relation for r = 0 ; series terminates at k = 1
ak+1 = ak (k+2)(k−1)
2(k+1)2
chapter 2 . book solved problems
•
•
•
•
59
Apply recursion relation for k = 0
a1 = −a0
Terminating series solution of the ODE for r = 0 . Use reduction of order to find the second linea
y(u) = a0 · (1 − u)
Revert the change of variables u = x + 1
[y(x) = −a0 x]
Use this particular
the order
solution to reduce of the ODE and find a complete solution
y(x) = −x C 1 − ln(x+1)
+ ln(x−1)
+ x1 + C 2
2
2
3 Mathematica. Time used: 0.001 (sec). Leaf size: 25
ode=(1-x^2)*D[y[x],{x,2}]-2*x*D[y[x],x]+2*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
x4
2
− − x + 1 + c2 x
3
3 Sympy. Time used: 0.244 (sec). Leaf size: 20
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-2*x*Derivative(y(x), x) + (1 - x**2)*Derivative(y(x), (x, 2)) + 2*y(x)
,0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
4
x
2
y(x) = C2 − − x + 1 + C1 x + O x6
3
chapter 2 . book solved problems
2.1.8
60
Problem 13
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
65
65
Internal problem ID [8573]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 13
Date solved : Tuesday, November 11, 2025 at 12:03:43 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 + x2 + 1 y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
dx2
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.11)
(2.12)
(2)
(3)
chapter 2 . book solved problems
61
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = − x2 + 1 y
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −2yx − x2 + 1 y 0
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= −4y 0 x + y x4 + 2x2 − 1
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= 8yx x2 + 1 + x4 + 2x2 − 5 y 0
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= 12 x3 + x y 0 − y x6 + 3x4 − 27x2 − 13
(7)
chapter 2 . book solved problems
62
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = −y(0)
F1 = −y 0 (0)
F2 = −y(0)
F3 = −5y 0 (0)
F4 = 13y(0)
Substituting all the above in (7) and simplifying gives the solution as
1 2
1 4
1 3
1 5 0
y = 1 − x − x y(0) + x − x − x y (0) + O x6
2
24
6
24
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
n−2
2
n
n(n − 1) an x
+ x +1
an x
=0
n=2
(1)
n=0
Which simplifies to
∞
X
∞
X
!
n(n − 1) an xn−2
+
n=2
!
xn+2 an
+
n=0
∞
X
!
an x n
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
∞
X
x
n+2
an =
n =0
∞
X
an−2 xn
n=2
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
n=0
!
(n + 2) an+2 (n + 1) xn
+
∞
X
!
an−2 xn
n=2
n = 0 gives
2a2 + a0 = 0
+
∞
X
n=0
!
an x n
=0
(3)
chapter 2 . book solved problems
63
a2 = −
a0
2
n = 1 gives
6a3 + a1 = 0
Which after substituting earlier equations, simplifies to
a3 = −
a1
6
For 2 ≤ n, the recurrence equation is
(n + 2) an+2 (n + 1) + an−2 + an = 0
(4)
Solving for an+2 , gives
an+2 = −
=−
an−2 + an
(n + 2) (n + 1)
an
an−2
−
(n + 2) (n + 1) (n + 2) (n + 1)
For n = 2 the recurrence equation gives
12a4 + a0 + a2 = 0
Which after substituting the earlier terms found becomes
a4 = −
a0
24
For n = 3 the recurrence equation gives
20a5 + a1 + a3 = 0
Which after substituting the earlier terms found becomes
a5 = −
a1
24
For n = 4 the recurrence equation gives
30a6 + a2 + a4 = 0
Which after substituting the earlier terms found becomes
a6 =
13a0
720
For n = 5 the recurrence equation gives
42a7 + a3 + a5 = 0
Which after substituting the earlier terms found becomes
a7 =
5a1
1008
(5)
chapter 2 . book solved problems
64
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
1
1
1
1
y = a0 + a1 x − a0 x 2 − a1 x 3 − a0 x 4 − a1 x 5 + . . .
2
6
24
24
Collecting terms, the solution becomes
1 2
1 4
1 3
1 5
y = 1 − x − x a0 + x − x − x a1 + O x 6
2
24
6
24
(3)
At x = 0 the solution above becomes
1 2
1 4
1 3
1 5 0
y = 1 − x − x y(0) + x − x − x y (0) + O x6
2
24
6
24
3 Maple. Time used: 0.005 (sec). Leaf size: 39
Order:=6;
ode:=diff(diff(y(x),x),x)+(x^2+1)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2
1 4
1 3
1 5 0
y = 1 − x − x y(0) + x − x − x y (0) + O x6
2
24
6
24
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Whittaker
-> hyper3: Equivalence to 1F1 under a power @ Moebius
<- hyper3 successful: received ODE is equivalent to the 1F1 ODE
<- Whittaker successful
<- special function solution successful
Maple step by step
•
Let’s solve
d d
y(x) + (x2 + 1) y(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
chapter 2 . book solved problems
65
d d
y(x)
dx dx
•
Assume series solution for y(x)
∞
P
y(x) =
ak x k
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..2
∞
P
xm · y(x) =
ak xk+m
k=0
◦
k=max(0,−m)
◦
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk
k=max(0,−m)+m
d d
Convert dx dx y(x) to series expansion
∞
P
d d
y(x)
=
ak k(k − 1) xk−2
dx dx
k=2
Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=0
Rewrite ODE with series expansions
∞
P
k
2a2 + a0 + (6a3 + a1 ) x +
(ak+2 (k + 2) (k + 1) + ak + ak−2 ) x = 0
k=2
•
•
•
•
•
The coefficients of each power of x must be 0
[2a2 + a0 = 0, 6a3 + a1 = 0]
Solve for the dependent coefficient(s)
a2 = − a20 , a3 = − a61
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 + ak + ak−2 = 0
Shift index using k− >k + 2
(k + 2)2 + 3k + 8 ak+4 + ak+2 + ak = 0
Recursion
relation that defines the series solution to the ODE
∞
P
+ak
y(x) =
ak xk , ak+4 = − ka2k+2
, a2 = − a20 , a3 = − a61
+7k+12
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 42
ode=D[y[x],{x,2}]+(1+x^2)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
4
x5 x3
x
x2
− −
+ x + c1 − −
+1
24
6
24
2
3 Sympy. Time used: 0.210 (sec). Leaf size: 37
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((x**2 + 1)*y(x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
4
x5 r(3)
x
x2
x4
y(x) = −
+ C2 − −
+ 1 + C1 x 1 −
+ O x6
20
24
2
20
chapter 2 . book solved problems
2.1.9
66
Problem 14
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
72
72
Internal problem ID [8574]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 14
Date solved : Tuesday, November 11, 2025 at 12:03:43 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 − 4y 0 x + 4x2 − 2 y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
dx2
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.14)
(2.15)
(2)
(3)
chapter 2 . book solved problems
67
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
68
To find y(x) series solution around x = 0. Hence
F0 = −4yx2 + 4y 0 x + 2y
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −16yx3 + 12y 0 x2 + 6y 0
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
0
1
3
4
2
= 32x + 48x y − 48y x + x −
4
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= −128yx5 + 80y 0 x4 − 320yx3 + 240y 0 x2 + 60y 0
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
0
9 4 9 2 3
5
3
6
= 192x + 960x + 720x y − 320 x + x + x −
y
2
4
8
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = 2y(0)
F1 = 6y 0 (0)
F2 = 12y(0)
F3 = 60y 0 (0)
F4 = 120y(0)
Substituting all the above in (7) and simplifying gives the solution as
1 4
1 5 0
2
3
y = 1 + x + x y(0) + x + x + x y (0) + O x6
2
2
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
y 00 =
∞
X
n=1
∞
X
nan xn−1
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
!
∞
∞
∞
X
X
X
n(n − 1) an xn−2 − 4
nan xn−1 x + 4x2 − 2
an x n = 0
n=2
n=1
n=0
(1)
chapter 2 . book solved problems
69
Which simplifies to
∞
X
!
n(n − 1) an x
n−2
+
n=2
∞
X
n
(−4n x an ) +
n =1
∞
X
!
4x
n+2
+
an
n=0
∞
X
(−2an xn ) = 0
(2)
n =0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
∞
X
4x
n+2
an =
n =0
∞
X
4an−2 xn
n=2
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
(n + 2) an+2 (n + 1) x
n
∞
X
+
n=0
n
(−4n x an ) +
∞
X
n =1
!
4an−2 x
n
+
n=2
∞
X
(−2an xn ) = 0 (3)
n =0
n = 0 gives
2a2 − 2a0 = 0
a2 = a0
n = 1 gives
6a3 − 6a1 = 0
Which after substituting earlier equations, simplifies to
a3 = a1
For 2 ≤ n, the recurrence equation is
(n + 2) an+2 (n + 1) − 4nan + 4an−2 − 2an = 0
(4)
Solving for an+2 , gives
an+2 =
=
4nan + 2an − 4an−2
(n + 2) (n + 1)
2(2n + 1) an
4an−2
−
(n + 2) (n + 1) (n + 2) (n + 1)
For n = 2 the recurrence equation gives
12a4 − 10a2 + 4a0 = 0
Which after substituting the earlier terms found becomes
a4 =
a0
2
(5)
chapter 2 . book solved problems
70
For n = 3 the recurrence equation gives
20a5 − 14a3 + 4a1 = 0
Which after substituting the earlier terms found becomes
a5 =
a1
2
For n = 4 the recurrence equation gives
30a6 − 18a4 + 4a2 = 0
Which after substituting the earlier terms found becomes
a6 =
a0
6
For n = 5 the recurrence equation gives
42a7 − 22a5 + 4a3 = 0
Which after substituting the earlier terms found becomes
a7 =
a1
6
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
1
1
y = a0 + a1 x + a0 x 2 + a1 x 3 + a0 x 4 + a1 x 5 + . . .
2
2
Collecting terms, the solution becomes
1 4
1 5
2
3
y = 1 + x + x a0 + x + x + x a1 + O x 6
2
2
(3)
At x = 0 the solution above becomes
1 4
1 5 0
2
3
y = 1 + x + x y(0) + x + x + x y (0) + O x6
2
2
3 Maple. Time used: 0.007 (sec). Leaf size: 35
Order:=6;
ode:=diff(diff(y(x),x),x)-4*diff(y(x),x)*x+(4*x^2-2)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 4
1 5 0
2
3
y = 1 + x + x y(0) + x + x + x y (0) + O x6
2
2
Maple trace
chapter 2 . book solved problems
71
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
y(x) − 4x dx
y(x) + (4x2 − 2) y(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Assume series solution for y(x)
∞
P
y(x) =
ak x k
•
•
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..2
∞
P
xm · y(x) =
ak xk+m
k=max(0,−m)
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk
k=max(0,−m)+m
d
Convert x · dx
y(x) to series expansion
∞
P
d
x · dx
y(x) =
ak k x k
k=0
◦
d d
Convert dx
y(x) to series expansion
dx
∞
P
d d
y(x) =
ak k(k − 1) xk−2
dx dx
◦
Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=2
k=0
Rewrite ODE with series expansions
∞
P
k
2a2 − 2a0 + (6a3 − 6a1 ) x +
(ak+2 (k + 2) (k + 1) − 2ak (2k + 1) + 4ak−2 ) x = 0
k=2
•
•
•
•
•
The coefficients of each power of x must be 0
[2a2 − 2a0 = 0, 6a3 − 6a1 = 0]
Solve for the dependent coefficient(s)
{a2 = a0 , a3 = a1 }
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 − 4ak k − 2ak + 4ak−2 = 0
Shift index using k− >k + 2
(k + 2)2 + 3k + 8 ak+4 − 4ak+2 (k + 2) − 2ak+2 + 4ak = 0
Recursion
relation that defines the series solution to the ODE
∞
P
2(2kak+2 −2ak +5ak+2 )
k
y(x) =
ak x , ak+4 =
, a 2 = a0 , a 3 = a1
k2 +7k+12
k=0
chapter 2 . book solved problems
72
3 Mathematica. Time used: 0.001 (sec). Leaf size: 34
ode=D[y[x],{x,2}]-4*x*D[y[x],x]+(4*x^2-2)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
5
4
x
x
3
2
y(x) → c2
+ x + x + c1
+x +1
2
2
3 Sympy. Time used: 0.219 (sec). Leaf size: 37
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-4*x*Derivative(y(x), x) + (4*x**2 - 2)*y(x) + Derivative(y(x), (x, 2))
,0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
4
7x5 r(3)
x
x4
2
y(x) =
+ C2
+ x + 1 + C1 x 1 −
+ O x6
10
2
5
chapter 2 . book solved problems
2.1.10
73
Problem 16
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
77
78
78
Internal problem ID [8575]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 16
Date solved : Tuesday, November 11, 2025 at 12:03:44 PM
CAS classification : [_quadrature]
y 0 + 4y = 1
y(0) =
5
4
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving first order ode. Let
y 0 = f (x, y)
Where f (x, y) is analytic at expansion point x0 . We can always shift to x0 = 0 if x0 is not
zero. So from now we assume x0 = 0 . Assume also that y(x0 ) = y0 . Using Taylor series
(x − x0 )2 00
(x − x0 )3 00
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2 df
x3 d 2 f
= y0 + xf +
+
+ ···
2 dx x0 ,y0 3! dx2 x0 ,y0
∞
X
xn+1 dn f
= y0 +
(n + 1) ! dxn x0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
But
df
∂f
∂f
=
+
f
dx
∂x ∂y
d2 f
d df
=
dx2
dx dx
∂ df
∂ df
=
+
f
∂x dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
=
+
f
∂x dx2
∂y dx2
..
.
(1)
(2)
(3)
And so on. Hence if we name F0 = f (x, y) then the above can be written as
F0 = f (x, y)
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
=
Fn−1 +
F0
∂x
∂y
(4)
(5)
chapter 2 . book solved problems
74
For example, for n = 1 we see that
d
(F0 )
dx
∂
∂F0
=
F0 +
F0
∂x
∂y
∂f
∂f
=
+
f
∂x ∂y
F1 =
Which is (1). And when n = 2
d
(F1 )
dx
∂
∂F1
=
F1 +
F0
∂x
∂y
∂ ∂f
∂f
∂ ∂f
∂f
=
+
f +
+
f f
∂x ∂x ∂y
∂y ∂x ∂y
∂ df
∂ df
=
+
f
∂x dx
∂y dx
F2 =
Which is (2) and so on. Therefore (4,5) can be used from now on along with
y(x) = y0 +
∞
X
xn+1
Fn |x0 ,y0
(n
+
1)
!
n=0
Hence
F0 = −4y + 1
dF0
F1 =
dx
∂F0 ∂F0
=
+
F0
∂x
∂y
= 16y − 4
dF1
F2 =
dx
∂F1 ∂F1
=
+
F1
∂x
∂y
= −64y + 16
dF2
F3 =
dx
∂F2 ∂F2
=
+
F2
∂x
∂y
= 256y − 64
dF3
F4 =
dx
∂F3 ∂F3
=
+
F3
∂x
∂y
= −1024y + 256
And so on. Evaluating all the above at initial conditions x(0) = 0 and y(0) = 54 gives
F0 = −4
F1 = 16
F2 = −64
F3 = 256
F4 = −1024
Substituting all the above in (6) and simplifying gives the solution as
y=
5
32x3 32x4 128x5
− 4x + 8x2 −
+
−
+ O x6
4
3
3
15
(6)
chapter 2 . book solved problems
75
Since x = 0 is also an ordinary point, then standard power series can also be used. Writing
the ODE as
y 0 + q(x)y = p(x)
y 0 + 4y = 1
Where
q(x) = 4
p(x) = 1
Next, the type of the expansion point x = 0 is determined. This point can be an ordinary
point, a regular singular point (also called removable singularity), or irregular singular point
(also called non-removable singularity or essential singularity). When x = 0 is an ordinary
point, then the standard power series is used. If the point is a regular singular point, Frobenius
series is used instead. Irregular singular point requires more advanced methods (asymptotic
methods) and is not supported now. Hopefully this will be added in the future. x = 0 is
called an ordinary point q(x) has a Taylor series expansion around the point x = 0. x = 0
is called a regular singular point if q(x) is not not analytic at x = 0 but xq(x) has Taylor
series expansion. And finally, x = 0 is an irregular singular point if the point is not ordinary
and not regular singular. This is the most complicated case. Now the expansion point x = 0
is checked to see if it is an ordinary point or not. Let the solution be represented as power
series of the form
∞
X
y=
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
Substituting the above back into the ode gives
!
!
∞
∞
X
X
nan xn−1 + 4
an x n = 1
n=1
(1)
n=0
Which simplifies to
∞
X
∞
X
!
nan xn−1
+
n=1
!
4an xn
=1
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
nan x
n =1
n−1
=
∞
X
(n + 1) an+1 xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
(n + 1) an+1 xn
n=0
+
∞
X
!
4an xn
=1
(3)
n=0
For 0 ≤ n, the recurrence equation is
((n + 1) an+1 + 4an ) xn = 1
(4)
chapter 2 . book solved problems
For n = 0 the recurrence equation gives
(a1 + 4a0 ) 1 = 1
a1 + 4a0 = 1
Which after substituting the earlier terms found becomes
a1 = 1 − 4a0
For n = 1 the recurrence equation gives
(2a2 + 4a1 ) x = 0
2a2 + 4a1 = 0
Which after substituting the earlier terms found becomes
a2 = −2 + 8a0
For n = 2 the recurrence equation gives
(3a3 + 4a2 ) x2 = 0
3a3 + 4a2 = 0
Which after substituting the earlier terms found becomes
a3 =
8 32a0
−
3
3
For n = 3 the recurrence equation gives
(4a4 + 4a3 ) x3 = 0
4a4 + 4a3 = 0
Which after substituting the earlier terms found becomes
8 32a0
a4 = − +
3
3
For n = 4 the recurrence equation gives
(5a5 + 4a4 ) x4 = 0
5a5 + 4a4 = 0
Which after substituting the earlier terms found becomes
a5 =
32 128a0
−
15
15
For n = 5 the recurrence equation gives
(6a6 + 4a5 ) x5 = 0
6a6 + 4a5 = 0
76
chapter 2 . book solved problems
77
Which after substituting the earlier terms found becomes
a6 = −
64 256a0
+
45
45
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
8 32a0
y = a0 +(1−4a0 ) x+(−2+8a0 ) x +
−
3
3
2
8 32a0
32 128a0
4
x + − +
x +
−
x5 +. . .
3
3
15
15
3
Collecting terms, the solution becomes
32 3 32 4 128 5
8x3 8x4 32x5
y = 1 − 4x + 8x − x + x −
x a0 + x − 2x2 +
−
+
+ O x6 (3)
3
3
15
3
3
15
2
At x = 0 the solution above becomes
y(0) = a0
Therefore the solution in Eq(3) now can be written as
y=
5
32x3 32x4 128x5
− 4x + 8x2 −
+
−
+ O x6
4
3
3
15
(a) Solutions plot
(b) Slope field y 0 + 4y = 1
3 Maple. Time used: 0.004 (sec). Leaf size: 20
Order:=6;
ode:=diff(y(x),x)+4*y(x) = 1;
ic:=[y(0) = 5/4];
dsolve([ode,op(ic)],y(x),type='series',x=0);
y=
Maple trace
5
32
32
128 5
− 4x + 8x2 − x3 + x4 −
x + O x6
4
3
3
15
chapter 2 . book solved problems
78
Methods for first order ODEs:
--- Trying classification methods --trying a quadrature
trying 1st order linear
<- 1st order linear successful
Maple step by step
•
•
•
Let’s solve
d
y(x) + 4y(x) = 1, y(0) = 54
dx
Highest derivative means the order of the ODE is 1
d
y(x)
dx
Solve for the highest derivative
d
y(x) = −4y(x) + 1
dx
Separate variables
d
y(x)
dx
=1
Integrate both sides with respect to x
R dxd y(x)
R
dx = 1dx + C 1
−4y(x)+1
Evaluate integral
− ln(−4y(x)+1)
= x + C1
4
Solve for y(x)
−4x−4C 1
y(x) = − e 4
+ 14
Redefine the integration constant(s)
y(x) = C 1 e−4x + 14
Use initial condition y(0) = 54
5
= C 1 + 14
4
Solve for _C1
C1 = 1
Substitute _C1 = 1 into general solution and simplify
y(x) = e−4x + 14
Solution to the IVP
y(x) = e−4x + 14
−4y(x)+1
•
•
•
•
•
•
•
•
3 Mathematica. Time used: 0.002 (sec). Leaf size: 36
ode=D[y[x],x]+4*y[x]==1;
ic={y[0]==125/100};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → −
128x5 32x4 32x3
5
+
−
+ 8x2 − 4x +
15
3
3
4
3 Sympy. Time used: 0.149 (sec). Leaf size: 37
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(4*y(x) + Derivative(y(x), x) - 1,0)
ics = {y(0): 5/4}
dsolve(ode,func=y(x),ics=ics,hint="1st_power_series",x0=0,n=6)
y(x) =
5
32x3 32x4 128x5
− 4x + 8x2 −
+
−
+ O x6
4
3
3
15
chapter 2 . book solved problems
2.1.11
79
Problem 17
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
84
85
Internal problem ID [8576]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 17
Date solved : Tuesday, November 11, 2025 at 12:03:45 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 + 3y 0 x + 2y = 0
y(0) = 1
y 0 (0) = 1
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.17)
(2.18)
(2)
(3)
chapter 2 . book solved problems
80
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = −3y 0 x − 2y
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= 9y 0 x2 + 6yx − 5y 0
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= −27y 0 x3 − 18yx2 + 39y 0 x + 16y
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= 81x4 − 216x2 + 55 y 0 + 54x3 − 114x y
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= −243x5 + 1026x3 − 711x y 0 + −162x4 + 594x2 − 224 y
(7)
chapter 2 . book solved problems
81
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = 1 and y 0 (0) = 1
gives
F0 = −2
F1 = −5
F2 = 16
F3 = 55
F4 = −224
Substituting all the above in (7) and simplifying gives the solution as
5x3 2x4 11x5
+
+
+ O x6
6
3
24
y = −x2 + x + 1 −
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
∞
X
0
y =
nan xn−1
n=1
∞
X
y 00 =
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
!
∞
∞
∞
X
X
X
n(n − 1) an xn−2 + 3
nan xn−1 x + 2
an x n = 0
n=2
n=1
(1)
n=0
Which simplifies to
∞
X
!
n(n − 1) an xn−2
∞
X
+
n=2
!
3n xn an
+
n=1
∞
X
!
2an xn
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
n=0
!
(n + 2) an+2 (n + 1) xn
+
∞
X
!
3n xn an
n=1
n = 0 gives
2a2 + 2a0 = 0
a2 = −a0
+
∞
X
n=0
!
2an xn
=0
(3)
chapter 2 . book solved problems
82
For 1 ≤ n, the recurrence equation is
(n + 2) an+2 (n + 1) + 3nan + 2an = 0
(4)
Solving for an+2 , gives
an+2 = −
an (3n + 2)
(n + 2) (n + 1)
For n = 1 the recurrence equation gives
6a3 + 5a1 = 0
Which after substituting the earlier terms found becomes
a3 = −
5a1
6
For n = 2 the recurrence equation gives
12a4 + 8a2 = 0
Which after substituting the earlier terms found becomes
2a0
3
a4 =
For n = 3 the recurrence equation gives
20a5 + 11a3 = 0
Which after substituting the earlier terms found becomes
a5 =
11a1
24
For n = 4 the recurrence equation gives
30a6 + 14a4 = 0
Which after substituting the earlier terms found becomes
a6 = −
14a0
45
For n = 5 the recurrence equation gives
42a7 + 17a5 = 0
Which after substituting the earlier terms found becomes
a7 = −
187a1
1008
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
(5)
chapter 2 . book solved problems
83
Substituting the values for an found above, the solution becomes
5
2
11
y = a0 + a1 x − a0 x2 − a1 x3 + a0 x4 + a1 x5 + . . .
6
3
24
Collecting terms, the solution becomes
2 4
5 3 11 5
2
y = 1 − x + x a0 + x − x + x a1 + O x 6
3
6
24
(3)
At x = 0 the solution above becomes
2 4
5 3 11 5 0
2
y = 1 − x + x y(0) + x − x + x y (0) + O x6
3
6
24
Using initial conditions the above becomes
y = −x2 + x + 1 −
5x3 2x4 11x5
+
+
+ O x6
6
3
24
Figure 2.5: Solutions plot
3 Maple. Time used: 0.004 (sec). Leaf size: 20
Order:=6;
ode:=diff(diff(y(x),x),x)+3*diff(y(x),x)*x+2*y(x) = 0;
ic:=[y(0) = 1, D(y)(0) = 1];
dsolve([ode,op(ic)],y(x),type='series',x=0);
5
2
11
y = 1 + x − x2 − x3 + x4 + x5 + O x6
6
3
24
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
chapter 2 . book solved problems
84
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Kummer
-> hyper3: Equivalence to 1F1 under a power @ Moebius
<- hyper3 successful: received ODE is equivalent to the 1F1 ODE
<- Kummer successful
<- special function solution successful
Maple step by step
Let’s
solve
d d
d
y(x)
+
3x
y(x)
+ 2y(x) = 0, y(0) = 1,
dx dx
dx
•
d
y(x)
dx
=1
{x=0}
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Assume series solution for y(x)
∞
P
y(x) =
ak x k
•
k=0
◦
Rewrite DE with series expansions
d
Convert x · dx
y(x) to series expansion
∞
P
d
x · dx
y(x) =
ak k x k
k=0
◦
d d
Convert dx
y(x) to series expansion
dx
∞
P
d d
y(x) =
ak k(k − 1) xk−2
dx dx
◦
Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=2
k=0
Rewrite DE with series expansions
∞
P
(ak+2 (k + 2) (k + 1) + ak (3k + 2)) xk = 0
k=0
•
•
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 + 3ak k + 2ak = 0
Recursion
relation that defines theseries solution to the ODE
∞
P
(3k+2)
y(x) =
ak xk , ak+2 = − akk2 +3k+2
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 32
ode=D[y[x],{x,2}]+3*x*D[y[x],x]+2*y[x]==0;
ic={y[0]==1,Derivative[1][y][0] ==1};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) →
11x5 2x4 5x3
+
−
− x2 + x + 1
24
3
6
chapter 2 . book solved problems
3 Sympy. Time used: 0.201 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(3*x*Derivative(y(x), x) + 2*y(x) + Derivative(y(x), (x, 2)),0)
ics = {y(0): 1, Subs(Derivative(y(x), x), x, 0): 1}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
y(x) = C2
2x4
5x2
2
− x + 1 + C1 x 1 −
+ O x6
3
6
85
chapter 2 . book solved problems
2.1.12
86
Problem 18
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
93
93
Internal problem ID [8577]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 18
Date solved : Tuesday, November 11, 2025 at 12:03:45 PM
CAS classification : [_Gegenbauer]
−x2 + 1 y 00 − 2y 0 x + 30y = 0
y(0) = 0
y 0 (0) =
15
8
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X
x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.19)
(2.20)
(2)
(3)
chapter 2 . book solved problems
87
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
88
To find y(x) series solution around x = 0. Hence
2(y 0 x − 15y)
x2 − 1
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
36y 0 x2 − 120yx − 28y 0
=
(x2 − 1)2
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
24(11y 0 x3 − 60yx2 − 9y 0 x + 30y)
=−
(x2 − 1)3
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
(2760x4 − 2880x2 + 504) y 0 + (−13680x3 + 7920x) y
=
(x2 − 1)4
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
1
(−30240x5 + 33600x3 − 7200x) y 0 + 151200 x4 − 23 x2 + 21
y
=
(x2 − 1)5
F0 = −
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = 0 and y 0 (0) = 15
8
gives
F0 = 0
F1 = −
105
2
F2 = 0
F3 = 945
F4 = 0
Substituting all the above in (7) and simplifying gives the solution as
y=
15x 35x3 63x5
−
+
+ O x6
8
4
8
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
−x2 + 1 y 00 − 2y 0 x + 30y = 0
Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n=2
n(n − 1) an xn−2
chapter 2 . book solved problems
89
Substituting the above back into the ode gives
!
!
!
∞
∞
∞
X
X
X
−x2 + 1
n(n − 1) an xn−2 − 2
nan xn−1 x + 30
an x n = 0
(1)
n=2
n=1
n=0
Which simplifies to
∞
X
(−xn an n(n − 1)) +
n =2
∞
X
!
n(n − 1) an xn−2
+
n=2
∞
X
(−2nan xn ) +
n =1
∞
X
!
30an xn
= 0 (2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
n =2
=
∞
X
(n + 2) an+2 (n + 1) xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
(−xn an n(n − 1)) +
n =2
∞
X
!
(n + 2) an+2 (n + 1) xn
+
n=0
∞
X
(−2nan xn ) +
n =1
∞
X
!
30an xn
=0
n=0
(3)
n = 0 gives
2a2 + 30a0 = 0
a2 = −15a0
n = 1 gives
6a3 + 28a1 = 0
Which after substituting earlier equations, simplifies to
a3 = −
14a1
3
For 2 ≤ n, the recurrence equation is
−nan (n − 1) + (n + 2) an+2 (n + 1) − 2nan + 30an = 0
(4)
Solving for an+2 , gives
an (n2 + n − 30)
an+2 =
(n + 2) (n + 1)
For n = 2 the recurrence equation gives
24a2 + 12a4 = 0
Which after substituting the earlier terms found becomes
a4 = 30a0
(5)
chapter 2 . book solved problems
90
For n = 3 the recurrence equation gives
18a3 + 20a5 = 0
Which after substituting the earlier terms found becomes
a5 =
21a1
5
For n = 4 the recurrence equation gives
10a4 + 30a6 = 0
Which after substituting the earlier terms found becomes
a6 = −10a0
For n = 5 the recurrence equation gives
42a7 = 0
Which after substituting the earlier terms found becomes
a7 = 0
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
y = a0 + a1 x − 15a0 x2 −
14
21
a1 x3 + 30a0 x4 + a1 x5 + . . .
3
5
Collecting terms, the solution becomes
14 3 21 5
y = 30x − 15x + 1 a0 + x − x + x a1 + O x6
3
5
4
2
At x = 0 the solution above becomes
14 3 21 5 0
y = 30x − 15x + 1 y(0) + x − x + x y (0) + O x6
3
5
4
2
Using initial conditions the above becomes
y=
15x 35x3 63x5
−
+
+ O x6
8
4
8
(3)
chapter 2 . book solved problems
91
Figure 2.6: Solutions plot
3 Maple. Time used: 0.004 (sec). Leaf size: 14
Order:=6;
ode:=(-x^2+1)*diff(diff(y(x),x),x)-2*diff(y(x),x)*x+30*y(x) = 0;
ic:=[y(0) = 0, D(y)(0) = 15/8];
dsolve([ode,op(ic)],y(x),type='series',x=0);
y=
15
35
63
x − x3 + x5 + O x6
8
4
8
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful
Maple step by step
Let’s
solve
(−x2 + 1)
•
d d
y(x)
dx dx
− 2x
d
y(x)
dx
+ 30y(x) = 0, y(0) = 0,
d
y(x)
dx
{x=0}
= 15
8
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative •
d d
y(x) = 30y(x)
−
dx dx
x2 −1
2x
d
y(x)
dx
x2 −1
•
Group terms with
y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
− 30y(x)
=0
x2 −1
Check to see if x0 is a regular singular point
Define functions
d d
y(x) +
dx dx
◦
2x
d
y(x)
dx
x2 −1
chapter 2 . book solved problems
◦
P2 (x) = x22x−1 , P3 (x) = − x230−1
(x + 1) · P2 (x) is analytic at x = −1
((x + 1) · P2 (x))
=1
x=−1
◦
(x + 1)2 · P3 (x) is analytic at x = −1
(x + 1)2 · P3 (x)
=0
◦
x = −1is a regular singular point
Check to see if x0 is a regular singular point
x0 = −1
Multiply by denominators
d d
d
(x2 − 1) dx
y(x) + 2x dx
y(x) − 30y(x) = 0
dx
Change variables using x = u − 1 so that the regular singular point is at u = 0
d d
d
(u2 − 2u) du
y(u) + (2u − 2) du
y(u) − 30y(u) = 0
du
Assume series solution for y(u)
∞
P
y(u) =
ak uk+r
x=−1
•
•
•
k=0
◦
Rewrite ODE with series expansions
d
Convert um · du
y(u) to series expansion for m = 0..1
∞
P
d
um · du
y(u) =
ak (k + r) uk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
um · du
y(u) =
ak+1−m (k + 1 − m + r) uk+r
k=−1+m
d d
Convert um · du
y(u) to series expansion for m = 1..2
du
∞
P
d d
um · du
y(u) =
ak (k + r) (k + r − 1) uk+r−2+m
du
k=0
◦
Shift index using k− >k + 2 − m
∞
P
d d
um · du
y(u)
=
ak+2−m (k + 2 − m + r) (k + 1 − m + r) uk+r
du
k=−2+m
Rewrite ODE with
∞series expansions
P
2
−2a0 r2 u−1+r +
−2ak+1 (k + 1 + r) + ak (k + r + 6) (k + r − 5) uk+r = 0
k=0
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
−2r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term in the series must be 0, giving the recursion relation
−2ak+1 (k + 1)2 + ak (k + 6) (k − 5) = 0
Recursion relation that defines series solution to ODE
ak+1 = ak (k+6)(k−5)
2(k+1)2
•
Recursion relation for r = 0 ; series terminates at k = 5
ak+1 = ak (k+6)(k−5)
2(k+1)2
•
Apply recursion relation for k = 0
a1 = −15a0
Apply recursion relation for k = 1
a2 = − 7a21
Express in terms of a0
0
a2 = 105a
2
Apply recursion relation for k = 2
a3 = − 4a32
Express in terms of a0
a3 = −70a0
Apply recursion relation for k = 3
3
a4 = − 9a
16
Express in terms of a0
•
•
•
•
•
•
92
chapter 2 . book solved problems
•
•
•
•
•
93
0
a4 = 315a
8
Apply recursion relation for k = 4
a5 = − a54
Express in terms of a0
0
a5 = − 63a
8
Terminating series solution of the ODE for r = 0 . Use reduction of order to find the second linea
y(u) = a0 · 1 − 15u + 105
u2 − 70u3 + 315
u4 − 63
u5
2
8
8
Revert the change of variables u = x + 1
y(x) = a0 − 15
x + 35
x3 − 63
x5
8
4
8
Use this particular solution toreduce
solution
the order of the ODE and find3 a935complete
7 −23x + 63 x
ln(x−1)
ln(x+1)
15
35 3
63 5
1
y(x) = − 8 x + 4 x − 8 x
C1
− 128 + 225x − 14400 x4 − 10 x2 + 5 + C 2
128
9
21
3 Mathematica. Time used: 0.001 (sec). Leaf size: 23
ode=(1-x^2)*D[y[x],{x,2}]-2*x*D[y[x],x]+30*y[x]==0;
ic={y[0]==0,Derivative[1][y][0] ==1875/1000};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) →
63x5 35x3 15x
−
+
8
4
8
3 Sympy. Time used: 0.257 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-2*x*Derivative(y(x), x) + (1 - x**2)*Derivative(y(x), (x, 2)) + 30*y(x)
,0)
ics = {y(0): 0, Subs(Derivative(y(x), x), x, 0): 15/8}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
14x2
4
2
y(x) = C2 30x − 15x + 1 + C1 x 1 −
+ O x6
3
chapter 2 . book solved problems
2.1.13
94
Problem 19
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
99
99
Internal problem ID [8578]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page 174
Problem number : 19
Date solved : Tuesday, November 11, 2025 at 12:03:46 PM
CAS classification : [_separable]
(x − 2) y 0 = yx
y(0) = 4
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving first order ode. Let
y 0 = f (x, y)
Where f (x, y) is analytic at expansion point x0 . We can always shift to x0 = 0 if x0 is not
zero. So from now we assume x0 = 0 . Assume also that y(x0 ) = y0 . Using Taylor series
(x − x0 )2 00
(x − x0 )3 00
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2 df
x3 d 2 f
= y0 + xf +
+
+ ···
2 dx x0 ,y0 3! dx2 x0 ,y0
∞
X
xn+1 dn f
= y0 +
(n + 1) ! dxn x0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
But
df
∂f
∂f
=
+
f
dx
∂x ∂y
d2 f
d df
=
dx2
dx dx
∂ df
∂ df
=
+
f
∂x dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
=
+
f
∂x dx2
∂y dx2
..
.
(1)
(2)
(3)
And so on. Hence if we name F0 = f (x, y) then the above can be written as
F0 = f (x, y)
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
=
Fn−1 +
F0
∂x
∂y
(4)
(5)
chapter 2 . book solved problems
95
For example, for n = 1 we see that
d
(F0 )
dx
∂
∂F0
=
F0 +
F0
∂x
∂y
∂f
∂f
=
+
f
∂x ∂y
F1 =
Which is (1). And when n = 2
d
(F1 )
dx
∂
∂F1
=
F1 +
F0
∂x
∂y
∂ ∂f
∂f
∂ ∂f
∂f
=
+
f +
+
f f
∂x ∂x ∂y
∂y ∂x ∂y
∂ df
∂ df
=
+
f
∂x dx
∂y dx
F2 =
Which is (2) and so on. Therefore (4,5) can be used from now on along with
y(x) = y0 +
∞
X
xn+1
Fn |x0 ,y0
(n + 1) !
n=0
Hence
yx
x−2
dF0
F1 =
dx
∂F0 ∂F0
=
+
F0
∂x
∂y
y(x2 − 2)
=
(x − 2)2
dF1
F2 =
dx
∂F1 ∂F1
=
+
F1
∂x
∂y
y(x2 + 2x − 2)
=
(x − 2)2
dF2
F3 =
dx
∂F2 ∂F2
=
+
F2
∂x
∂y
yx(x + 4)
=
(x − 2)2
dF3
F4 =
dx
∂F3 ∂F3
=
+
F3
∂x
∂y
y(x2 + 6x + 4)
=
(x − 2)2
F0 =
And so on. Evaluating all the above at initial conditions x(0) = 0 and y(0) = 4 gives
F0 = 0
F1 = −2
F2 = −2
F3 = 0
F4 = 4
(6)
chapter 2 . book solved problems
96
Substituting all the above in (6) and simplifying gives the solution as
y = −x2 + 4 −
x3 x5
+
+ O x6
3
30
Since x = 0 is also an ordinary point, then standard power series can also be used. Writing
the ODE as
y 0 + q(x)y = p(x)
yx
y0 −
=0
x−2
Where
q(x) = −
x
x−2
p(x) = 0
Next, the type of the expansion point x = 0 is determined. This point can be an ordinary
point, a regular singular point (also called removable singularity), or irregular singular point
(also called non-removable singularity or essential singularity). When x = 0 is an ordinary
point, then the standard power series is used. If the point is a regular singular point, Frobenius
series is used instead. Irregular singular point requires more advanced methods (asymptotic
methods) and is not supported now. Hopefully this will be added in the future. x = 0 is
called an ordinary point q(x) has a Taylor series expansion around the point x = 0. x = 0
is called a regular singular point if q(x) is not not analytic at x = 0 but xq(x) has Taylor
series expansion. And finally, x = 0 is an irregular singular point if the point is not ordinary
and not regular singular. This is the most complicated case. Now the expansion point x = 0
is checked to see if it is an ordinary point or not. Let the solution be represented as power
series of the form
∞
X
y=
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
Substituting the above back into the ode gives
(x − 2)
∞
X
!
nan xn−1
=
n=1
∞
X
!
an x n x
(1)
n=0
Which simplifies to
∞
X
!
nan x
n
n=1
+
∞
X
−2nan x
n−1
n =1
+
∞
X
−x1+n an = 0
(2)
n =0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
−2nan x
n =1
n−1
=
∞
X
(−2(1 + n) a1+n xn )
n=0
∞
X
∞
X
−x1+n an =
(−an−1 xn )
n =0
n=1
chapter 2 . book solved problems
97
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
nan x
n=1
n
+
∞
X
n
(−2(1 + n) a1+n x ) +
n =0
∞
X
(−an−1 xn ) = 0
(3)
n =1
For 1 ≤ n, the recurrence equation is
nan − 2(1 + n) a1+n − an−1 = 0
(4)
Solving for a1+n , gives
a1+n =
nan − an−1
2 + 2n
For n = 1 the recurrence equation gives
a1 − 4a2 − a0 = 0
Which after substituting the earlier terms found becomes
a2 = −
a0
4
For n = 2 the recurrence equation gives
2a2 − 6a3 − a1 = 0
Which after substituting the earlier terms found becomes
a3 = −
a0
12
For n = 3 the recurrence equation gives
3a3 − 8a4 − a2 = 0
Which after substituting the earlier terms found becomes
a4 = 0
For n = 4 the recurrence equation gives
4a4 − 10a5 − a3 = 0
Which after substituting the earlier terms found becomes
a5 =
a0
120
For n = 5 the recurrence equation gives
5a5 − 12a6 − a4 = 0
Which after substituting the earlier terms found becomes
a6 =
a0
288
(5)
chapter 2 . book solved problems
98
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
1
1
1
y = a0 − a0 x2 − a0 x3 +
a0 x 5 + . . .
4
12
120
Collecting terms, the solution becomes
1 2
1 3
1 5
y = 1− x − x +
x a0 + O x 6
4
12
120
(3)
At x = 0 the solution above becomes
y(0) = a0
Therefore the solution in Eq(3) now can be written as
y = −x2 + 4 −
x3 x5
+
+ O x6
3
30
(a) Solutions plot
(b) Slope field (x − 2) y 0 = yx
3 Maple. Time used: 0.004 (sec). Leaf size: 16
Order:=6;
ode:=(x-2)*diff(y(x),x) = x*y(x);
ic:=[y(0) = 4];
dsolve([ode,op(ic)],y(x),type='series',x=0);
1
1
y = 4 − x2 − x 3 + x5 + O x6
3
30
Maple trace
Methods for first order ODEs:
--- Trying classification methods --trying a quadrature
trying 1st order linear
<- 1st order linear successful
chapter 2 . book solved problems
99
Maple step by step
•
•
•
Let’s solve
d
(x − 2) dx
y(x) = xy(x) , y(0) = 4
Highest derivative means the order of the ODE is 1
d
y(x)
dx
Solve for the highest derivative
d
y(x) = xy(x)
dx
x−2
Separate variables
d
y(x)
dx
x
= x−2
Integrate both sides with respect to x
R dxd y(x)
R x
dx = x−2
dx + C 1
y(x)
Evaluate integral
ln (y(x)) = x + 2 ln (x − 2) + C 1
Solve for y(x)
y(x) = ex+C 1 (x − 2)2
Redefine the integration constant(s)
y(x) = C 1 ex (x − 2)2
Use initial condition y(0) = 4
4 = 4C 1
Solve for _C1
C1 = 1
Substitute _C1 = 1 into general solution and simplify
y(x) = ex (x − 2)2
Solution to the IVP
y(x) = ex (x − 2)2
y(x)
•
•
•
•
•
•
•
•
3 Mathematica. Time used: 0.001 (sec). Leaf size: 24
ode=(x-2)*D[y[x],x]==x*y[x];
ic={y[0]==4};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
x5 x3
y(x) →
−
− x2 + 4
30
3
3 Sympy. Time used: 0.195 (sec). Leaf size: 20
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-x*y(x) + (x - 2)*Derivative(y(x), x),0)
ics = {y(0): 4}
dsolve(ode,func=y(x),ics=ics,hint="1st_power_series",x0=0,n=6)
y(x) = 4 − x2 −
x3 x5
+
+ O x6
3
30
chapter 2 . book solved problems
2.2
100
Chapter 5. Series Solutions of ODEs. Special
Functions. Problem set 5.3. Extended Power
Series Method: Frobenius Method page 186
Local contents
2.2.1 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.3 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.4 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.5 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.6 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.7 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.8 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.9 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.10 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.11 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.12 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.13 Problem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.14 Problem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.15 Problem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.16 Problem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.17 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.18 Problem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
101
107
116
126
133
144
150
157
166
175
184
193
200
209
218
227
235
243
chapter 2 . book solved problems
2.2.1
101
Problem 2
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
Internal problem ID [8579]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 2
Date solved : Tuesday, November 11, 2025 at 12:03:46 PM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]
(x − 2)2 y 00 + (x + 2) y 0 − y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X
x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.21)
(2.22)
(2)
(3)
chapter 2 . book solved problems
102
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
103
To find y(x) series solution around x = 0. Hence
y 0 x + 2y 0 − y
(x − 2)2
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
(y 0 x + 2y 0 − y) (3x − 2)
=
(x − 2)4
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
4(y 0 x + 2y 0 − y) x(3x − 4)
=−
(x − 2)6
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
60((x + 2) y 0 − y) x3 − 2x2 + 16
15
=
(x − 2)8
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
8(y 0 x + 2y 0 − y) (45x4 − 120x3 + 192x − 112)
=−
(x − 2)10
F0 = −
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
y(0) y 0 (0)
−
4
2
y(0) y 0 (0)
F1 =
−
8
4
F2 = 0
y 0 (0) y(0)
F3 =
−
2
4
7y(0) 7y 0 (0)
F4 = −
+
8
4
F0 =
Substituting all the above in (7) and simplifying gives the solution as
y=
1 2
1 3
1 5
1 2
1 3
1 5 0
1+ x + x −
x y(0) + x − x − x +
x y (0) + O x6
8
48
480
4
24
240
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
x2 − 4x + 4 y 00 + (x + 2) y 0 − y = 0
Let the solution be represented as power series of the form
y=
∞
X
n=0
an x n
chapter 2 . book solved problems
104
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
x2 − 4x + 4
n(n − 1) an xn−2 + (x + 2)
nan xn−1 −
n=2
n=1
∞
X
!
an x n
=0
(1)
n=0
Which simplifies to
∞
X
!
xn an n(n − 1)
+
n=2
+
∞
X
!
nan x
n
+
∞
X
−4n xn−1 an (n − 1) +
n =2
∞
X
n=1
∞
X
!
4n(n − 1) an xn−2
(2)
n=2
!
2nan x
n−1
+
n=1
∞
X
(−an xn ) = 0
n =0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
∞
X
−4n xn−1 an (n − 1) =
(−4(n + 1) an+1 n xn )
n =2
n=1
∞
X
4n(n − 1) an x
n−2
=
n =2
∞
X
4(n + 2) an+2 (n + 1) xn
n=0
∞
X
2nan x
n−1
n =1
=
∞
X
2(n + 1) an+1 xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
xn an n(n − 1)
+
n=2
+
∞
X
n=1
!
nan x
n
+
∞
X
(−4(n + 1) an+1 n xn ) +
n =1
∞
X
∞
X
!
4(n + 2) an+2 (n + 1) xn
n=0
!
2(n + 1) an+1 x
n
+
n=0
∞
X
n =0
n = 0 gives
8a2 + 2a1 − a0 = 0
a2 =
a0 a1
−
8
4
n = 1 gives
−4a2 + 24a3 = 0
Which after substituting earlier equations, simplifies to
−
a0
+ a1 + 24a3 = 0
2
(−an xn ) = 0
(3)
chapter 2 . book solved problems
105
Or
a3 =
a0
a1
−
48 24
For 2 ≤ n, the recurrence equation is
nan (n − 1) − 4(n + 1) an+1 n + 4(n + 2) an+2 (n + 1) + nan + 2(n + 1) an+1 − an = 0 (4)
Solving for an+2 , gives
an+2 = −
=−
nan − 4nan+1 − an + 2an+1
4 (n + 2)
(n − 1) an (−4n + 2) an+1
−
4 (n + 2)
4 (n + 2)
For n = 2 the recurrence equation gives
3a2 − 18a3 + 48a4 = 0
Which after substituting the earlier terms found becomes
a4 = 0
For n = 3 the recurrence equation gives
8a3 − 40a4 + 80a5 = 0
Which after substituting the earlier terms found becomes
a5 = −
a0
a1
+
480 240
For n = 4 the recurrence equation gives
15a4 − 70a5 + 120a6 = 0
Which after substituting the earlier terms found becomes
a6 = −
7a0
7a1
+
5760 2880
For n = 5 the recurrence equation gives
24a5 − 108a6 + 168a7 = 0
Which after substituting the earlier terms found becomes
a7 = −
13a0
13a1
+
26880 13440
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
y = a0 + a1 x +
a
0
8
−
a1 2 a0
a1 3 a0
a1 5
x +
−
x + −
+
x + ...
4
48 24
480 240
(5)
chapter 2 . book solved problems
106
Collecting terms, the solution becomes
y=
1 2
1 3
1 5
1 2
1 3
1 5
1+ x + x −
x a0 + x − x − x +
x a1 + O x 6
8
48
480
4
24
240
(3)
At x = 0 the solution above becomes
y=
1 2
1 3
1 5
1 2
1 3
1 5 0
1+ x + x −
x y(0) + x − x − x +
x y (0) + O x6
8
48
480
4
24
240
3 Maple. Time used: 0.009 (sec). Leaf size: 49
Order:=6;
ode:=(x-2)^2*diff(diff(y(x),x),x)+(2+x)*diff(y(x),x)-y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
y=
1 2
1 3
1 5
1 2
1 3
1 5 0
1+ x + x −
x y(0) + x − x − x +
x y (0) + O x6
8
48
480
4
24
240
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
<- linear_1 successful
3 Mathematica. Time used: 0.001 (sec). Leaf size: 56
ode=(x-2)^2*D[y[x],{x,2}]+(x+2)*D[y[x],x]-y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
5
x5
x3 x2
x
x3 x2
−
+
+
+ 1 + c2
−
−
+x
480 48
8
240 24
4
3 Sympy. Time used: 0.292 (sec). Leaf size: 32
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((x - 2)**2*Derivative(y(x), (x, 2)) + (x + 2)*Derivative(y(x), x) - y(x)
,0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
3
2
x
x2
x
x
y(x) = C2
+
+ 1 + C 1 x − − + 1 + O x6
48
8
24 4
chapter 2 . book solved problems
2.2.2
107
Problem 3
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
Internal problem ID [8580]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 3
Date solved : Tuesday, November 11, 2025 at 12:03:47 PM
CAS classification : [_Lienard]
xy 00 + 2y 0 + yx = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + 2y 0 + yx = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
2
x
q(x) = 1
p(x) =
Table 2.14: Table p(x), q(x) singularites.
p(x) = x2
singularity
type
x=0
“regular”
q(x) = 1
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + 2y 0 + yx = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
108
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x + 2
n=0
!
(n + r) an xn+r−1
∞
X
+
n=0
!
an xn+r
x = 0 (1)
n=0
Which simplifies to
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
!
2(n + r) an xn+r−1
+
n=0
∞
X
!
x1+n+r an
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
x
1+n+r
an =
n =0
∞
X
an−2 xn+r−1
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
!
2(n + r) an xn+r−1 +
n=0
∞
X
!
an−2 xn+r−1
n=2
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + 2(n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + 2ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + 2r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
r x−1+r (1 + r) = 0
Since the above is true for all x then the indicial equation becomes
r(1 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = −1
Since a0 6= 0 then the indicial equation becomes
r x−1+r (1 + r) = 0
= 0 (2B)
chapter 2 . book solved problems
109
Solving for r gives the roots of the indicial equation as [0, −1].
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) + xr2
!
bn x n
n=0
Or
y1 (x) =
∞
X
an x n
n=0
∞
P
y2 (x) = Cy1 (x) ln (x) +
bn x n
n=0
x
Or
y1 (x) =
∞
X
an xn
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn xn−1
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + 2an (n + r) + an−2 = 0
(3)
Solving for an from recursive equation (4) gives
an−2
an = − 2
n + 2nr + r2 + n + r
(4)
Which for the root r = 0 becomes
an = −
an−2
n (1 + n)
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
1
a2 = − 2
r + 5r + 6
Which for the root r = 0 becomes
a2 = −
And the table now becomes
1
6
chapter 2 . book solved problems
n
a0
a1
a2
110
an,r
1
0
1
− r2 +5r+6
an
1
0
− 16
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
1
− r2 +5r+6
0
an
1
0
− 16
0
For n = 4, using the above recursive equation gives
a4 =
1
r4 + 14r3 + 71r2 + 154r + 120
Which for the root r = 0 becomes
a4 =
1
120
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
1
− r2 +5r+6
0
an
1
0
− 16
0
1
r4 +14r3 +71r2 +154r+120
1
120
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
1
− r2 +5r+6
0
an
1
0
− 16
0
1
r4 +14r3 +71r2 +154r+120
1
120
0
0
Using the above table, then the solution y1 (x) is
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x2
x4
=1−
+
+ O x6
6
120
Now the second solution y2 (x) is found. Let
r1 − r2 = N
chapter 2 . book solved problems
111
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=0
Therefore
lim 0 = lim 0
r→r2
r→−1
=0
The limit is 0. Since the limit exists then the log term is not needed and we can set C = 0.
Therefore the second solution has the form
y2 (x) =
=
∞
X
n=0
∞
X
bn xn+r
bn xn−1
n=0
Eq (3) derived above is used to find all bn coefficients. The case n = 0 is skipped since
it was used to find the roots of the indicial equation. b0 is arbitrary and taken as b0 = 1.
Substituting n = 1 in Eq(3) gives
b1 = 0
For 2 ≤ n the recursive equation is
bn (n + r) (n + r − 1) + 2(n + r) bn + bn−2 = 0
(4)
Which for for the root r = −1 becomes
bn (n − 1) (n − 2) + 2(n − 1) bn + bn−2 = 0
(4A)
Solving for bn from the recursive equation (4) gives
bn−2
bn = − 2
n + 2nr + r2 + n + r
(5)
Which for the root r = −1 becomes
bn−2
bn = − 2
n −n
(6)
At this point, it is a good idea to keep track of bn in a table both before substituting r = −1
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
0
bn
1
0
For n = 2, using the above recursive equation gives
1
b2 = − 2
r + 5r + 6
Which for the root r = −1 becomes
b2 = −
And the table now becomes
1
2
chapter 2 . book solved problems
n
b0
b1
b2
112
bn,r
1
0
1
− r2 +5r+6
bn
1
0
− 12
For n = 3, using the above recursive equation gives
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
0
1
− r2 +5r+6
0
bn
1
0
− 12
0
For n = 4, using the above recursive equation gives
b4 =
1
(r2 + 5r + 6) (r2 + 9r + 20)
Which for the root r = −1 becomes
b4 =
1
24
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
1
− r2 +5r+6
0
bn
1
0
− 12
0
1
r4 +14r3 +71r2 +154r+120
1
24
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
0
1
− r2 +5r+6
0
bn
1
0
− 12
0
1
r4 +14r3 +71r2 +154r+120
1
24
0
0
Using the above table, then the solution y2 (x) is
y2 (x) = 1 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
2
4
1 − x2 + x24 + O(x6 )
=
x
chapter 2 . book solved problems
113
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
c 1 − x2 + x4 + O(x6 )
2
4
2
2
24
x
x
= c1 1 −
+
+ O x6 +
6
120
x
Hence the final solution is
y = yh
2
= c1 1 −
4
x
x
+
+ O x6 +
6
120
2
4
c2 1 − x2 + x24 + O(x6 )
x
3 Maple. Time used: 0.029 (sec). Leaf size: 32
Order:=6;
ode:=diff(diff(y(x),x),x)*x+2*diff(y(x),x)+x*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2
1 4
6
c
1
−
x
+
x
+
O
(x
)
1 2
1 4
2
2
24
y = c1 1 − x +
x + O x6 +
6
120
x
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Group is reducible or imprimitive
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x dx
y(x) + 2 dx
y(x) + xy(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative •
•
d d
y(x) = −y(x) −
dx dx
•
2
d
y(x)
dx
x
Group termswith y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(x) +
dx dx
◦
◦
d
y(x)
dx
+ y(x) = 0
Check to see if x0 = 0 is a regular singular point
Define functions
P2 (x) = x2 , P3 (x) = 1
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
2
x
=2
x=0
x2 · P3 (x) is analytic at x = 0
chapter 2 . book solved problems
(x2 · P3 (x))
◦
•
•
114
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x) + 2 dx
y(x) + xy(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert x · y(x) to series expansion
∞
P
x · y(x) =
ak xk+r+1
k=0
◦
Shift index using k− >k − 1
∞
P
x · y(x) =
ak−1 xk+r
◦
d
Convert dx
y(x) to series expansion
∞
P
d
y(x) =
ak (k + r) xk+r−1
dx
k=1
k=0
◦
◦
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + r + 1) xk+r
dx
k=−1
d d
Convert x · dx
y(x)
to series expansion
dx
∞
P
d d
x · dx
y(x) =
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + r + 1) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
a0 r(1 + r) x
−1+r
r
+ a1 (1 + r) (2 + r) x +
∞
P
(ak+1 (k + r + 1) (k + 2 + r) + ak−1 ) x
k+r
=0
k=1
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r(1 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {−1, 0}
Each term must be 0
a1 (1 + r) (2 + r) = 0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + r + 1) (k + 2 + r) + ak−1 = 0
Shift index using k− >k + 1
ak+2 (k + 2 + r) (k + 3 + r) + ak = 0
Recursion relation that defines series solution to ODE
ak
ak+2 = − (k+2+r)(k+3+r)
Recursion relation for r = −1
ak
ak+2 = − (k+1)(k+2)
Solution
for r = −1
∞
P
ak
k−1
y(x) =
ak x , ak+2 = − (k+1)(k+2) , 0 = 0
k=0
•
•
Recursion relation for r = 0
ak
ak+2 = − (k+2)(k+3)
Solution
for r = 0
∞
P
ak
k
y(x) =
ak x , ak+2 = − (k+2)(k+3) , 2a1 = 0
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
ak
bk
k−1
k
y(x) =
ak x
+
bk x , ak+2 = − (k+1)(k+2) , 0 = 0, bk+2 = − (k+2)(k+3) , 2b1 = 0
k=0
k=0
chapter 2 . book solved problems
3 Mathematica. Time used: 0.006 (sec). Leaf size: 42
ode=x*D[y[x],{x,2}]+2*D[y[x],x]+x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
115
3
4
x
x 1
x
x2
y(x) → c1
− +
+ c2
−
+1
24 2 x
120
6
3 Sympy. Time used: 0.264 (sec). Leaf size: 39
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*y(x) + x*Derivative(y(x), (x, 2)) + 2*Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C2
C − x6 + x4 − x2 + 1
1
720
24
2
x
x
−
+1 +
+ O x6
120
6
x
4
2
chapter 2 . book solved problems
2.2.3
116
Problem 4
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Internal problem ID [8581]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 4
Date solved : Tuesday, November 11, 2025 at 12:03:47 PM
CAS classification : [[_Emden, _Fowler]]
xy 00 + y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
p(x) = 0
1
q(x) =
x
Table 2.16: Table p(x), q(x) singularites.
q(x) = x1
singularity
type
x=0
“regular”
p(x) = 0
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
117
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
(n + r) (n + r − 1) an xn+r−2 x +
∞
X
n=0
!
an xn+r
=0
(1)
=0
(2A)
n=0
Which simplifies to
∞
X
!
xn+r−1 an (n + r) (n + r − 1)
∞
X
+
n=0
!
an xn+r
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
an x
n+r
n =0
=
∞
X
an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n + r) (n + r − 1)
+
n=0
∞
X
!
an−1 xn+r−1
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) = 0
Or
x−1+r a0 r(−1 + r) = 0
Since a0 6= 0 then the above simplifies to
x−1+r r(−1 + r) = 0
Since the above is true for all x then the indicial equation becomes
r(−1 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 1
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r(−1 + r) = 0
=0
(2B)
chapter 2 . book solved problems
118
Solving for r gives the roots of the indicial equation as [1, 0].
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) + xr2
!
bn x n
n=0
Or
∞
X
y1 (x) = x
!
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) +
!
bn x n
n=0
Or
y1 (x) =
∞
X
an xn+1
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) +
!
bn x n
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. For 1 ≤ n the recursive equation is
an (n + r) (n + r − 1) + an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an = −
an−1
(n + r) (n + r − 1)
(4)
an−1
(n + 1) n
(5)
Which for the root r = 1 becomes
an = −
At this point, it is a good idea to keep track of an in a table both before substituting r = 1
and after as more terms are found using the above recursive equation.
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 = −
1
(1 + r) r
Which for the root r = 1 becomes
a1 = −
1
2
And the table now becomes
n
a0
a1
an,r
1
1
− (1+r)r
an
1
− 12
chapter 2 . book solved problems
119
For n = 2, using the above recursive equation gives
a2 =
1
(1 + r) r (2 + r)
2
Which for the root r = 1 becomes
1
12
a2 =
And the table now becomes
n
a0
a1
a2
an,r
1
1
− (1+r)r
an
1
− 12
1
(1+r)2 r(2+r)
1
12
For n = 3, using the above recursive equation gives
a3 = −
1
(1 + r) r (2 + r)2 (3 + r)
2
Which for the root r = 1 becomes
a3 = −
1
144
And the table now becomes
n
a0
a1
a2
a3
an,r
1
1
− (1+r)r
an
1
− 12
1
(1+r)2 r(2+r)
1
− (1+r)2 r(2+r)
2
(3+r)
1
12
1
− 144
For n = 4, using the above recursive equation gives
a4 =
1
(1 + r) r (2 + r)2 (3 + r)2 (4 + r)
2
Which for the root r = 1 becomes
a4 =
1
2880
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
1
− (1+r)r
an
1
− 12
1
(1+r)2 r(2+r)
1
− (1+r)2 r(2+r)
2
(3+r)
1
(1+r)2 r(2+r)2 (3+r)2 (4+r)
1
12
1
− 144
1
2880
For n = 5, using the above recursive equation gives
a5 = −
1
(1 + r) r (2 + r) (3 + r)2 (4 + r)2 (5 + r)
2
2
Which for the root r = 1 becomes
a5 = −
And the table now becomes
1
86400
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
120
an,r
1
1
− (1+r)r
an
1
− 12
1
(1+r)2 r(2+r)
1
− (1+r)2 r(2+r)
2
(3+r)
1
2
2
(1+r) r(2+r) (3+r)2 (4+r)
1
− (1+r)2 r(2+r)2 (3+r)
2
(4+r)2 (5+r)
1
12
1
− 144
1
2880
1
− 86400
Using the above table, then the solution y1 (x) is
y1 (x) = x a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x x2
x3
x4
x5
6
=x 1− +
−
+
−
+O x
2 12 144 2880 86400
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=−
1
(1 + r) r
Therefore
lim −
r→r2
1
1
= lim −
(1 + r) r r→0 (1 + r) r
= undefined
Since the limit does not exist then the log term is needed. Therefore the second solution has
the form
!
∞
X
y2 (x) = Cy1 (x) ln (x) +
bn xn+r2
n=0
Therefore
d
Cy1 (x)
y2 (x) = Cy10 (x) ln (x) +
+
dx
x
Cy1 (x)
= Cy10 (x) ln (x) +
+
x
∞
X
bn xn+r2 (n + r2 )
x
n=0
∞
X
!
x
−1+n+r2
bn (n + r2 )
n=0
∞
d2
2Cy10 (x) Cy1 (x) X
00
y2 (x) = Cy1 (x) ln (x) +
−
+
dx2
x
x2
n=0
2Cy10 (x) Cy1 (x)
= Cy100 (x) ln (x) +
−
+
2
x
!
x
∞
X
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!
x
−2+n+r2
bn (n + r2 ) (−1 + n + r2 )
n=0
Substituting these back into the given ode xy 00 + y = 0 gives
∞
2Cy10 (x) Cy1 (x) X
00
Cy1 (x) ln (x) +
−
+
x
x2
n=0
!
∞
X
+ Cy1 (x) ln (x) +
bn xn+r2 = 0
n=0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!!
x
!
chapter 2 . book solved problems
121
Which can be written as
0
2y1 (x) y1 (x)
00
(y1 (x) x + y1 (x)) ln (x) + x
− 2
C
x
x
!!
∞
X
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
+
−
x+
x2
x2
n=0
(7)
∞
X
!
bn xn+r2
=0
n=0
But since y1 (x) is a solution to the ode, then
y100 (x) x + y1 (x) = 0
Eq (7) simplifes to
2y10 (x) y1 (x)
x
− 2
x
x
C+
∞
X
n=0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
∞
X
!!
x+
!
bn xn+r2
=0
n=0
(8)
Substituting y1 =
∞
P
an xn+r1 into the above gives
n=0
∞
∞
P −1+n+r1
P
n+r1
2
x
an (n + r1 ) x −
an x
C
n=0
n=0
(9)
x
+
∞
X
!
x
−2+n+r2
bn (n + r2 ) (−1 + n + r2 ) x +
n=0
∞
X
!
bn xn+r2
=0
n=0
Since r1 = 1 and r2 = 0 then the above becomes
∞
∞
P n
P
n+1
2
x an (n + 1) x −
an x
C
n=0
n=0
∞
X
+
x
!
x−2+n bn n(n − 1) x +
n=0
∞
X
!
bn x n
=0
n=0
(10)
Which simplifies to
∞
X
n=0
!
2xn an (n + 1) C
+
∞
X
∞
X
(−C xn an ) +
n =0
!
n xn−1 bn (n − 1)
+
n=0
∞
X
!
bn x n
= 0 (2A)
n=0
The next step is to make all powers of x be n − 1 in each summation term. Going over each
summation term above with power of x in it which is not already xn−1 and adjusting the
power and the corresponding index gives
∞
X
2xn an (n + 1) C =
n =0
∞
X
2Can−1 n xn−1
n=1
∞
X
n
(−C x an ) =
n =0
∞
X
−Can−1 xn−1
n=1
∞
X
n =0
n
bn x =
∞
X
n=1
bn−1 xn−1
chapter 2 . book solved problems
122
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n − 1.
∞
X
n=1
!
2Can−1 n xn−1
+
∞
X
−Can−1 x
n−1
n =1
+
∞
X
!
n xn−1 bn (n − 1)
n=0
+
∞
X
!
bn−1 xn−1
=0
n=1
(2B)
For n = 0 in Eq. (2B), we choose arbitray value for b0 as b0 = 1. For n = N , where N = 1
which is the difference between the two roots, we are free to choose b1 = 0. Hence for n = 1,
Eq (2B) gives
C +1=0
Which is solved for C. Solving for C gives
C = −1
For n = 2, Eq (2B) gives
3Ca1 + b1 + 2b2 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
3
2b2 + = 0
2
Solving the above for b2 gives
3
b2 = −
4
For n = 3, Eq (2B) gives
5Ca2 + b2 + 6b3 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
7
6b3 − = 0
6
Solving the above for b3 gives
7
b3 =
36
For n = 4, Eq (2B) gives
7Ca3 + b3 + 12b4 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
35
12b4 +
=0
144
Solving the above for b4 gives
35
b4 = −
1728
For n = 5, Eq (2B) gives
9Ca4 + b4 + 20b5 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
101
20b5 −
=0
4320
Solving the above for b5 gives
101
b5 =
86400
Now that we found all bn and C, we can calculate the second solution from
!
∞
X
y2 (x) = Cy1 (x) ln (x) +
bn xn+r2
n=0
chapter 2 . book solved problems
123
Using the above value found for C = −1 and all bn , then the second solution becomes
x x2
x3
x4
x5
6
y2 (x) = (−1) x 1 − +
−
+
−
+O x
ln (x)
2 12 144 2880 86400
3x2 7x3 35x4 101x5
+1−
+
−
+
+ O x6
4
36
1728 86400
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x x2
x3
x4
x5
6
= c1 x 1 − +
−
+
−
+O x
2 12 144 2880 86400
x x2
x3
x4
x5
3x2
6
+ c2 (−1) x 1 − +
−
+
−
+O x
ln (x) + 1 −
2 12 144 2880 86400
4
3
4
5
7x
35x
101x
6
+
−
+
+O x
36
1728 86400
Hence the final solution is
y = yh
x x2
x3
x4
x5
6
= c1 x 1 − +
−
+
−
+O x
2 12 144 2880 86400
x x2
x3
x4
x5
3x2 7x3 35x4
6
+ c2 −x 1 − +
−
+
−
+O x
ln (x) + 1 −
+
−
2 12 144 2880 86400
4
36
1728
101x5
6
+
+O x
86400
3 Maple. Time used: 0.027 (sec). Leaf size: 58
Order:=6;
ode:=y(x)+diff(diff(y(x),x),x)*x = 0;
dsolve(ode,y(x),type='series',x=0);
1
1 2
1 3
1 4
1
5
6
y = c1 x 1 − x + x −
x +
x −
x +O x
2
12
144
2880
86400
1 2
1 3
1 4
1 5
6
+ c2 ln (x) −x + x − x +
x −
x +O x
2
12
144
2880
3 2
7 3
35 4
101 5
6
+ 1− x + x −
x +
x +O x
4
36
1728
86400
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
chapter 2 . book solved problems
124
Maple step by step
•
•
•
◦
◦
Let’s solve
d d
x dx
y(x) + y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
d d
y(x) = − y(x)
dx dx
x
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(x) + y(x)
=0
dx dx
x
Check to see if x0 = 0 is a regular singular point
Define functions
P2 (x) = 0, P3 (x) = x1
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
=0
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
x dx
y(x) + y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series
∞ expansions
P
a0 r(−1 + r) x−1+r +
(ak+1 (k + 1 + r) (k + r) + ak ) xk+r = 0
k=0
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r(−1 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {0, 1}
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1 + r) (k + r) + ak = 0
Recursion relation that defines series solution to ODE
ak
ak+1 = − (k+1+r)(k+r)
Recursion relation for r = 0
ak
ak+1 = − (k+1)k
Solution
for r = 0
∞
P
ak
k
y(x) =
ak x , ak+1 = − (k+1)k
k=0
•
•
Recursion relation for r = 1
ak
ak+1 = − (k+2)(k+1)
Solution
for r = 1
∞
P
ak
k+1
y(x) =
ak x , ak+1 = − (k+2)(k+1)
k=0
•
Combine solutions and rename parameters
chapter 2 . book solved problems
y(x) =
∞
P
k=0
ak x
k
+
∞
P
k=0
bk x
k+1
125
ak
bk
, ak+1 = − (k+1)k
, bk+1 = − (k+2)(k+1)
3 Mathematica. Time used: 0.011 (sec). Leaf size: 85
ode=x*D[y[x],{x,2}]+y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
1
−47x4 + 480x3 − 2160x2 + 1728x + 1728
y(x) → c1
x x3 − 12x2 + 72x − 144 log(x) +
144
1728
5
4
3
2
x
x
x
x
+ c2
−
+
−
+x
2880 144 12
2
3 Sympy. Time used: 0.172 (sec). Leaf size: 29
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*Derivative(y(x), (x, 2)) + y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
x4
x3
x2 x
y(x) = C1 x
−
+
− + 1 + O x6
2880 144 12 2
chapter 2 . book solved problems
2.2.4
126
Problem 5
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
Internal problem ID [8582]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 5
Date solved : Tuesday, November 11, 2025 at 12:03:48 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
xy 00 + (2x + 1) y 0 + (x + 1) y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + (2x + 1) y 0 + (x + 1) y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
2x + 1
x
x+1
q(x) =
x
p(x) =
Table 2.18: Table p(x), q(x) singularites.
q(x) = x+1
x
singularity
type
x=0
“regular”
p(x) = 2x+1
x
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + (2x + 1) y 0 + (x + 1) y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
127
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
(n + r) (n + r − 1) an xn+r−2 x
(1)
n=0
+ (2x + 1)
!
∞
X
(n + r) an xn+r−1
+ (x + 1)
n=0
∞
X
!
an xn+r
=0
n=0
Which simplifies to
!
∞
X
xn+r−1 an (n + r) (n + r − 1)
+
∞
X
n=0
+
!
2xn+r an (n + r)
(2A)
n=0
∞
X
∞
X
!
(n + r) an x
n+r−1
+
n=0
∞
X
!
x
1+n+r
+
an
n=0
!
an xn+r
=0
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
2x
n+r
an (n + r) =
n =0
∞
X
2an−1 (n + r − 1) xn+r−1
n=1
∞
X
x
1+n+r
an =
n =0
∞
X
∞
X
an−2 xn+r−1
n=2
an x
n+r
=
n =0
∞
X
an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n + r) (n + r − 1)
n=0
+
+
∞
X
!
2an−1 (n + r − 1) xn+r−1
(2B)
n=1
∞
X
!
(n + r) an xn+r−1
n=0
+
∞
X
n=2
!
an−2 xn+r−1
+
∞
X
!
an−1 xn+r−1
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + (n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + r x−1+r a0 = 0
=0
chapter 2 . book solved problems
128
Since a0 6= 0 then the above simplifies to
x−1+r r2 = 0
Since the above is true for all x then the indicial equation becomes
r2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r2 = 0
Solving for r gives the roots of the indicial equation as [0, 0].
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
y1 (x) =
∞
X
an xn+r
(1A)
n=0
Now the second solution y2 is found using
y2 (x) = y1 (x) ln (x) +
∞
X
!
bn xn+r
(1B)
n=1
Then the general solution will be
y = c1 y1 (x) + c2 y2 (x)
In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), a0 is never zero, and is arbitrary
and is typically taken as a0 = 1, and {c1 , c2 } are two arbitray constants of integration which
can be found from initial conditions. We start by finding the first solution y1 (x). Eq (2B)
derived above is now used to find all an coefficients. The case n = 0 is skipped since it
was used to find the roots of the indicial equation. a0 is arbitrary and taken as a0 = 1.
Substituting n = 1 in Eq. (2B) gives
a1 =
−2r − 1
(1 + r)2
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + 2an−1 (n + r − 1) + an (n + r) + an−2 + an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an = −
2nan−1 + 2ran−1 + an−2 − an−1
n2 + 2nr + r2
(4)
−2nan−1 − an−2 + an−1
n2
(5)
Which for the root r = 0 becomes
an =
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
−2r−1
(1+r)2
an
1
−1
chapter 2 . book solved problems
129
For n = 2, using the above recursive equation gives
a2 =
3r2 + 6r + 2
(1 + r)2 (r + 2)2
Which for the root r = 0 becomes
1
2
a2 =
And the table now becomes
n
a0
a1
an,r
1
−2r−1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
a2
an
1
−1
1
2
For n = 3, using the above recursive equation gives
a3 =
−4r3 − 18r2 − 22r − 6
(1 + r)2 (r + 2)2 (r + 3)2
Which for the root r = 0 becomes
a3 = −
1
6
And the table now becomes
n
a0
a1
a2
a3
an,r
1
−2r−1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
−4r3 −18r2 −22r−6
(1+r)2 (r+2)2 (r+3)2
an
1
−1
1
2
− 16
For n = 4, using the above recursive equation gives
5r4 + 40r3 + 105r2 + 100r + 24
a4 =
(1 + r)2 (r + 2)2 (r + 3)2 (r + 4)2
Which for the root r = 0 becomes
a4 =
1
24
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
−2r−1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
−4r3 −18r2 −22r−6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
an
1
−1
1
2
− 16
1
24
For n = 5, using the above recursive equation gives
a5 =
−6r5 − 75r4 − 340r3 − 675r2 − 548r − 120
(1 + r)2 (r + 2)2 (r + 3)2 (r + 4)2 (r + 5)2
Which for the root r = 0 becomes
a5 = −
And the table now becomes
1
120
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
130
an,r
1
−2r−1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
−4r3 −18r2 −22r−6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
−6r5 −75r4 −340r3 −675r2 −548r−120
(1+r)2 (r+2)2 (r+3)2 (r+4)2 (r+5)2
an
1
−1
1
2
− 16
1
24
1
− 120
Using the above table, then the first solution y1 (x) becomes
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
=1−x+
x2 x3 x4
x5
−
+
−
+ O x6
2
6
24 120
Now the second solution is found. The second solution is given by
!
∞
X
y2 (x) = y1 (x) ln (x) +
bn xn+r
n=1
Where bn is found using
d
an,r
dr
And the above is then evaluated at r = 0. The above table for an,r is used for this purpose.
Computing the derivatives gives the following table
bn =
n
b0
b1
b2
b3
b4
b5
bn,r
1
−2r−1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
−4r3 −18r2 −22r−6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
−6r5 −75r4 −340r3 −675r2 −548r−120
(1+r)2 (r+2)2 (r+3)2 (r+4)2 (r+5)2
an
1
−1
1
2
− 16
1
24
1
− 120
d
bn,r = dr
an,r
N/A since bn starts from 1
2r
(1+r)3
−6r3 −18r2 −14r
(1+r)3 (r+2)3
12r r4 +8r3 + 47
r2 +30r+ 85
2
6
(1+r)3 (r+2)3 (r+3)3
20 r6 +15r5 + 183
r4 +290r3 + 5031
r2 +453r+166 r
2
10
−
(1+r)3 (r+2)3 (r+3)3 (r+4)3
30 r8 +24r7 + 739
r6 +1410r5 +4915r4 +10668r3 +14063r2 +10290r+ 48076
r
3
15
3
3
3
3
3
(1+r) (r+2) (r+3) (r+4) (r+5)
The above table gives all values of bn needed. Hence the second solution is
y2 (x) = y1 (x) ln (x) + b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x2 x3 x4
x5
6
= 1−x+
−
+
−
+O x
ln (x) + O x6
2
6
24 120
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x2 x3 x4
x5
6
= c1 1 − x +
−
+
−
+O x
2
6
24 120
x2 x3 x4
x5
6
6
+ c2
1−x+
−
+
−
+O x
ln (x) + O x
2
6
24 120
Hence the final solution is
y = yh
x2 x3 x 4
x5
6
= c1 1 − x +
−
+
−
+O x
2
6
24 120
x2 x3 x4
x5
6
6
+ c2
1−x+
−
+
−
+O x
ln (x) + O x
2
6
24 120
bn (r =
N/A
0
0
0
0
0
chapter 2 . book solved problems
3 Maple. Time used: 0.025 (sec). Leaf size: 41
Order:=6;
ode:=diff(diff(y(x),x),x)*x+(2*x+1)*diff(y(x),x)+(x+1)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
131
1 2 1 3
1 4
1 5
y = 1−x+ x − x + x −
x (ln (x) c2 + c1 ) + O x6
2
6
24
120
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x dx
y(x)
+
(2x
+
1)
y(x)
+ (x + 1) y(x) = 0
dx
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d d
y(x) = − (x+1)y(x)
−
dx dx
x
•
(2x+1)
d
y(x)
dx
x
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
◦
◦
(x · P2 (x))
◦
•
•
d
y(x)
dx
x
=1
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
(2x+1)
+ (x+1)y(x)
=0
x
Check to see if x0 = 0 is a regular singular point
Define functions
x+1
P2 (x) = 2x+1
,
P
(x)
=
3
x
x
x · P2 (x) is analytic at x = 0
d d
y(x) +
dx dx
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x)
+
(2x
+
1)
y(x)
+ (x + 1) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..1
∞
P
xm · y(x) =
ak xk+r+m
k=0
chapter 2 . book solved problems
◦
◦
132
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert xm · dx
y(x) to series expansion for m = 0..1
∞
P
d
xm · dx
y(x) =
ak (k + r) xk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
xm · dx
y(x) =
ak+1−m (k + 1 − m + r) xk+r
k=−1+m
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
∞
P
2
2
a0 r2 x−1+r + a1 (1 + r) + a0 (1 + 2r) xr +
ak+1 (k + 1 + r) + ak (2k + 2r + 1) + ak−1 xk+r
k=1
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term must be 0
a1 (1 + r)2 + a0 (1 + 2r) = 0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1)2 + 2ak k + ak + ak−1 = 0
Shift index using k− >k + 1
ak+2 (k + 2)2 + 2ak+1 (k + 1) + ak+1 + ak = 0
Recursion relation that defines series solution to ODE
+ak +3ak+1
ak+2 = − 2kak+1(k+2)
2
•
Recursion relation for r = 0
+ak +3ak+1
ak+2 = − 2kak+1(k+2)
2
•
Solution
for r = 0
∞
P
2kak+1 +ak +3ak+1
k
y(x) =
ak x , ak+2 = −
, a 1 + a0 = 0
(k+2)2
k=0
3 Mathematica. Time used: 0.004 (sec). Leaf size: 78
ode=x*D[y[x],{x,2}]+(2*x+1)*D[y[x],x]+(x+1)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
x5
x4 x3 x2
x5
x4 x3 x2
−
+
−
+
− x + 1 + c2 −
+
−
+
− x + 1 log(x)
120 24
6
2
120 24
6
2
3 Sympy. Time used: 0.238 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*Derivative(y(x), (x, 2)) + (x + 1)*y(x) + (2*x + 1)*Derivative(y(x), x
),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
x5
x4 x3 x2
y(x) = C1 −
+
−
+
− x + 1 + O x6
120 24
6
2
chapter 2 . book solved problems
2.2.5
133
Problem 6
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
Internal problem ID [8583]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 6
Date solved : Tuesday, November 11, 2025 at 12:03:49 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
xy 00 + 2y 0 x3 + x2 − 2 y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + 2y 0 x3 + x2 − 2 y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
p(x) = 2x2
x2 − 2
q(x) =
x
Table 2.20: Table p(x), q(x) singularites.
2
q(x) = x x−2
singularity
type
x=0
“regular”
x=∞
“regular”
x = −∞ “regular”
2
p(x) = 2x
singularity
type
x=∞
“regular”
x = −∞ “regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [∞, −∞, 0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + 2y 0 x3 + x2 − 2 y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
134
Then
∞
X
0
y =
(n + r) an xn+r−1
n=0
∞
X
00
y =
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
(n + r) (n + r − 1) an xn+r−2 x
(1)
n=0
∞
X
+2
!
(n + r) an xn+r−1
∞
X
x3 + x2 − 2
n=0
!
an xn+r
=0
n=0
Which simplifies to
!
∞
X
xn+r−1 an (n + r) (n + r − 1)
+
∞
X
n=0
+
!
2x2+n+r an (n + r)
(2A)
n=0
∞
X
!
x
2+n+r
an
+
n=0
∞
X
−2an xn+r = 0
n =0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
2x
2+n+r
an (n + r) =
n =0
∞
X
2an−3 (n + r − 3) xn+r−1
n=3
∞
X
x
2+n+r
an =
n =0
∞
X
∞
X
an−3 xn+r−1
n=3
−2an x
n+r
n =0
=
∞
X
−2an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n + r) (n + r − 1)
n=0
+
+
∞
X
!
2an−3 (n + r − 3) xn+r−1
n=3
∞
X
!
an−3 xn+r−1
n=3
+
∞
X
−2an−1 xn+r−1 = 0
n =1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) = 0
Or
x−1+r a0 r(−1 + r) = 0
(2B)
chapter 2 . book solved problems
135
Since a0 6= 0 then the above simplifies to
x−1+r r(−1 + r) = 0
Since the above is true for all x then the indicial equation becomes
r(−1 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 1
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r(−1 + r) = 0
Solving for r gives the roots of the indicial equation as [1, 0].
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = Cy1 (x) ln (x) + xr2
∞
X
!
bn x n
n=0
Or
∞
X
y1 (x) = x
!
an x n
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
an xn+1
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x n
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 =
2
r (1 + r)
Substituting n = 2 in Eq. (2B) gives
a2 =
4
(1 + r) r (2 + r)
2
For 3 ≤ n the recursive equation is
an (n + r) (n + r − 1) + 2an−3 (n + r − 3) + an−3 − 2an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an = −
2nan−3 + 2ran−3 − 5an−3 − 2an−1
(n + r) (n + r − 1)
(4)
chapter 2 . book solved problems
136
Which for the root r = 1 becomes
an =
−2nan−3 + 3an−3 + 2an−1
(n + 1) n
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 1
and after as more terms are found using the above recursive equation.
n
a0
a1
a2
an,r
1
2
r(1+r)
4
(1+r)2 r(2+r)
an
1
1
1
3
For n = 3, using the above recursive equation gives
a3 =
−2r5 − 9r4 − 14r3 − 9r2 − 2r + 8
(1 + r)2 r (2 + r)2 (3 + r)
Which for the root r = 1 becomes
a3 = −
7
36
And the table now becomes
n
a0
a1
a2
an,r
1
a3
2
r(1+r)
4
(1+r)2 r(2+r)
−2r5 −9r4 −14r3 −9r2 −2r+8
(1+r)2 r(2+r)2 (3+r)
an
1
1
1
3
7
− 36
For n = 4, using the above recursive equation gives
a4 =
−8r5 − 56r4 − 168r3 − 268r2 − 220r − 56
(1 + r)2 r (2 + r)2 (3 + r)2 (4 + r)
Which for the root r = 1 becomes
a4 = −
97
360
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
2
r(1+r)
4
(1+r)2 r(2+r)
−2r5 −9r4 −14r3 −9r2 −2r+8
(1+r)2 r(2+r)2 (3+r)
−8r5 −56r4 −168r3 −268r2 −220r−56
(1+r)2 r(2+r)2 (3+r)2 (4+r)
an
1
1
1
3
7
− 36
97
− 360
For n = 5, using the above recursive equation gives
a5 =
−24r5 − 228r4 − 1000r3 − 2412r2 − 3056r − 1552
(1 + r)2 r (2 + r)2 (3 + r)2 (4 + r)2 (5 + r)
Which for the root r = 1 becomes
a5 = −
And the table now becomes
517
5400
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
137
an,r
1
2
r(1+r)
4
(1+r)2 r(2+r)
−2r5 −9r4 −14r3 −9r2 −2r+8
(1+r)2 r(2+r)2 (3+r)
−8r5 −56r4 −168r3 −268r2 −220r−56
(1+r)2 r(2+r)2 (3+r)2 (4+r)
−24r5 −228r4 −1000r3 −2412r2 −3056r−1552
(1+r)2 r(2+r)2 (3+r)2 (4+r)2 (5+r)
an
1
1
1
3
7
− 36
97
− 360
517
− 5400
Using the above table, then the solution y1 (x) is
y1 (x) = x a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x2 7x3 97x4 517x5
6
=x 1+x+
−
−
−
+O x
3
36
360
5400
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=
2
r (1 + r)
Therefore
2
2
= lim
r→r2 r (1 + r)
r→0 r (1 + r)
lim
= undefined
Since the limit does not exist then the log term is needed. Therefore the second solution has
the form
!
∞
X
n+r2
y2 (x) = Cy1 (x) ln (x) +
bn x
n=0
Therefore
d
Cy1 (x)
y2 (x) = Cy10 (x) ln (x) +
+
dx
x
Cy1 (x)
= Cy10 (x) ln (x) +
+
x
∞
X
bn xn+r2 (n + r2 )
n=0
∞
X
!
x
!
x−1+n+r2 bn (n + r2 )
n=0
∞
d2
2Cy10 (x) Cy1 (x) X
00
y
(x)
=
Cy
(x)
ln
(x)
+
−
+
2
1
dx2
x
x2
n=0
2Cy10 (x) Cy1 (x)
= Cy100 (x) ln (x) +
−
+
x
x2
∞
X
n=0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!
x−2+n+r2 bn (n + r2 ) (−1 + n + r2 )
!
chapter 2 . book solved problems
138
Substituting these back into the given ode xy 00 + 2y 0 x3 + (x2 − 2) y = 0 gives
Cy1 (x)
2 Cy10 (x) ln (x) +
+
x
+
Cy1 (x) ln (x) +
∞
X
∞
X
bn xn+r2 (n + r2 )
!!
x
n=0
x3
!!
bn x
n+r2
x2
n=0
∞
2Cy10 (x) Cy1 (x) X
+ Cy100 (x) ln (x) +
−
+
x
x2
n=0
!
∞
X
− 2Cy1 (x) ln (x) − 2
bn xn+r2 = 0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!!
x
n=0
Which can be written as
2y10 (x) x3 + y1 (x) x2 + y100 (x) x − 2y1 (x) ln (x) + 2y1 (x) x2
!
!
0
∞
∞
X
X
2y1 (x) y1 (x)
bn xn+r2 (n + r2 )
+x
− 2
C +2
x3 +
bn xn+r2 x2
x
x
x
n=0
n=0
!!
!
∞
∞
2
n+r2
n+r2
X
X
bn x
(n + r2 )
bn x
(n + r2 )
+
−
x−2
bn xn+r2 = 0
2
2
x
x
n=0
n=0
(7)
But since y1 (x) is a solution to the ode, then
y100 (x) x + 2y10 (x) x3 + x2 − 2 y1 (x) = 0
Eq (7) simplifes to
2y10 (x) x3 + y1 (x) x2 + y100 (x) x − 2y1 (x) ln (x) + 2y1 (x) x2
!
!
0
∞
∞
X
X
2y1 (x) y1 (x)
bn xn+r2 (n + r2 )
+x
− 2
C +2
x3 +
bn xn+r2 x2
x
x
x
n=0
n=0
!!
!
∞
∞
2
n+r2
n+r2
X
X
bn x
(n + r2 )
bn x
(n + r2 )
+
−
x−2
bn xn+r2 = 0
2
2
x
x
n=0
n=0
Substituting y1 =
∞
P
(8)
an xn+r1 into the above gives
n=0
∞
P
+
∞
P
x
an (n + r1 ) (−1 + n + r1 ) ln (x) x + 2x(ln (x) x + 1)
x
an (n + r1 ) + ((x3 − 2
n=0
n=0
(9)
x
∞
∞
∞
P −2+n+r2
P −1+n+r2
P
2
4
n+r2
x
bn (n + r2 ) (−1 + n + r2 ) x + 2
x
bn (n + r2 ) x +
bn x
x(x2 − 2)
−2+n+r1
n=0
2
3
n=0
−1+n+r1
n=0
x
=0
Since r1 = 1 and r2 = 0 then the above becomes
∞
P
∞
∞
P n
P
2
3
3
3
x an (n + 1) n ln (x) x + 2x(ln (x) x + 1)
x an (n + 1) + ((x − 2x) ln (x) + 2x − 1)
n=0
n=0
n=0
(10)
x
∞
∞
∞
P −2+n
P n−1
P
2
4
n
x
bn n(n − 1) x + 2
x bn n x +
bn x x(x2 − 2)
n=0
n=0
+ n=0
=0
x
n−1
chapter 2 . book solved problems
139
Which simplifies to
∞
X
!
n
Cn x an ln (x) (n + 1)
+
∞
X
n=0
+
!
2C x
n+3
an ln (x) (n + 1)
n=0
∞
X
!
∞
X
2xn an (n + 1) C
+
∞
X
∞
X
n=0
+
C xn+3 an ln (x)
+
n=0
!
2C xn+3 an
+
n=0
+
!
∞
X
2n x2+n bn
+
n=0
−2C xn+1 an ln (x)
n =0
(−an xn C) +
n =0
∞
X
!
∞
X
∞
X
(2A)
!
n xn−1 bn (n − 1)
n=0
!
x2+n bn
∞
X
+
n=0
(−2bn xn ) = 0
n =0
The next step is to make all powers of x be n − 1 in each summation term. Going over each
summation term above with power of x in it which is not already xn−1 and adjusting the
power and the corresponding index gives
∞
X
n
Cn x an ln (x) (n + 1) =
n =0
∞
X
∞
X
C(n − 1) an−1 n ln (x) xn−1
n=1
2C x
n+3
an ln (x) (n + 1) =
n =0
∞
X
2Can−4 (n − 3) ln (x) xn−1
n=4
∞
X
2xn an (n + 1) C =
n =0
∞
X
2Can−1 n xn−1
n=1
Cx
n+3
an ln (x) =
n =0
∞
X
∞
X
∞
X
Can−4 ln (x) xn−1
n=4
−2C x
n+1
∞
X
an ln (x) =
−2Ca−2+n ln (x) xn−1
n =0
n=2
∞
X
2C xn+3 an =
n =0
∞
X
∞
X
2Can−4 xn−1
n=4
n
(−an x C) =
n =0
∞
X
−Can−1 xn−1
n=1
∞
X
2n x
2+n
bn =
n =0
x2+n bn =
n =0
n =0
2bn−3 (n − 3) xn−1
n=3
∞
X
∞
X
∞
X
∞
X
bn−3 xn−1
n=3
n
(−2bn x ) =
∞
X
−2bn−1 xn−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
chapter 2 . book solved problems
140
are the same and equal to n − 1.
∞
X
!
C(n − 1) an−1 n ln (x) xn−1
+
n=1
+
+
∞
X
!
2Can−4 (n − 3) ln (x) xn−1
n=4
∞
X
!
2Can−1 n xn−1
+
n=1
∞
X
∞
X
!
Can−4 ln (x) xn−1
n=4
−2Ca−2+n ln (x) x
n−1
∞
X
+
n =2
+
+
!
(2B)
2Can−4 xn−1
n=4
∞
X
−Can−1 x
n =1
∞
X
n−1
+
∞
X
!
n xn−1 bn (n − 1)
n=0
!
2bn−3 (n − 3) x
n−1
∞
X
+
n=3
!
bn−3 x
n=3
n−1
+
∞
X
−2bn−1 xn−1 = 0
n =1
For n = 0 in Eq. (2B), we choose arbitray value for b0 as b0 = 1. For n = N , where N = 1
which is the difference between the two roots, we are free to choose b1 = 0. Hence for n = 1,
Eq (2B) gives
C −2=0
Which is solved for C. Solving for C gives
C=2
For n = 2, Eq (2B) gives
−2C(a0 − a1 ) ln (x) + 3Ca1 − 2b1 + 2b2 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
6 + 2b2 = 0
Solving the above for b2 gives
b2 = −3
For n = 3, Eq (2B) gives
−2C(a1 − 3a2 ) ln (x) + 5Ca2 + b0 − 2b2 + 6b3 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
31
+ 6b3 = 0
3
Solving the above for b3 gives
31
b3 = −
18
For n = 4, Eq (2B) gives
2a2
3C a0 −
+ 4a3 ln (x) + (2a0 + 7a3 ) C + 3b1 − 2b3 + 12b4 = 0
3
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
85
+ 12b4 = 0
18
Solving the above for b4 gives
85
b4 = −
216
For n = 5, Eq (2B) gives
2a3
5 a1 −
+ 4a4 C ln (x) + (2a1 + 9a4 ) C + 5b2 − 2b4 + 20b5 = 0
5
chapter 2 . book solved problems
141
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
4067
−
+ 20b5 = 0
270
Solving the above for b5 gives
4067
b5 =
5400
Now that we found all bn and C, we can calculate the second solution from
!
∞
X
n+r2
y2 (x) = Cy1 (x) ln (x) +
bn x
n=0
Using the above value found for C = 2 and all bn , then the second solution becomes
x2 7x3 97x4 517x5
6
y2 (x) = 2 x 1 + x +
−
−
−
+O x
ln (x)
3
36
360
5400
31x3 85x4 4067x5
+ 1 − 3x2 −
−
+
+ O x6
18
216
5400
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x2 7x3 97x4 517x5
6
= c1 x 1 + x +
−
−
−
+O x
3
36
360
5400
x2 7x3 97x4 517x5
31x3
6
+ c2 2 x 1 + x +
−
−
−
+O x
ln (x) + 1 − 3x2 −
3
36
360
5400
18
4
5
85x
4067x
6
−
+
+O x
216
5400
Hence the final solution is
y = yh
x2 7x3 97x4 517x5
6
= c1 x 1 + x +
−
−
−
+O x
3
36
360
5400
x2 7x3 97x4 517x5
31x3 85x4
6
+ c2 2x 1 + x +
−
−
−
+O x
ln (x) + 1 − 3x2 −
−
3
36
360
5400
18
216
4067x5
6
+
+O x
5400
3 Maple. Time used: 0.022 (sec). Leaf size: 58
Order:=6;
ode:=diff(diff(y(x),x),x)*x+2*x^3*diff(y(x),x)+(x^2-2)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2
7 3
97 4
517 5
6
y = c1 x 1 + x + x − x −
x −
x +O x
3
36
360
5400
2 3
7 4
97 5
2
6
+ c2 ln (x) 2x + 2x + x − x −
x +O x
3
18
180
31 3
85 4 4067 5
2
6
+ 1 − 3x − x −
x +
x +O x
18
216
5400
Maple trace
chapter 2 . book solved problems
142
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Kummer
-> hyper3: Equivalence to 1F1 under a power @ Moebius
-> hypergeometric
-> heuristic approach
-> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius
-> Mathieu
-> Equivalence to the rational form of Mathieu ODE under a power @ Moebi\
us
trying a solution in terms of MeijerG functions
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \
@ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x),\
dx)) * 2F1([a1, a2], [b1], f)
trying a symmetry of the form [xi=0, eta=F(x)]
trying differential order: 2; exact nonlinear
trying symmetries linear in x and y(x)
trying to convert to a linear ODE with constant coefficients
trying 2nd order, integrating factor of the form mu(x,y)
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Kummer
-> hyper3: Equivalence to 1F1 under a power @ Moebius
-> hypergeometric
-> heuristic approach
-> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius
-> Mathieu
-> Equivalence to the rational form of Mathieu ODE under a power @ Mo\
ebius
trying 2nd order exact linear
trying symmetries linear in x and y(x)
trying to convert to a linear ODE with constant coefficients
trying to convert to an ODE of Bessel type
trying to convert to an ODE of Bessel type
-> trying reduction of order to Riccati
trying Riccati sub-methods:
-> trying a symmetry pattern of the form [F(x)*G(y), 0]
-> trying a symmetry pattern of the form [0, F(x)*G(y)]
-> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)]
--- Trying Lie symmetry methods, 2nd order ---
chapter 2 . book solved problems
143
-> Computing symmetries using: way = 3
[0, y]
<- successful computation of symmetries.
-> Computing symmetries using: way = 5
3 Mathematica. Time used: 0.013 (sec). Leaf size: 83
ode=x*D[y[x],{x,2}]+2*x^3*D[y[x],x]+(x^2-2)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
1
1
4
3
2
3
2
y(x) → c1
−x − 516x − 1080x − 432x + 216 − x 7x − 12x − 36x − 36 log(x)
216
18
5
4
3
97x
7x
x
+ c2 −
−
+
+ x2 + x
360
36
3
3 Sympy. Time used: 0.270 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(2*x**3*Derivative(y(x), x) + x*Derivative(y(x), (x, 2)) + (x**2 - 2)*y(x
),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
97x4 7x3 x2
y(x) = C1 x −
−
+
+ x + 1 + O x6
360
36
3
chapter 2 . book solved problems
2.2.6
144
Problem 7
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
Internal problem ID [8584]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 7
Date solved : Tuesday, November 11, 2025 at 12:03:50 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 + y(x − 1) = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.24)
(2.25)
(2)
(3)
chapter 2 . book solved problems
145
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = −y(x − 1)
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −(x − 1) y 0 − y
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= −2y 0 + (x − 1)2 y
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= ((x − 1) y 0 + 4y) (x − 1)
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= −y(x − 1)3 + (6x − 6) y 0 + 4y
(7)
chapter 2 . book solved problems
146
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = y(0)
F1 = y 0 (0) − y(0)
F2 = −2y 0 (0) + y(0)
F3 = y 0 (0) − 4y(0)
F4 = 5y(0) − 6y 0 (0)
Substituting all the above in (7) and simplifying gives the solution as
1 2 1 3
1 4
1 5
1 3
1 4
1 5 0
y = 1 + x − x + x − x y(0) + x + x − x +
x y (0) + O x6
2
6
24
30
6
12
120
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
n−2
n
n(n − 1) an x
+
an x (x − 1) = 0
n=2
(1)
n=0
Which simplifies to
∞
X
∞
X
!
n(n − 1) an x
n−2
+
n=2
!
x
1+n
an
+
n=0
∞
X
(−an xn ) = 0
(2)
n =0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (1 + n) xn
n=0
∞
X
x
1+n
an =
n =0
∞
X
an−1 xn
n=1
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
n=0
!
(n + 2) an+2 (1 + n) x
n
+
∞
X
!
an−1 x
n=1
n = 0 gives
2a2 − a0 = 0
n
+
∞
X
n =0
(−an xn ) = 0
(3)
chapter 2 . book solved problems
147
a2 =
a0
2
For 1 ≤ n, the recurrence equation is
(n + 2) an+2 (1 + n) + an−1 − an = 0
(4)
Solving for an+2 , gives
an+2 =
=
−an−1 + an
(n + 2) (1 + n)
an
an−1
−
(n + 2) (1 + n) (n + 2) (1 + n)
For n = 1 the recurrence equation gives
6a3 + a0 − a1 = 0
Which after substituting the earlier terms found becomes
a3 = −
a0 a1
+
6
6
For n = 2 the recurrence equation gives
12a4 + a1 − a2 = 0
Which after substituting the earlier terms found becomes
a4 = −
a1
a0
+
12 24
For n = 3 the recurrence equation gives
20a5 + a2 − a3 = 0
Which after substituting the earlier terms found becomes
a5 = −
a0
a1
+
30 120
For n = 4 the recurrence equation gives
30a6 + a3 − a4 = 0
Which after substituting the earlier terms found becomes
a6 =
a0
a1
−
144 120
For n = 5 the recurrence equation gives
42a7 + a4 − a5 = 0
Which after substituting the earlier terms found becomes
a7 =
11a1
a0
−
5040 560
(5)
chapter 2 . book solved problems
148
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
a0 x2 a0 a1 3 a1
a0 4 a0
a1 5
y = a0 + a1 x +
+ − +
x + − +
x + − +
x + ...
2
6
6
12 24
30 120
Collecting terms, the solution becomes
1 2 1 3
1 4
1 5
1 3
1 4
1 5
y = 1 + x − x + x − x a0 + x + x − x +
x a1 + O x6 (3)
2
6
24
30
6
12
120
At x = 0 the solution above becomes
1 2 1 3
1 4
1 5
1 3
1 4
1 5 0
y = 1 + x − x + x − x y(0) + x + x − x +
x y (0) + O x6
2
6
24
30
6
12
120
3 Maple. Time used: 0.007 (sec). Leaf size: 54
Order:=6;
ode:=diff(diff(y(x),x),x)+(x-1)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2 1 3
1 4
1 5
1 3
1 4
1 5 0
y = 1 + x − x + x − x y(0) + x + x − x +
x y (0) + O x6
2
6
24
30
6
12
120
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
•
Let’s solve
d d
y(x) + (x − 1) y(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Assume series solution for y(x)
∞
P
y(x) =
ak x k
k=0
Rewrite ODE with series expansions
chapter 2 . book solved problems
149
◦ Convert xm · y(x) to series expansion for m = 0..1
∞
P
xm · y(x) =
ak xk+m
k=max(0,−m)
◦ Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk
k=max(0,−m)+m
d d
◦ Convert dx
y(x) to series expansion
dx
∞
P
d d
y(x) =
ak k(k − 1) xk−2
dx dx
k=2
◦ Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=0
Rewrite ODE
∞with series expansions
P
k
2a2 − a0 +
(ak+2 (k + 2) (k + 1) − ak + ak−1 ) x = 0
k=1
•
•
•
•
Each term must be 0
2a2 − a0 = 0
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 − ak + ak−1 = 0
Shift index using k− >k + 1
(k + 1)2 + 3k + 5 ak+3 − ak+1 + ak = 0
Recursion
relation that defines the series solution to the ODE
∞
P
k+1 +ak
y(x) =
ak xk , ak+3 = − −a
, 2a2 − a0 = 0
k2 +5k+6
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 63
ode=D[y[x],{x,2}]+(x-1)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
5
x5
x4 x3
x
x4 x3 x2
−
+
+ x + c1 − +
−
+
+1
120 12
6
30 24
6
2
3 Sympy. Time used: 0.214 (sec). Leaf size: 39
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((x - 1)*y(x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
4
3
x
x3 x2
x
x2
y(x) = C2
−
+
+ 1 + C1 x − +
+ 1 + O x6
24
6
2
12
6
chapter 2 . book solved problems
2.2.7
150
Problem 8
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
Internal problem ID [8585]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 8
Date solved : Tuesday, November 11, 2025 at 12:03:51 PM
CAS classification : [_Lienard]
xy 00 + y 0 + yx = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + y 0 + yx = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
q(x) = 1
p(x) =
Table 2.22: Table p(x), q(x) singularites.
p(x) = x1
singularity
type
x=0
“regular”
q(x) = 1
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + y 0 + yx = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
151
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x +
n=0
!
(n + r) an xn+r−1
∞
X
+
n=0
!
an xn+r
x = 0 (1)
n=0
Which simplifies to
∞
X
!
xn+r−1 an (n + r) (n + r − 1)
+
n=0
∞
X
!
(n + r) an xn+r−1
+
n=0
∞
X
!
x1+n+r an
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
x
1+n+r
an =
n =0
∞
X
an−2 xn+r−1
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
!
(n + r) an xn+r−1 +
n=0
∞
X
!
an−2 xn+r−1
n=2
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + (n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
x−1+r r2 = 0
Since the above is true for all x then the indicial equation becomes
r2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r2 = 0
= 0 (2B)
chapter 2 . book solved problems
152
Solving for r gives the roots of the indicial equation as [0, 0].
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
y1 (x) =
∞
X
an xn+r
(1A)
n=0
Now the second solution y2 is found using
∞
X
y2 (x) = y1 (x) ln (x) +
!
bn xn+r
(1B)
n=1
Then the general solution will be
y = c1 y1 (x) + c2 y2 (x)
In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), a0 is never zero, and is arbitrary
and is typically taken as a0 = 1, and {c1 , c2 } are two arbitray constants of integration which
can be found from initial conditions. We start by finding the first solution y1 (x). Eq (2B)
derived above is now used to find all an coefficients. The case n = 0 is skipped since it
was used to find the roots of the indicial equation. a0 is arbitrary and taken as a0 = 1.
Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + an (n + r) + an−2 = 0
(3)
Solving for an from recursive equation (4) gives
an−2
an = − 2
n + 2nr + r2
Which for the root r = 0 becomes
(4)
an−2
(5)
n2
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
an = −
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
a2 = −
1
(r + 2)2
Which for the root r = 0 becomes
a2 = −
1
4
And the table now becomes
n
a0
a1
a2
an,r
1
0
1
− (r+2)
2
For n = 3, using the above recursive equation gives
a3 = 0
an
1
0
− 14
chapter 2 . book solved problems
153
And the table now becomes
n
a0
a1
a2
an,r
1
0
1
− (r+2)
2
an
1
0
− 14
a3
0
0
For n = 4, using the above recursive equation gives
a4 =
1
(r + 2) (r + 4)2
2
Which for the root r = 0 becomes
a4 =
1
64
And the table now becomes
n
a0
a1
a2
an,r
1
0
1
− (r+2)
2
an
1
0
− 14
a3
a4
0
0
1
(r+2)2 (r+4)2
1
64
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
an,r
1
0
1
− (r+2)
2
an
1
0
− 14
a3
a4
0
0
1
(r+2)2 (r+4)2
1
64
a5
0
0
Using the above table, then the first solution y1 (x) becomes
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
=1−
x2 x4
+
+ O x6
4
64
Now the second solution is found. The second solution is given by
!
∞
X
y2 (x) = y1 (x) ln (x) +
bn xn+r
n=1
Where bn is found using
d
an,r
dr
And the above is then evaluated at r = 0. The above table for an,r is used for this purpose.
Computing the derivatives gives the following table
bn =
chapter 2 . book solved problems
154
n
b0
b1
b2
bn,r
1
0
1
− (r+2)
2
an
1
0
− 14
d
bn,r = dr
an,r
N/A since bn starts from 1
0
bn (r = 0)
N/A
0
2
(r+2)3
1
4
b3
b4
0
0
0
1
(r+2)2 (r+4)2
1
64
−4r−12
(r+2)3 (r+4)3
0
3
− 128
b5
0
0
0
0
The above table gives all values of bn needed. Hence the second solution is
y2 (x) = y1 (x) ln (x) + b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x2 x4
x2 3x4
6
= 1−
+
+O x
ln (x) +
−
+ O x6
4
64
4
128
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x2 x4
x2 x4
x2 3x4
6
6
6
= c1 1 − +
+O x
+ c2
1− +
+O x
ln (x) + −
+O x
4
64
4
64
4
128
Hence the final solution is
y = yh
x2 x4
x2 x4
x2 3x4
6
6
6
= c1 1 −
+
+O x
+ c2
1−
+
+O x
ln (x) +
−
+O x
4
64
4
64
4
128
3 Maple. Time used: 0.029 (sec). Leaf size: 32
Order:=6;
ode:=diff(diff(y(x),x),x)*x+diff(y(x),x)+x*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2
1 4
1 2
3 4
6
6
y = (c2 ln (x) + c1 ) 1 − x + x + O x
+
x −
x +O x
c2
4
64
4
128
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
chapter 2 . book solved problems
155
Let’s solve
d d
d
x dx
y(x) + dx
y(x) + xy(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
d d
dx
y(x)
=
−y(x)
−
dx dx
x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
d d
dx
y(x)
+
+ y(x) = 0
dx dx
x
◦
◦
Check to see if x0 = 0 is a regular singular point
Define functions
P2 (x) = x1 , P3 (x) = 1
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
=1
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x)
+ dx
y(x) + xy(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert x · y(x) to series expansion
∞
P
x · y(x) =
ak xk+r+1
k=0
◦
Shift index using k− >k − 1
∞
P
x · y(x) =
ak−1 xk+r
◦
d
Convert dx
y(x) to series expansion
∞
P
d
y(x) =
ak (k + r) xk+r−1
dx
◦
Shift index using k− >k + 1
∞
P
d
y(x) =
ak+1 (k + r + 1) xk+r
dx
k=−1
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x) =
ak (k + r) (k + r − 1) xk+r−1
dx
k=1
k=0
◦
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x) =
ak+1 (k + r + 1) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
∞
k+r
P
2 r
2
2 −1+r
a0 r x
+ a1 (1 + r) x +
ak+1 (k + r + 1) + ak−1 x
=0
k=1
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term must be 0
a1 (1 + r)2 = 0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1)2 + ak−1 = 0
Shift index using k− >k + 1
chapter 2 . book solved problems
•
156
ak+2 (k + 2)2 + ak = 0
Recursion relation that defines series solution to ODE
ak
ak+2 = − (k+2)
2
•
Recursion relation for r = 0
ak
ak+2 = − (k+2)
2
•
Solution
for r = 0
∞
P
ak
k
y(x) =
ak x , ak+2 = − (k+2)2 , a1 = 0
k=0
3 Mathematica. Time used: 0.002 (sec). Leaf size: 60
ode=x*D[y[x],{x,2}]+D[y[x],x]+x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
4
4
x
x2
3x4 x2
x
x2
y(x) → c1
−
+ 1 + c2 −
+
+
−
+ 1 log(x)
64
4
128
4
64
4
3 Sympy. Time used: 0.207 (sec). Leaf size: 19
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*y(x) + x*Derivative(y(x), (x, 2)) + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
4
x
x2
y(x) = C1
−
+ 1 + O x6
64
4
chapter 2 . book solved problems
2.2.8
157
Problem 9
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
Internal problem ID [8586]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 9
Date solved : Tuesday, November 11, 2025 at 12:03:51 PM
CAS classification : [_Jacobi]
2x(x − 1) y 00 − (1 + x) y 0 + y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
2x2 − 2x y 00 + (−x − 1) y 0 + y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
x+1
2x (x − 1)
1
q(x) =
2x (x − 1)
p(x) = −
Table 2.24: Table p(x), q(x) singularites.
x+1
p(x) = − 2x(x−1)
singularity
type
x=0
“regular”
x=1
“regular”
1
q(x) = 2x(x−1)
singularity
type
x=0
“regular”
x=1
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0, 1, ∞]
Irregular singular points : []
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
2x(x − 1) y 00 + (−x − 1) y 0 + y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
158
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
2x(x − 1)
!
(n + r) (n + r − 1) an xn+r−2
(1)
n=0
∞
X
+ (−x − 1)
!
(n + r) an xn+r−1
+
n=0
∞
X
!
an xn+r
=0
n=0
Which simplifies to
!
∞
X
2xn+r an (n + r) (n + r − 1)
+
n=0
∞
X
−2xn+r−1 an (n + r) (n + r − 1)
n =0
+
∞
X
−xn+r an (n + r) +
n =0
∞
X
−(n + r) an xn+r−1 +
n =0
∞
X
(2A)
!
an xn+r
=0
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
2x
n+r
an (n + r) (n + r − 1) =
n =0
∞
X
2an−1 (n + r − 1) (n + r − 2) xn+r−1
n=1
∞
X
−x
n+r
∞
X
an (n + r) =
−an−1 (n + r − 1) xn+r−1
n =0
n=1
∞
X
an x
n+r
=
n =0
∞
X
an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
2an−1 (n + r − 1) (n + r − 2) xn+r−1
n=1
∞
X
+
−2x
n+r−1
∞
X
an (n + r) (n + r − 1) +
−an−1 (n + r − 1) xn+r−1
n =0
+
∞
X
n =1
−(n + r) an x
n =0
n+r−1
+
∞
X
!
an−1 xn+r−1
=0
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
−2xn+r−1 an (n + r) (n + r − 1) − (n + r) an xn+r−1 = 0
When n = 0 the above becomes
−2x−1+r a0 r(−1 + r) − ra0 x−1+r = 0
(2B)
chapter 2 . book solved problems
159
Or
−2x−1+r r(−1 + r) − r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
−2r2 + r x−1+r = 0
Since the above is true for all x then the indicial equation becomes
−2r2 + r = 0
Solving for r gives the roots of the indicial equation as
1
2
r2 = 0
r1 =
Since a0 6= 0 then the indicial equation becomes
−2r2 + r x−1+r = 0
Solving for r gives the roots of the indicial equation as 12 , 0 .
Since r1 − r2 = 12 is not an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
y2 (x) = xr2
n=0
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
1
an xn+ 2
n=0
y2 (x) =
∞
X
bn x n
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. For 1 ≤ n the recursive equation is
2an−1 (n + r − 1) (n + r − 2) − 2an (n + r) (n + r − 1) − an−1 (n + r − 1) − an (n + r) + an−1 = 0 (3)
Solving for an from recursive equation (4) gives
an =
an−1 (2n2 + 4nr + 2r2 − 7n − 7r + 6)
2n2 + 4nr + 2r2 − n − r
(4)
Which for the root r = 12 becomes
an =
an−1 (2n2 − 5n + 3)
2n2 + n
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 12
and after as more terms are found using the above recursive equation.
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 =
2r2 − 3r + 1
2r2 + 3r + 1
chapter 2 . book solved problems
160
Which for the root r = 12 becomes
a1 = 0
And the table now becomes
n
a0
a1
an,r
1
2r2 −3r+1
2r2 +3r+1
an
1
0
For n = 2, using the above recursive equation gives
a2 =
2r3 − 3r2 + r
2r3 + 9r2 + 13r + 6
Which for the root r = 12 becomes
a2 = 0
And the table now becomes
n
a0
a1
a2
an,r
1
2r2 −3r+1
2r2 +3r+1
2r3 −3r2 +r
2r3 +9r2 +13r+6
an
1
0
0
For n = 3, using the above recursive equation gives
a3 =
2r3 − 3r2 + r
2r3 + 15r2 + 37r + 30
Which for the root r = 12 becomes
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
2r2 −3r+1
2r2 +3r+1
2r3 −3r2 +r
2r3 +9r2 +13r+6
2r3 −3r2 +r
2r3 +15r2 +37r+30
an
1
0
0
0
For n = 4, using the above recursive equation gives
2r3 − 3r2 + r
a4 = 3
2r + 21r2 + 73r + 84
Which for the root r = 12 becomes
a4 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
2r2 −3r+1
2r2 +3r+1
2r3 −3r2 +r
2r3 +9r2 +13r+6
2r3 −3r2 +r
2r3 +15r2 +37r+30
2r3 −3r2 +r
2r3 +21r2 +73r+84
an
1
0
0
0
0
For n = 5, using the above recursive equation gives
a5 =
2r3 − 3r2 + r
2r3 + 27r2 + 121r + 180
chapter 2 . book solved problems
161
Which for the root r = 12 becomes
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
2r2 −3r+1
2r2 +3r+1
2r3 −3r2 +r
2r3 +9r2 +13r+6
2r3 −3r2 +r
2r3 +15r2 +37r+30
2r3 −3r2 +r
2r3 +21r2 +73r+84
2r3 −3r2 +r
2r3 +27r2 +121r+180
an
1
0
0
0
0
0
Using the above table, then the solution y1 (x) is
√
x a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + a6 x 6 . . .
√
= x 1 + O x6
y1 (x) =
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. For 1 ≤ n the recursive equation is
2bn−1 (n + r − 1) (n + r − 2) − 2bn (n + r) (n + r − 1) − bn−1 (n + r − 1) − (n + r) bn + bn−1 = 0 (3)
Solving for bn from recursive equation (4) gives
bn =
bn−1 (2n2 + 4nr + 2r2 − 7n − 7r + 6)
2n2 + 4nr + 2r2 − n − r
(4)
Which for the root r = 0 becomes
bn =
bn−1 (2n2 − 7n + 6)
2n2 − n
(5)
At this point, it is a good idea to keep track of bn in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
b0
bn,r
1
bn
1
For n = 1, using the above recursive equation gives
b1 =
2r2 − 3r + 1
2r2 + 3r + 1
Which for the root r = 0 becomes
b1 = 1
And the table now becomes
n
b0
b1
bn,r
1
2r2 −3r+1
2r2 +3r+1
bn
1
1
For n = 2, using the above recursive equation gives
b2 =
2r3 − 3r2 + r
2r3 + 9r2 + 13r + 6
chapter 2 . book solved problems
162
Which for the root r = 0 becomes
b2 = 0
And the table now becomes
n
b0
b1
b2
bn,r
1
2r2 −3r+1
2r2 +3r+1
2r3 −3r2 +r
2r3 +9r2 +13r+6
bn
1
1
0
For n = 3, using the above recursive equation gives
b3 =
2r3 − 3r2 + r
2r3 + 15r2 + 37r + 30
Which for the root r = 0 becomes
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
2r2 −3r+1
2r2 +3r+1
2r3 −3r2 +r
2r3 +9r2 +13r+6
2r3 −3r2 +r
2r3 +15r2 +37r+30
bn
1
1
0
0
For n = 4, using the above recursive equation gives
b4 =
2r3 − 3r2 + r
2r3 + 21r2 + 73r + 84
Which for the root r = 0 becomes
b4 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
2r2 −3r+1
2r2 +3r+1
2r3 −3r2 +r
2r3 +9r2 +13r+6
2r3 −3r2 +r
2r3 +15r2 +37r+30
2r3 −3r2 +r
2r3 +21r2 +73r+84
bn
1
1
0
0
0
For n = 5, using the above recursive equation gives
b5 =
2r3 − 3r2 + r
2r3 + 27r2 + 121r + 180
Which for the root r = 0 becomes
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
2r2 −3r+1
2r2 +3r+1
2r3 −3r2 +r
2r3 +9r2 +13r+6
2r3 −3r2 +r
2r3 +15r2 +37r+30
2r3 −3r2 +r
2r3 +21r2 +73r+84
2r3 −3r2 +r
2r3 +27r2 +121r+180
bn
1
1
0
0
0
0
chapter 2 . book solved problems
163
Using the above table, then the solution y2 (x) is
y2 (x) = b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
= 1 + x + O x6
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
√
= c1 x 1 + O x 6 + c2 1 + x + O x 6
Hence the final solution is
y = yh
√
= c1 x 1 + O x 6 + c2 1 + x + O x 6
3 Maple. Time used: 0.030 (sec). Leaf size: 26
Order:=6;
ode:=2*x*(x-1)*diff(diff(y(x),x),x)-(x+1)*diff(y(x),x)+y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
√
y = c1 x 1 + O x 6 + c2 1 + x + O x 6
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Reducible group (found another exponential solution)
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
2x(x − 1) dx
y(x) − (x + 1) dx
y(x) + y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
(x+1) dx
y(x)
y(x)
d d
y(x)
=
−
+
dx dx
2x(x−1)
2x(x−1)
•
Group terms with
y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
◦
(x+1)
d
y(x)
dx
y(x)
+ 2x(x−1)
=0
Check to see if x0 is a regular singular point
Define
functions
h
i
d d
y(x) −
dx dx
2x(x−1)
x+1
1
P2 (x) = − 2x(x−1)
, P3 (x) = 2x(x−1)
◦
x · P2 (x) is analytic at x = 0
chapter 2 . book solved problems
(x · P2 (x))
◦
•
•
= 12
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
x=0
164
=0
x=0
x = 0is a regular singular point
Check to see if x0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
2x(x − 1) dx
y(x) + (−x − 1) dx
y(x) + y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d
Convert xm · dx
y(x) to series expansion for m = 0..1
∞
P
d
xm · dx
y(x) =
ak (k + r) xk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
xm · dx
y(x) =
ak+1−m (k + 1 − m + r) xk+r
k=−1+m
d d
Convert xm · dx
y(x) to series expansion for m = 1..2
dx
∞
P
d d
xm · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−2+m
dx
k=0
◦
Shift index using k− >k + 2 − m
∞
P
d d
xm · dx
y(x)
=
ak+2−m (k + 2 − m + r) (k + 1 − m + r) xk+r
dx
k=−2+m
Rewrite ODE with series
expansions
∞
P
−1+r
k+r
−a0 r(−1 + 2r) x
+
(−ak+1 (k + 1 + r) (2k + 1 + 2r) + ak (2k + 2r − 1) (k + r − 1)) x
k=0
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
−r(−1 + 2r) = 0
Values of r that satisfy the indicial equation
r ∈ 0, 12
Each term in the series must be 0, giving the recursion relation
−2(k + 1 + r) k + 12 + r ak+1 + 2ak k + r − 12 (k + r − 1) = 0
Recursion relation that defines series solution to ODE
(2k+2r−1)(k+r−1)
ak+1 = ak(k+1+r)(2k+1+2r)
Recursion relation for r = 0 ; series terminates at k = 1
(2k−1)(k−1)
ak+1 = ak(k+1)(2k+1)
Apply recursion relation for k = 0
a1 = a0
Terminating series solution of the ODE for r = 0 . Use reduction of order to find the second linea
y(x) = a0 · (x + 1)
Recursion relation
for r = 12
2ak k k− 12
k+ 32 (2k+2)
Solution
for r = 12
∞
P
1
y(x) =
ak xk+ 2 , ak+1 =
k=0
ak+1 =
•
•
2ak k k− 12
k+ 32 (2k+2)
Combine
solutions andrename parameters
∞
P
k+ 12
y(x) = a0 · (x + 1) +
bk x
, bk+1 =
k=0
2bk k k− 12
k+ 32 (2k+2)
chapter 2 . book solved problems
165
3 Mathematica. Time used: 0.004 (sec). Leaf size: 18
ode=2*x*(x-1)*D[y[x],{x,2}]-(x+1)*D[y[x],x]+y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
√
y(x) → c1 x + c2 (x + 1)
3 Sympy. Time used: 0.348 (sec). Leaf size: 14
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(2*x*(x - 1)*Derivative(y(x), (x, 2)) - (x + 1)*Derivative(y(x), x) + y(x
),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
√
y(x) = C2 x + C1 + O x6
chapter 2 . book solved problems
2.2.9
166
Problem 10
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
Internal problem ID [8587]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 10
Date solved : Tuesday, November 11, 2025 at 12:03:52 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
xy 00 + 2y 0 + 4yx = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + 2y 0 + 4yx = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
2
x
q(x) = 4
p(x) =
Table 2.26: Table p(x), q(x) singularites.
p(x) = x2
singularity
type
x=0
“regular”
q(x) = 4
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + 2y 0 + 4yx = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
167
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x + 2
n=0
!
∞
X
(n + r) an xn+r−1 + 4
n=0
!
an xn+r
x = 0 (1)
n=0
Which simplifies to
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
!
2(n + r) an xn+r−1 +
n=0
∞
X
!
4x1+n+r an
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
4x
1+n+r
n =0
an =
∞
X
4an−2 xn+r−1
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n+r) (n+r −1) +
n=0
∞
X
!
2(n+r) an xn+r−1 +
n=0
∞
X
!
4an−2 xn+r−1
n=2
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + 2(n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + 2ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + 2r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
r x−1+r (1 + r) = 0
Since the above is true for all x then the indicial equation becomes
r(1 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = −1
Since a0 6= 0 then the indicial equation becomes
r x−1+r (1 + r) = 0
= 0 (2B)
chapter 2 . book solved problems
168
Solving for r gives the roots of the indicial equation as [0, −1].
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) + xr2
!
bn x n
n=0
Or
y1 (x) =
∞
X
an x n
n=0
∞
P
y2 (x) = Cy1 (x) ln (x) +
bn x n
n=0
x
Or
y1 (x) =
∞
X
an xn
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn xn−1
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + 2an (n + r) + 4an−2 = 0
(3)
Solving for an from recursive equation (4) gives
4an−2
an = − 2
n + 2nr + r2 + n + r
(4)
Which for the root r = 0 becomes
an = −
4an−2
n (1 + n)
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
4
a2 = − 2
r + 5r + 6
Which for the root r = 0 becomes
a2 = −
And the table now becomes
2
3
chapter 2 . book solved problems
n
a0
a1
a2
169
an,r
1
0
4
− r2 +5r+6
an
1
0
− 23
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
4
− r2 +5r+6
0
an
1
0
− 23
0
For n = 4, using the above recursive equation gives
a4 =
16
(r2 + 5r + 6) (r2 + 9r + 20)
Which for the root r = 0 becomes
a4 =
2
15
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
4
− r2 +5r+6
0
an
1
0
− 23
0
16
(r2 +5r+6)(r2 +9r+20)
2
15
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
4
− r2 +5r+6
0
an
1
0
− 23
0
16
(r2 +5r+6)(r2 +9r+20)
2
15
0
0
Using the above table, then the solution y1 (x) is
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
2x2 2x4
=1−
+
+ O x6
3
15
Now the second solution y2 (x) is found. Let
r1 − r2 = N
chapter 2 . book solved problems
170
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=0
Therefore
lim 0 = lim 0
r→r2
r→−1
=0
The limit is 0. Since the limit exists then the log term is not needed and we can set C = 0.
Therefore the second solution has the form
y2 (x) =
∞
X
bn xn+r
n=0
=
∞
X
bn xn−1
n=0
Eq (3) derived above is used to find all bn coefficients. The case n = 0 is skipped since
it was used to find the roots of the indicial equation. b0 is arbitrary and taken as b0 = 1.
Substituting n = 1 in Eq(3) gives
b1 = 0
For 2 ≤ n the recursive equation is
bn (n + r) (n + r − 1) + 2(n + r) bn + 4bn−2 = 0
(4)
Which for for the root r = −1 becomes
bn (n − 1) (n − 2) + 2(n − 1) bn + 4bn−2 = 0
(4A)
Solving for bn from the recursive equation (4) gives
4bn−2
bn = − 2
n + 2nr + r2 + n + r
(5)
Which for the root r = −1 becomes
4bn−2
bn = − 2
n −n
(6)
At this point, it is a good idea to keep track of bn in a table both before substituting r = −1
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
0
bn
1
0
For n = 2, using the above recursive equation gives
4
b2 = − 2
r + 5r + 6
Which for the root r = −1 becomes
b2 = −2
And the table now becomes
chapter 2 . book solved problems
n
b0
b1
b2
171
bn,r
1
0
4
− r2 +5r+6
bn
1
0
−2
For n = 3, using the above recursive equation gives
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
0
4
− r2 +5r+6
0
bn
1
0
−2
0
For n = 4, using the above recursive equation gives
b4 =
16
(r2 + 5r + 6) (r2 + 9r + 20)
Which for the root r = −1 becomes
b4 =
2
3
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
4
− r2 +5r+6
0
bn
1
0
−2
0
16
(r2 +5r+6)(r2 +9r+20)
2
3
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
0
4
− r2 +5r+6
0
bn
1
0
−2
0
16
(r2 +5r+6)(r2 +9r+20)
2
3
0
0
Using the above table, then the solution y2 (x) is
y2 (x) = 1 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
4
1 − 2x2 + 2x3 + O(x6 )
=
x
chapter 2 . book solved problems
172
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
c 1 − 2x2 + 2x4 + O(x6 )
2
4
2
3
2x
2x
= c1 1 −
+
+ O x6 +
3
15
x
Hence the final solution is
y = yh
2
= c1 1 −
4
2x
2x
+
+ O x6 +
3
15
4
c2 1 − 2x2 + 2x3 + O(x6 )
x
3 Maple. Time used: 0.028 (sec). Leaf size: 32
Order:=6;
ode:=diff(diff(y(x),x),x)*x+2*diff(y(x),x)+4*x*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
2 4
2
6
c
1
−
2x
+
x
+
O
(x
)
2 2
2 4
2
3
y = c1 1 − x + x + O x 6 +
3
15
x
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Group is reducible or imprimitive
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x dx
y(x) + 2 dx
y(x) + 4xy(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative •
•
d d
y(x) = −4y(x) −
dx dx
•
2
d
y(x)
dx
x
Group termswith y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(x) +
dx dx
◦
◦
d
y(x)
dx
+ 4y(x) = 0
Check to see if x0 = 0 is a regular singular point
Define functions
P2 (x) = x2 , P3 (x) = 4
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
2
x
=2
x=0
x2 · P3 (x) is analytic at x = 0
chapter 2 . book solved problems
(x2 · P3 (x))
◦
•
•
173
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x) + 2 dx
y(x) + 4xy(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert x · y(x) to series expansion
∞
P
x · y(x) =
ak xk+r+1
k=0
◦
Shift index using k− >k − 1
∞
P
x · y(x) =
ak−1 xk+r
◦
d
Convert dx
y(x) to series expansion
∞
P
d
y(x) =
ak (k + r) xk+r−1
dx
k=1
k=0
◦
◦
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + r + 1) xk+r
dx
k=−1
d d
Convert x · dx
y(x)
to series expansion
dx
∞
P
d d
x · dx
y(x) =
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + r + 1) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
a0 r(1 + r) x
−1+r
r
+ a1 (1 + r) (2 + r) x +
∞
P
(ak+1 (k + r + 1) (k + 2 + r) + 4ak−1 ) x
k+r
=0
k=1
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r(1 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {−1, 0}
Each term must be 0
a1 (1 + r) (2 + r) = 0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + r + 1) (k + 2 + r) + 4ak−1 = 0
Shift index using k− >k + 1
ak+2 (k + 2 + r) (k + 3 + r) + 4ak = 0
Recursion relation that defines series solution to ODE
4ak
ak+2 = − (k+2+r)(k+3+r)
Recursion relation for r = −1
4ak
ak+2 = − (k+1)(k+2)
Solution
for r = −1
∞
P
4ak
k−1
y(x) =
ak x , ak+2 = − (k+1)(k+2) , 0 = 0
k=0
•
•
Recursion relation for r = 0
4ak
ak+2 = − (k+2)(k+3)
Solution
for r = 0
∞
P
4ak
k
y(x) =
ak x , ak+2 = − (k+2)(k+3) , 2a1 = 0
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
4ak
4bk
k−1
k
y(x) =
ak x
+
bk x , ak+2 = − (k+1)(k+2) , 0 = 0, bk+2 = − (k+2)(k+3) , 2b1 = 0
k=0
k=0
chapter 2 . book solved problems
174
3 Mathematica. Time used: 0.006 (sec). Leaf size: 40
ode=x*D[y[x],{x,2}]+2*D[y[x],x]+4*x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
2x3
1
− 2x +
3
x
+ c2
2x4 2x2
−
+1
15
3
3 Sympy. Time used: 0.261 (sec). Leaf size: 46
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(4*x*y(x) + x*Derivative(y(x), (x, 2)) + 2*Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C2
C − 4x6 + 2x4 − 2x2 + 1
1
45
3
2x
2x
−
+1 +
+ O x6
15
3
x
4
2
chapter 2 . book solved problems
2.2.10
175
Problem 11
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
Internal problem ID [8588]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 11
Date solved : Tuesday, November 11, 2025 at 12:03:53 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
xy 00 + (2 − 2x) y 0 + (x − 2) y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + (2 − 2x) y 0 + (x − 2) y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
2(x − 1)
x
x−2
q(x) =
x
p(x) = −
Table 2.28: Table p(x), q(x) singularites.
p(x) = − 2(x−1)
x
singularity
type
x=0
“regular”
q(x) = x−2
x
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + (2 − 2x) y 0 + (x − 2) y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
176
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
(n + r) (n + r − 1) an xn+r−2 x
(1)
n=0
+ (2 − 2x)
∞
X
!
(n + r) an xn+r−1
∞
X
+ (x − 2)
n=0
!
an xn+r
=0
n=0
Which simplifies to
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
+
n=0
+
∞
X
2(n + r) an x
−2xn+r an (n + r)
n =0
∞
X
1+n+r
!
n+r−1
∞
X
+
x
n=0
!
an
+
n=0
∞
X
(2A)
−2an xn+r = 0
n =0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
−2x
n+r
∞
X
an (n + r) =
−2an−1 (n + r − 1) xn+r−1
n =0
n=1
∞
X
x
1+n+r
an =
n =0
∞
X
∞
X
an−2 xn+r−1
n=2
−2an x
n+r
=
n =0
∞
X
−2an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
n=0
+
∞
X
!
2(n + r) an xn+r−1
n=0
+
+
∞
X
n =1
∞
X
−2an−1 (n + r − 1) xn+r−1
an−2 xn+r−1
n=2
!
+
∞
X
−2an−1 xn+r−1 = 0
n =1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + 2(n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + 2ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + 2r x−1+r a0 = 0
(2B)
chapter 2 . book solved problems
177
Since a0 6= 0 then the above simplifies to
r x−1+r (1 + r) = 0
Since the above is true for all x then the indicial equation becomes
r(1 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = −1
Since a0 6= 0 then the indicial equation becomes
r x−1+r (1 + r) = 0
Solving for r gives the roots of the indicial equation as [0, −1].
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = Cy1 (x) ln (x) + xr2
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
an x n
n=0
∞
P
y2 (x) = Cy1 (x) ln (x) + n=0
bn x n
x
Or
y1 (x) =
∞
X
an xn
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x
n−1
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 =
2
r+2
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) − 2an−1 (n + r − 1) + 2an (n + r) + an−2 − 2an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an =
2nan−1 + 2ran−1 − an−2
n2 + 2nr + r2 + n + r
(4)
2nan−1 − an−2
n (1 + n)
(5)
Which for the root r = 0 becomes
an =
chapter 2 . book solved problems
178
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
an
1
1
2
r+2
For n = 2, using the above recursive equation gives
3
r2 + 5r + 6
a2 =
Which for the root r = 0 becomes
a2 =
1
2
And the table now becomes
n
a0
a1
a2
an,r
1
2
r+2
3
r2 +5r+6
an
1
1
1
2
For n = 3, using the above recursive equation gives
a3 =
4
(r + 4) (r + 3) (r + 2)
Which for the root r = 0 becomes
a3 =
1
6
And the table now becomes
n
a0
a1
a2
a3
an,r
1
2
r+2
3
r2 +5r+6
4
(r+4)(r+3)(r+2)
an
1
1
1
2
1
6
For n = 4, using the above recursive equation gives
a4 =
5
(r2 + 9r + 20) (r2 + 5r + 6)
Which for the root r = 0 becomes
a4 =
1
24
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
2
r+2
3
r2 +5r+6
4
(r+4)(r+3)(r+2)
5
(r2 +9r+20)(r2 +5r+6)
an
1
1
1
2
1
6
1
24
For n = 5, using the above recursive equation gives
a5 =
6
(r + 6) (r + 5) (r + 4) (r + 3) (r + 2)
chapter 2 . book solved problems
179
Which for the root r = 0 becomes
a5 =
1
120
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
2
r+2
3
r2 +5r+6
4
(r+4)(r+3)(r+2)
5
(r2 +9r+20)(r2 +5r+6)
6
(r+6)(r+5)(r+4)(r+3)(r+2)
an
1
1
1
2
1
6
1
24
1
120
Using the above table, then the solution y1 (x) is
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x2 x 3 x4
x5
=1+x+
+
+
+
+ O x6
2
6
24 120
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=
2
r+2
Therefore
2
2
= lim
r→r2 r + 2
r→−1 r + 2
=2
lim
The limit is 2. Since the limit exists then the log term is not needed and we can set C = 0.
Therefore the second solution has the form
y2 (x) =
∞
X
bn xn+r
n=0
=
∞
X
bn xn−1
n=0
Eq (3) derived above is used to find all bn coefficients. The case n = 0 is skipped since
it was used to find the roots of the indicial equation. b0 is arbitrary and taken as b0 = 1.
Substituting n = 1 in Eq(3) gives
2
b1 =
r+2
For 2 ≤ n the recursive equation is
bn (n + r) (n + r − 1) − 2bn−1 (n + r − 1) + 2(n + r) bn + bn−2 − 2bn−1 = 0
(4)
Which for for the root r = −1 becomes
bn (n − 1) (n − 2) − 2bn−1 (n − 2) + 2(n − 1) bn + bn−2 − 2bn−1 = 0
(4A)
chapter 2 . book solved problems
180
Solving for bn from the recursive equation (4) gives
bn =
2nbn−1 + 2rbn−1 − bn−2
n2 + 2nr + r2 + n + r
(5)
2nbn−1 − bn−2 − 2bn−1
n2 − n
(6)
Which for the root r = −1 becomes
bn =
At this point, it is a good idea to keep track of bn in a table both before substituting r = −1
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
bn
1
2
2
r+2
For n = 2, using the above recursive equation gives
3
b2 =
r2 + 5r + 6
Which for the root r = −1 becomes
b2 =
3
2
And the table now becomes
n
b0
b1
b2
bn,r
1
2
r+2
3
r2 +5r+6
bn
1
2
3
2
For n = 3, using the above recursive equation gives
b3 =
4
(r2 + 7r + 12) (r + 2)
Which for the root r = −1 becomes
b3 =
2
3
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
2
r+2
3
r2 +5r+6
4
(r+4)(r+3)(r+2)
bn
1
2
3
2
2
3
For n = 4, using the above recursive equation gives
b4 =
5
(r2 + 9r + 20) (r2 + 5r + 6)
Which for the root r = −1 becomes
b4 =
5
24
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
2
r+2
3
r2 +5r+6
4
(r+4)(r+3)(r+2)
5
(r2 +9r+20)(r2 +5r+6)
bn
1
2
3
2
2
3
5
24
chapter 2 . book solved problems
181
For n = 5, using the above recursive equation gives
b5 =
6
(r2 + 7r + 12) (r + 2) (r2 + 11r + 30)
Which for the root r = −1 becomes
b5 =
1
20
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
2
r+2
3
r2 +5r+6
4
(r+4)(r+3)(r+2)
5
(r2 +9r+20)(r2 +5r+6)
6
(r+6)(r+5)(r+4)(r+3)(r+2)
bn
1
2
3
2
2
3
5
24
1
20
Using the above table, then the solution y2 (x) is
y2 (x) = 1 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
2
3
4
5
1 + 2x + 3x2 + 2x3 + 5x
+ x20 + O(x6 )
24
=
x
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
c 1 + 2x + 3x2 + 2x3 + 5x4 + x5 + O(x6 )
2
3
4
5
2
2
3
24
20
x
x
x
x
= c1 1+x+ + + +
+O x6 +
2
6 24 120
x
Hence the final solution is
y = yh
c 1 + 2x + 3x2 + 2x3 + 5x4 + x5 + O(x6 )
2
3
4
5
2
2
3
24
20
x
x
x
x
= c1 1 + x +
+
+
+
+ O x6 +
2
6
24 120
x
3 Maple. Time used: 0.034 (sec). Leaf size: 44
Order:=6;
ode:=diff(diff(y(x),x),x)*x+(-2*x+2)*diff(y(x),x)+(x-2)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2 1 3
1 4
1 5
6
y = c1 1 + x + x + x + x +
x +O x
2
6
24
120
5 4
1 5
c2 1 + 2x + 32 x2 + 23 x3 + 24
x + 20
x + O (x6 )
+
x
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
chapter 2 . book solved problems
182
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x dx
y(x) + (2 − 2x) dx
y(x) + (x − 2) y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d d
y(x) = − (x−2)y(x)
+
dx dx
x
2(−1+x)
d
y(x)
dx
x
•
Group terms with y(x) on
the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
+ (x−2)y(x)
=0
x
Check to see if x0 = 0 is a regular singular point
Define
functions
h
i
2(−1+x)
d d
y(x) −
dx dx
◦
d
y(x)
dx
x
P2 (x) = − 2(−1+x)
, P3 (x) = x−2
x
x
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
=2
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x) + (2 − 2x) dx
y(x) + (x − 2) y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..1
∞
P
xm · y(x) =
ak xk+r+m
k=0
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert xm · dx
y(x) to series expansion for m = 0..1
∞
P
d
xm · dx
y(x) =
ak (k + r) xk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
xm · dx
y(x) =
ak+1−m (k + 1 − m + r) xk+r
k=−1+m
d d
Convert x · dx
y(x)
to series expansion
dx
∞
P
d d
x · dx
y(x) =
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
a0 r(1 + r) x
−1+r
r
+ (a1 (1 + r) (2 + r) − 2a0 (1 + r)) x +
∞
P
k=1
•
a0 cannot be 0 by assumption, giving the indicial equation
(ak+1 (k + 1 + r) (k + 2 + r) − 2ak (k +
chapter 2 . book solved problems
•
•
•
•
•
•
•
183
r(1 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {−1, 0}
Each term must be 0
a1 (1 + r) (2 + r) − 2a0 (1 + r) = 0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1 + r) (k + 2 + r) − 2ak k − 2ak r − 2ak + ak−1 = 0
Shift index using k− >k + 1
ak+2 (k + 2 + r) (k + 3 + r) − 2ak+1 (k + 1) − 2rak+1 − 2ak+1 + ak = 0
Recursion relation that defines series solution to ODE
+2rak+1 −ak +4ak+1
ak+2 = 2kak+1
(k+2+r)(k+3+r)
Recursion relation for r = −1
−ak +2ak+1
ak+2 = 2kak+1
(k+1)(k+2)
Solution
for r = −1
∞
P
2kak+1 −ak +2ak+1
k−1
y(x) =
ak x , ak+2 =
,0 = 0
(k+1)(k+2)
k=0
•
•
Recursion relation for r = 0
−ak +4ak+1
ak+2 = 2kak+1
(k+2)(k+3)
Solution
for r = 0
∞
P
2kak+1 −ak +4ak+1
k
y(x) =
ak x , ak+2 =
, 2a1 − 2a0 = 0
(k+2)(k+3)
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
−ak +2ak+1
−bk +4bk+1
k−1
k
y(x) =
ak x
+
bk x , ak+2 = 2kak+1
, 0 = 0, bk+2 = 2kbk+1
, 2b1 − 2b
(k+1)(k+2)
(k+2)(k+3)
k=0
k=0
3 Mathematica. Time used: 0.016 (sec). Leaf size: 58
ode=x*D[y[x],{x,2}]+(2-2*x)*D[y[x],x]+(x-2)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
3 Sympy. Time used: 0.312 (sec). Leaf size: 61
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*Derivative(y(x), (x, 2)) + (2 - 2*x)*Derivative(y(x), x) + (x - 2)*y(x
),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C2
C − x6 − x5 − x4 − x3 − x2 + 1
1
144
30
8
3
2
x
x
x
x
+
+
+
+x+1 +
+ O x6
120 24
6
2
x
5
4
4
5x3 2x2 3x 1
x
x3 x2
+
+
+ + 2 + c2
+
+
+x+1
24
3
2
x
24
6
2
3
2
chapter 2 . book solved problems
2.2.11
184
Problem 12
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192
Internal problem ID [8589]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 12
Date solved : Tuesday, November 11, 2025 at 12:03:53 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
x2 y 00 + 6y 0 x + 4x2 + 6 y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
x2 y 00 + 6y 0 x + 4x2 + 6 y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
6
x
4x2 + 6
q(x) =
x2
p(x) =
Table 2.30: Table p(x), q(x) singularites.
2
p(x) = x6
singularity
type
x=0
“regular”
q(x) = 4xx2+6
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
x2 y 00 + 6y 0 x + 4x2 + 6 y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
185
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
x2
∞
X
!
(n + r) (n + r − 1) an xn+r−2
n=0
∞
X
+6
!
(n + r) an x
n+r−1
(1)
∞
X
x + 4x + 6
an xn+r
!
2
n=0
=0
n=0
Which simplifies to
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
4xn+r+2 an
+
n=0
∞
X
∞
X
!
6xn+r an (n + r)
n=0
!
6an xn+r
(2A)
=0
n=0
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives
∞
X
4xn+r+2 an =
n =0
∞
X
4an−2 xn+r
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r.
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
4an−2 x
n+r
+
n=2
∞
X
∞
X
!
6xn+r an (n + r)
n=0
!
6an xn+r
=0
n=0
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r an (n + r) (n + r − 1) + 6xn+r an (n + r) + 6an xn+r = 0
When n = 0 the above becomes
xr a0 r(−1 + r) + 6xr a0 r + 6a0 xr = 0
Or
(xr r(−1 + r) + 6xr r + 6xr ) a0 = 0
Since a0 6= 0 then the above simplifies to
r2 + 5r + 6 xr = 0
Since the above is true for all x then the indicial equation becomes
r2 + 5r + 6 = 0
(2B)
chapter 2 . book solved problems
186
Solving for r gives the roots of the indicial equation as
r1 = −2
r2 = −3
Since a0 6= 0 then the indicial equation becomes
r2 + 5r + 6 xr = 0
Solving for r gives the roots of the indicial equation as [−2, −3].
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) + xr2
!
bn x n
n=0
Or
∞
P
y1 (x) =
an x n
n=0
x2
∞
P
y2 (x) = Cy1 (x) ln (x) +
bn x n
n=0
x3
Or
y1 (x) =
∞
X
an xn−2
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn xn−3
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + 6an (n + r) + 4an−2 + 6an = 0
(3)
Solving for an from recursive equation (4) gives
4an−2
an = − 2
n + 2nr + r2 + 5n + 5r + 6
(4)
Which for the root r = −2 becomes
an = −
4an−2
n (n + 1)
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = −2
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
chapter 2 . book solved problems
187
For n = 2, using the above recursive equation gives
4
a2 = − 2
r + 9r + 20
Which for the root r = −2 becomes
a2 = −
2
3
And the table now becomes
n
a0
a1
a2
an,r
1
0
4
− r2 +9r+20
an
1
0
− 23
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
4
− r2 +9r+20
0
an
1
0
− 23
0
For n = 4, using the above recursive equation gives
a4 =
16
(r + 5) (r + 4) (r + 7) (r + 6)
Which for the root r = −2 becomes
a4 =
2
15
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
4
− r2 +9r+20
0
an
1
0
− 23
0
16
(r+5)(r+4)(r+7)(r+6)
2
15
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
4
− r2 +9r+20
0
an
1
0
− 23
0
16
(r+5)(r+4)(r+7)(r+6)
2
15
0
0
chapter 2 . book solved problems
188
Using the above table, then the solution y1 (x) is
1
2
3
4
5
6
a
+
a
x
+
a
x
+
a
x
+
a
x
+
a
x
+
a
x
.
.
.
0
1
2
3
4
5
6
x2
2
4
1 − 2x3 + 2x
+ O(x6 )
15
=
x2
y1 (x) =
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=0
Therefore
lim 0 = lim 0
r→r2
r→−3
=0
The limit is 0. Since the limit exists then the log term is not needed and we can set C = 0.
Therefore the second solution has the form
y2 (x) =
∞
X
bn xn+r
n=0
=
∞
X
bn xn−3
n=0
Eq (3) derived above is used to find all bn coefficients. The case n = 0 is skipped since
it was used to find the roots of the indicial equation. b0 is arbitrary and taken as b0 = 1.
Substituting n = 1 in Eq(3) gives
b1 = 0
For 2 ≤ n the recursive equation is
bn (n + r) (n + r − 1) + 6bn (n + r) + 4bn−2 + 6bn = 0
(4)
Which for for the root r = −3 becomes
bn (n − 3) (n − 4) + 6bn (n − 3) + 4bn−2 + 6bn = 0
(4A)
Solving for bn from the recursive equation (4) gives
4bn−2
bn = − 2
n + 2nr + r2 + 5n + 5r + 6
(5)
Which for the root r = −3 becomes
4bn−2
bn = − 2
n −n
(6)
At this point, it is a good idea to keep track of bn in a table both before substituting r = −3
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
0
bn
1
0
chapter 2 . book solved problems
189
For n = 2, using the above recursive equation gives
4
b2 = − 2
r + 9r + 20
Which for the root r = −3 becomes
b2 = −2
And the table now becomes
n
b0
b1
b2
bn,r
1
0
4
− r2 +9r+20
bn
1
0
−2
For n = 3, using the above recursive equation gives
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
0
4
− r2 +9r+20
0
bn
1
0
−2
0
For n = 4, using the above recursive equation gives
b4 =
16
(r2 + 9r + 20) (r2 + 13r + 42)
Which for the root r = −3 becomes
b4 =
2
3
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
4
− r2 +9r+20
0
bn
1
0
−2
0
16
(r+5)(r+4)(r+7)(r+6)
2
3
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
0
4
− r2 +9r+20
0
bn
1
0
−2
0
16
(r+5)(r+4)(r+7)(r+6)
2
3
0
0
chapter 2 . book solved problems
190
Using the above table, then the solution y2 (x) is
1
b0 + b1 x + b2 x 2 + b3 x 3 + b4 x 4 + b5 x 5 + b 6 x 6 . . .
2
x
4
1 − 2x2 + 2x3 + O(x6 )
=
x3
y2 (x) =
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
2
4
2x4
6
2
6
c1 1 − 2x3 + 2x
+
O(x
)
c
1
−
2x
+
+
O(x
)
2
15
3
=
+
x2
x3
Hence the final solution is
y = yh
2
4
2x4
6
2
6
c1 1 − 2x3 + 2x
+
O(x
)
c
1
−
2x
+
+
O(x
)
2
15
3
=
+
2
3
x
x
3 Maple. Time used: 0.021 (sec). Leaf size: 34
Order:=6;
ode:=x^2*diff(diff(y(x),x),x)+6*diff(y(x),x)*x+(4*x^2+6)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
2 4
c1 1 − 23 x2 + 15
x + O (x6 ) x + c2 1 − 2x2 + 23 x4 + O (x6 )
y=
x3
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Group is reducible or imprimitive
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x2 dx
y(x) + 6x dx
y(x) + (4x2 + 6) y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
6
2 2x2 +3 y(x)
d d
y(x)
=
−
−
2
dx dx
x
•
d
y(x)
dx
x
Group termswith y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(x) +
dx dx
◦
6
d
y(x)
dx
2 2x2 +3 y(x)
+
=0
x
x2
Check to see if x0 = 0 is a regular singular point
Define functions
chapter 2 . book solved problems
h
◦
i
2 2x2 +3
6
P2 (x) = x , P3 (x) =
x2
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
=6
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
191
=6
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x2 dx
y(x)
+
6x
y(x)
+ (4x2 + 6) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..2
∞
P
xm · y(x) =
ak xk+r+m
k=0
◦
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert x · dx
y(x) to series expansion
∞
P
d
x · dx
y(x) =
ak (k + r) xk+r
k=0
d d
Convert x2 · dx
y(x) to series expansion
dx
∞
P
d d
x2 · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r
dx
k=0
Rewrite ODE with series expansions
r
a0 (3 + r) (2 + r) x + a1 (4 + r) (3 + r) x
1+r
+
∞
P
(ak (k + r + 3) (k + r + 2) + 4ak−2 ) x
k=2
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
(3 + r) (2 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {−3, −2}
Each term must be 0
a1 (4 + r) (3 + r) = 0
Solve for the dependent coefficient(s)
a1 = 0
Each term in the series must be 0, giving the recursion relation
ak (k + r + 3) (k + r + 2) + 4ak−2 = 0
Shift index using k− >k + 2
ak+2 (k + 5 + r) (k + 4 + r) + 4ak = 0
Recursion relation that defines series solution to ODE
4ak
ak+2 = − (k+5+r)(k+4+r)
Recursion relation for r = −3
4ak
ak+2 = − (k+2)(k+1)
Solution
for r = −3
∞
P
4ak
k−3
y(x) =
ak x , ak+2 = − (k+2)(k+1) , a1 = 0
k=0
•
•
Recursion relation for r = −2
4ak
ak+2 = − (k+3)(k+2)
Solution
for r = −2
∞
P
4ak
k−2
y(x) =
ak x , ak+2 = − (k+3)(k+2) , a1 = 0
•
Combine solutions and rename parameters
k=0
k+r
=0
chapter 2 . book solved problems
y(x) =
∞
P
ak x
k−3
+
∞
P
k=0
bk x
192
k−2
k=0
4ak
4bk
, ak+2 = − (k+2)(k+1)
, a1 = 0, bk+2 = − (k+3)(k+2)
, b1 = 0
3 Mathematica. Time used: 0.007 (sec). Leaf size: 38
ode=x^2*D[y[x],{x,2}]+6*x*D[y[x],x]+(4*x^2+6)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
1
2x 2
+
−
3
x
3
x
+ c2
2x2
1
2
+ 2−
15
x
3
3 Sympy. Time used: 0.321 (sec). Leaf size: 36
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x**2*Derivative(y(x), (x, 2)) + 6*x*Derivative(y(x), x) + (4*x**2 + 6)*y
(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
C1
y(x) =
6
4
2x8
− 4x
+ 2x3 − 2x2 + 1
315
45
x3
+ O(1)
chapter 2 . book solved problems
2.2.12
193
Problem 13
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
Internal problem ID [8590]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 13
Date solved : Tuesday, November 11, 2025 at 12:03:54 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
xy 00 + (1 − 2x) y 0 + y(x − 1) = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + (1 − 2x) y 0 + y(x − 1) = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
2x − 1
x
x−1
q(x) =
x
p(x) = −
Table 2.32: Table p(x), q(x) singularites.
q(x) = x−1
x
singularity
type
x=0
“regular”
p(x) = − 2x−1
x
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + (1 − 2x) y 0 + y(x − 1) = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
194
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
(n + r) (n + r − 1) an xn+r−2 x
(1)
n=0
∞
X
+ (1 − 2x)
!
(n + r) an xn+r−1
+
n=0
∞
X
!
an xn+r
(x − 1) = 0
n=0
Which simplifies to
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
+
n=0
∞
X
−2xn+r an (n + r)
(2A)
n =0
+
∞
X
∞
X
!
(n + r) an x
n+r−1
+
n=0
∞
X
!
x
1+n+r
an
+
n=0
−an xn+r = 0
n =0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
−2x
n+r
∞
X
an (n + r) =
−2an−1 (n + r − 1) xn+r−1
n =0
n=1
∞
X
x
1+n+r
∞
X
an =
n =0
∞
X
an−2 xn+r−1
n=2
−an x
n+r
=
n =0
∞
X
−an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
n=0
+
∞
X
!
(n + r) an xn+r−1
n=0
+
+
∞
X
n =1
∞
X
−2an−1 (n + r − 1) xn+r−1
an−2 xn+r−1
n=2
!
+
∞
X
−an−1 xn+r−1 = 0
n =1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + (n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + r x−1+r a0 = 0
(2B)
chapter 2 . book solved problems
195
Since a0 6= 0 then the above simplifies to
x−1+r r2 = 0
Since the above is true for all x then the indicial equation becomes
r2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r2 = 0
Solving for r gives the roots of the indicial equation as [0, 0].
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
y1 (x) =
∞
X
an xn+r
(1A)
n=0
Now the second solution y2 is found using
y2 (x) = y1 (x) ln (x) +
∞
X
!
bn x
n+r
(1B)
n=1
Then the general solution will be
y = c1 y1 (x) + c2 y2 (x)
In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), a0 is never zero, and is arbitrary
and is typically taken as a0 = 1, and {c1 , c2 } are two arbitray constants of integration which
can be found from initial conditions. We start by finding the first solution y1 (x). Eq (2B)
derived above is now used to find all an coefficients. The case n = 0 is skipped since it
was used to find the roots of the indicial equation. a0 is arbitrary and taken as a0 = 1.
Substituting n = 1 in Eq. (2B) gives
a1 =
2r + 1
(1 + r)2
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) − 2an−1 (n + r − 1) + an (n + r) + an−2 − an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an =
2nan−1 + 2ran−1 − an−2 − an−1
n2 + 2nr + r2
(4)
Which for the root r = 0 becomes
an =
(2n − 1) an−1 − an−2
n2
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
2r+1
(1+r)2
an
1
1
chapter 2 . book solved problems
196
For n = 2, using the above recursive equation gives
a2 =
3r2 + 6r + 2
(1 + r)2 (r + 2)2
Which for the root r = 0 becomes
a2 =
1
2
And the table now becomes
n
a0
a1
an,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
a2
an
1
1
1
2
For n = 3, using the above recursive equation gives
a3 =
4r3 + 18r2 + 22r + 6
(1 + r)2 (r + 2)2 (r + 3)2
Which for the root r = 0 becomes
a3 =
1
6
And the table now becomes
n
a0
a1
a2
a3
an,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
4r3 +18r2 +22r+6
(1+r)2 (r+2)2 (r+3)2
an
1
1
1
2
1
6
For n = 4, using the above recursive equation gives
5r4 + 40r3 + 105r2 + 100r + 24
a4 =
(1 + r)2 (r + 2)2 (r + 3)2 (r + 4)2
Which for the root r = 0 becomes
a4 =
1
24
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
4r3 +18r2 +22r+6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
an
1
1
1
2
1
6
1
24
For n = 5, using the above recursive equation gives
a5 =
6r5 + 75r4 + 340r3 + 675r2 + 548r + 120
(1 + r)2 (r + 2)2 (r + 3)2 (r + 4)2 (r + 5)2
Which for the root r = 0 becomes
a5 =
And the table now becomes
1
120
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
197
an,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
4r3 +18r2 +22r+6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
6r5 +75r4 +340r3 +675r2 +548r+120
(1+r)2 (r+2)2 (r+3)2 (r+4)2 (r+5)2
an
1
1
1
2
1
6
1
24
1
120
Using the above table, then the first solution y1 (x) becomes
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
=1+x+
x2 x3 x4
x5
+
+
+
+ O x6
2
6
24 120
Now the second solution is found. The second solution is given by
!
∞
X
y2 (x) = y1 (x) ln (x) +
bn xn+r
n=1
Where bn is found using
d
an,r
dr
And the above is then evaluated at r = 0. The above table for an,r is used for this purpose.
Computing the derivatives gives the following table
bn =
n
b0
b1
b2
b3
b4
b5
bn,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
4r3 +18r2 +22r+6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
6r5 +75r4 +340r3 +675r2 +548r+120
(1+r)2 (r+2)2 (r+3)2 (r+4)2 (r+5)2
an
1
1
d
bn,r = dr
an,r
N/A since bn starts from 1
2r
− (1+r)
3
bn (r = 0
N/A
0
1
2
−6r3 −18r2 −14r
(1+r)3 (r+2)3
12r r4 +8r3 + 47
r2 +30r+ 85
2
6
− (1+r)3 (r+2)
3
(r+3)3
4 +290r 3 + 5031 r 2 +453r+166 r
20 r6 +15r5 + 183
r
2
10
−
(1+r)3 (r+2)3 (r+3)3 (r+4)3
30 r8 +24r7 + 739
r6 +1410r5 +4915r4 +10668r3 +14063r2 +10290r+ 48076
r
3
15
−
3
3
3
3
3
(1+r) (r+2) (r+3) (r+4) (r+5)
0
1
6
1
24
1
120
The above table gives all values of bn needed. Hence the second solution is
y2 (x) = y1 (x) ln (x) + b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x2 x3 x 4
x5
6
= 1+x+
+
+
+
+O x
ln (x) + O x6
2
6
24 120
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x2 x3 x4
x5
6
= c1 1 + x +
+
+
+
+O x
2
6
24 120
x 2 x3 x 4
x5
6
6
+ c2
1+x+
+
+
+
+O x
ln (x) + O x
2
6
24 120
Hence the final solution is
y = yh
x2 x3 x4
x5
6
= c1 1 + x +
+
+
+
+O x
2
6
24 120
x2 x3 x4
x5
6
6
+ c2
1+x+
+
+
+
+O x
ln (x) + O x
2
6
24 120
0
0
0
chapter 2 . book solved problems
3 Maple. Time used: 0.025 (sec). Leaf size: 39
Order:=6;
ode:=diff(diff(y(x),x),x)*x+(1-2*x)*diff(y(x),x)+(x-1)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
198
1 2 1 3
1 4
1 5
y = 1+x+ x + x + x +
x (ln (x) c2 + c1 ) + O x6
2
6
24
120
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x dx
y(x)
+
(1
−
2x)
y(x)
+ (x − 1) y(x) = 0
dx
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d d
y(x) = − (x−1)y(x)
+
dx dx
x
•
(−1+2x)
d
y(x)
dx
x
Group terms with y(x) on
the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
◦
◦
(x · P2 (x))
◦
•
•
d
y(x)
dx
x
=1
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
(−1+2x)
+ (x−1)y(x)
=0
x
Check to see if x0 = 0 is a regular singular point
Define functions
x−1
P2 (x) = − −1+2x
,
P
(x)
=
3
x
x
x · P2 (x) is analytic at x = 0
d d
y(x) −
dx dx
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x)
+
(1
−
2x)
y(x)
+ (x − 1) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..1
∞
P
xm · y(x) =
ak xk+r+m
k=0
chapter 2 . book solved problems
◦
◦
199
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert xm · dx
y(x) to series expansion for m = 0..1
∞
P
d
xm · dx
y(x) =
ak (k + r) xk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
xm · dx
y(x) =
ak+1−m (k + 1 − m + r) xk+r
k=−1+m
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
∞
P
2
2
a0 r2 x−1+r + a1 (1 + r) − a0 (1 + 2r) xr +
ak+1 (k + 1 + r) − ak (2k + 2r + 1) + ak−1 xk+r
k=1
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term must be 0
a1 (1 + r)2 − a0 (1 + 2r) = 0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1)2 + (−2k − 1) ak + ak−1 = 0
Shift index using k− >k + 1
ak+2 (k + 2)2 + (−2k − 3) ak+1 + ak = 0
Recursion relation that defines series solution to ODE
−ak +3ak+1
ak+2 = 2kak+1(k+2)
2
•
Recursion relation for r = 0
−ak +3ak+1
ak+2 = 2kak+1(k+2)
2
•
Solution
for r = 0
∞
P
2kak+1 −ak +3ak+1
k
y(x) =
ak x , ak+2 =
, a 1 − a0 = 0
(k+2)2
k=0
3 Mathematica. Time used: 0.004 (sec). Leaf size: 74
ode=x*D[y[x],{x,2}]+(1-2*x)*D[y[x],x]+(x-1)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
5
x5
x4 x3 x2
x
x4 x3 x2
+
+
+
+ x + 1 + c2
+
+
+
+ x + 1 log(x)
120 24
6
2
120 24
6
2
3 Sympy. Time used: 0.247 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*Derivative(y(x), (x, 2)) + (1 - 2*x)*Derivative(y(x), x) + (x - 1)*y(x
),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C1
x5
x4 x 3 x2
+
+
+
+ x + 1 + O x6
120 24
6
2
chapter 2 . book solved problems
2.2.13
200
Problem 15
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
Internal problem ID [8591]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 15
Date solved : Tuesday, November 11, 2025 at 12:03:55 PM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]
2x(1 − x) y 00 − (1 + 6x) y 0 − 2y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
−2x2 + 2x y 00 + (−6x − 1) y 0 − 2y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1 + 6x
2x (x − 1)
1
q(x) =
x (x − 1)
p(x) =
Table 2.34: Table p(x), q(x) singularites.
1+6x
p(x) = 2x(x−1)
singularity
type
x=0
“regular”
x=1
“regular”
1
q(x) = x(x−1)
singularity
type
x=0
“regular”
x=1
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0, 1, ∞]
Irregular singular points : []
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
−2x(x − 1) y 00 + (−6x − 1) y 0 − 2y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
201
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
−2x(x − 1)
+ (−6x − 1)
!
(n + r) (n + r − 1) an xn+r−2
n=0
∞
X
!
(n + r) an xn+r−1
−2
n=0
(1)
∞
X
!
an xn+r
=0
n=0
Which simplifies to
∞
X
∞
X
−2xn+r an (n + r) (n + r − 1) +
n =0
!
2xn+r−1 an (n + r) (n + r − 1)
(2A)
n=0
+
∞
X
−6xn+r an (n + r) +
n =0
∞
X
−(n + r) an xn+r−1 +
n =0
∞
X
−2an xn+r = 0
n =0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
−2x
n+r
∞
X
an (n + r) (n + r − 1) =
−2an−1 (n + r − 1) (n + r − 2) xn+r−1
n =0
n=1
∞
X
−6x
n+r
∞
X
an (n + r) =
−6an−1 (n + r − 1) xn+r−1
n =0
n=1
∞
X
−2an x
n+r
n =0
=
∞
X
−2an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
−2an−1 (n + r − 1) (n + r − 2) xn+r−1
n =1
+
∞
X
!
2x
n+r−1
an (n + r) (n + r − 1)
n=0
+
∞
X
+
∞
X
−6an−1 (n + r − 1) xn+r−1
n =1
−(n + r) an xn+r−1 +
n =0
∞
X
−2an−1 xn+r−1 = 0
n =1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
2xn+r−1 an (n + r) (n + r − 1) − (n + r) an xn+r−1 = 0
When n = 0 the above becomes
2x−1+r a0 r(−1 + r) − ra0 x−1+r = 0
(2B)
chapter 2 . book solved problems
202
Or
2x−1+r r(−1 + r) − r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
r x−1+r (−3 + 2r) = 0
Since the above is true for all x then the indicial equation becomes
2r2 − 3r = 0
Solving for r gives the roots of the indicial equation as
3
2
r2 = 0
r1 =
Since a0 6= 0 then the indicial equation becomes
r x−1+r (−3 + 2r) = 0
Solving for r gives the roots of the indicial equation as
3
2
,0 .
Since r1 − r2 = 32 is not an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = xr2
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
3
an xn+ 2
n=0
y2 (x) =
∞
X
bn x n
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. For 1 ≤ n the recursive equation is
−2an−1 (n + r − 1) (n + r − 2) + 2an (n + r) (n + r − 1)
− 6an−1 (n + r − 1) − an (n + r) − 2an−1 = 0
(3)
Solving for an from recursive equation (4) gives
2(n + r) an−1
2n − 3 + 2r
(4)
n + 32 an−1
an =
n
(5)
an =
Which for the root r = 32 becomes
At this point, it is a good idea to keep track of an in a table both before substituting r = 32
and after as more terms are found using the above recursive equation.
n
a0
an,r
1
an
1
chapter 2 . book solved problems
203
For n = 1, using the above recursive equation gives
a1 =
2 + 2r
2r − 1
Which for the root r = 32 becomes
5
2
a1 =
And the table now becomes
n
a0
a1
an,r
1
an
1
2+2r
2r−1
5
2
For n = 2, using the above recursive equation gives
4r2 + 12r + 8
4r2 − 1
a2 =
Which for the root r = 32 becomes
a2 =
35
8
And the table now becomes
n
a0
a1
a2
an,r
1
an
1
2+2r
2r−1
4r2 +12r+8
4r2 −1
5
2
35
8
For n = 3, using the above recursive equation gives
a3 =
8r3 + 48r2 + 88r + 48
8r3 + 12r2 − 2r − 3
Which for the root r = 32 becomes
a3 =
105
16
And the table now becomes
n
a0
a1
a2
a3
an,r
1
an
1
2+2r
2r−1
4r2 +12r+8
4r2 −1
8r3 +48r2 +88r+48
8r3 +12r2 −2r−3
5
2
35
8
105
16
For n = 4, using the above recursive equation gives
a4 =
16r4 + 160r3 + 560r2 + 800r + 384
16r4 + 64r3 + 56r2 − 16r − 15
Which for the root r = 32 becomes
a4 =
1155
128
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
an
1
2+2r
2r−1
4r2 +12r+8
4r2 −1
8r3 +48r2 +88r+48
8r3 +12r2 −2r−3
16r4 +160r3 +560r2 +800r+384
16r4 +64r3 +56r2 −16r−15
5
2
35
8
105
16
1155
128
chapter 2 . book solved problems
204
For n = 5, using the above recursive equation gives
a5 =
32r5 + 480r4 + 2720r3 + 7200r2 + 8768r + 3840
32r5 + 240r4 + 560r3 + 360r2 − 142r − 105
Which for the root r = 32 becomes
a5 =
3003
256
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
an
1
2+2r
2r−1
4r2 +12r+8
4r2 −1
8r3 +48r2 +88r+48
8r3 +12r2 −2r−3
16r4 +160r3 +560r2 +800r+384
16r4 +64r3 +56r2 −16r−15
32r5 +480r4 +2720r3 +7200r2 +8768r+3840
32r5 +240r4 +560r3 +360r2 −142r−105
5
2
35
8
105
16
1155
128
3003
256
Using the above table, then the solution y1 (x) is
y1 (x) = x3/2 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
5x 35x2 105x3 1155x4 3003x5
3/2
6
=x
1+
+
+
+
+
+O x
2
8
16
128
256
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. For 1 ≤ n the recursive equation is
−2bn−1 (n + r − 1) (n + r − 2) + 2bn (n + r) (n + r − 1)
− 6bn−1 (n + r − 1) − (n + r) bn − 2bn−1 = 0
(3)
Solving for bn from recursive equation (4) gives
bn =
2(n + r) bn−1
2n − 3 + 2r
Which for the root r = 0 becomes
(4)
2nbn−1
(5)
2n − 3
At this point, it is a good idea to keep track of bn in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
bn =
n
b0
bn,r
1
bn
1
For n = 1, using the above recursive equation gives
b1 =
2 + 2r
2r − 1
Which for the root r = 0 becomes
b1 = −2
And the table now becomes
n
b0
b1
bn,r
1
2+2r
2r−1
bn
1
−2
chapter 2 . book solved problems
205
For n = 2, using the above recursive equation gives
b2 =
4r2 + 12r + 8
4r2 − 1
Which for the root r = 0 becomes
b2 = −8
And the table now becomes
n
b0
b1
b2
bn,r
1
2+2r
2r−1
4r2 +12r+8
4r2 −1
bn
1
−2
−8
For n = 3, using the above recursive equation gives
b3 =
8r3 + 48r2 + 88r + 48
8r3 + 12r2 − 2r − 3
Which for the root r = 0 becomes
b3 = −16
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
2+2r
2r−1
4r2 +12r+8
4r2 −1
8r3 +48r2 +88r+48
8r3 +12r2 −2r−3
bn
1
−2
−8
−16
For n = 4, using the above recursive equation gives
b4 =
16r4 + 160r3 + 560r2 + 800r + 384
16r4 + 64r3 + 56r2 − 16r − 15
Which for the root r = 0 becomes
b4 = −
128
5
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
2+2r
2r−1
4r2 +12r+8
4r2 −1
8r3 +48r2 +88r+48
8r3 +12r2 −2r−3
16r4 +160r3 +560r2 +800r+384
16r4 +64r3 +56r2 −16r−15
bn
1
−2
−8
−16
− 128
5
For n = 5, using the above recursive equation gives
b5 =
32r5 + 480r4 + 2720r3 + 7200r2 + 8768r + 3840
32r5 + 240r4 + 560r3 + 360r2 − 142r − 105
Which for the root r = 0 becomes
b5 = −
And the table now becomes
256
7
chapter 2 . book solved problems
n
b0
b1
b2
b3
b4
b5
206
bn,r
1
2+2r
2r−1
4r2 +12r+8
4r2 −1
8r3 +48r2 +88r+48
8r3 +12r2 −2r−3
16r4 +160r3 +560r2 +800r+384
16r4 +64r3 +56r2 −16r−15
32r5 +480r4 +2720r3 +7200r2 +8768r+3840
32r5 +240r4 +560r3 +360r2 −142r−105
bn
1
−2
−8
−16
− 128
5
− 256
7
Using the above table, then the solution y2 (x) is
y2 (x) = b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
= 1 − 2x − 8x2 − 16x3 −
128x4 256x5
−
+ O x6
5
7
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
5x 35x2 105x3 1155x4 3003x5
3/2
6
= c1 x
1+
+
+
+
+
+O x
2
8
16
128
256
128x4 256x5
2
3
6
+ c2 1 − 2x − 8x − 16x −
−
+O x
5
7
Hence the final solution is
y = yh
5x 35x2 105x3 1155x4 3003x5
6
= c1 x
1+
+
+
+
+
+O x
2
8
16
128
256
128x4 256x5
2
3
6
+ c2 1 − 2x − 8x − 16x −
−
+O x
5
7
3/2
3 Maple. Time used: 0.026 (sec). Leaf size: 44
Order:=6;
ode:=2*(1-x)*x*diff(diff(y(x),x),x)-(1+6*x)*diff(y(x),x)-2*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
5
35 2 105 3 1155 4 3003 5
6
y = c1 x
1+ x+ x +
x +
x +
x +O x
2
8
16
128
256
128 4 256 5
2
3
6
+ c2 1 − 2x − 8x − 16x −
x −
x +O x
5
7
3/2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
<- linear_1 successful
Maple step by step
Let’s solve
chapter 2 . book solved problems
207
d d
d
2x(1 − x) dx
y(x) − (1 + 6x) dx
y(x) − 2y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
(1+6x) dx
y(x)
y(x)
d d
y(x)
=
−
−
dx dx
x(−1+x)
2x(−1+x)
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
(1+6x)
◦
d
y(x)
dx
y(x)
+ x(−1+x)
=0
Check to see if x0 is a regular singular point
Define
functions
h
i
d d
y(x) +
dx dx
2x(−1+x)
1+6x
1
P2 (x) = 2x(−1+x)
, P3 (x) = x(−1+x)
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
= − 12
=0
x=0
x = 0is a regular singular point
Check to see if x0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
2x(−1 + x) dx
y(x)
+
(1
+
6x)
y(x)
+ 2y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d
Convert xm · dx
y(x) to series expansion for m = 0..1
∞
P
d
xm · dx
y(x) =
ak (k + r) xk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
xm · dx
y(x) =
ak+1−m (k + 1 − m + r) xk+r
k=−1+m
d d
Convert xm · dx
y(x) to series expansion for m = 1..2
dx
∞
P
d d
xm · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−2+m
dx
k=0
◦
Shift index using k− >k + 2 − m
∞
P
d d
xm · dx
y(x)
=
ak+2−m (k + 2 − m + r) (k + 1 − m + r) xk+r
dx
k=−2+m
Rewrite ODE with series
expansions
∞
P
2 k+r
−1+r
−a0 r(−3 + 2r) x
+
−ak+1 (k + r + 1) (2k − 1 + 2r) + 2ak (k + r + 1) x
=0
k=0
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
−r(−3 + 2r) = 0
Values of r that satisfy the indicial equation
r ∈ 0, 32
Each term in the series must be 0, giving the recursion relation
−2 k + r − 12 ak+1 (k + r + 1) + 2ak (k + r + 1)2 = 0
Recursion relation that defines series solution to ODE
k (k+r+1)
ak+1 = 2a2k−1+2r
Recursion relation for r = 0
k (k+1)
ak+1 = 2a2k−1
Solution
for r = 0
∞
P
2ak (k+1)
k
y(x) =
ak x , ak+1 = 2k−1
k=0
•
Recursion relation
for r = 32
ak+1 =
2ak k+ 52
2k+2
chapter 2 . book solved problems
208
•
Solution
for r = 32
∞
5
P
3
2a
k+
k
y(x) =
ak xk+ 2 , ak+1 = 2k+2 2
•
Combine
parameters
solutions
andrename
∞
∞
P
P
2bk k+ 52
2ak (k+1)
k+ 32
k
y(x) =
ak x +
bk x
, ak+1 = 2k−1 , bk+1 = 2k+2
k=0
k=0
k=0
3 Mathematica. Time used: 0.005 (sec). Leaf size: 79
ode=2*x*(1-x)*D[y[x],{x,2}]-(1+6*x)*D[y[x],x]-2*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
256x5 128x4
3
2
y(x) → c2 −
−
− 16x − 8x − 2x + 1
7
5
3003x5 1155x4 105x3 35x2 5x
+ c1
+
+
+
+
+ 1 x3/2
256
128
16
8
2
3 Sympy. Time used: 0.371 (sec). Leaf size: 14
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(2*x*(1 - x)*Derivative(y(x), (x, 2)) - (6*x + 1)*Derivative(y(x), x) 2*y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
3
y(x) = C2 x 2 + C1 + O x6
chapter 2 . book solved problems
2.2.14
209
Problem 16
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
Internal problem ID [8592]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 16
Date solved : Tuesday, November 11, 2025 at 12:03:56 PM
CAS classification : [_Jacobi]
00
x(1 − x) y +
1
+ 2x y 0 − 2y = 0
2
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
00
1
2
−x + x y +
+ 2x y 0 − 2y = 0
2
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
4x + 1
2x (x − 1)
2
q(x) =
x (x − 1)
p(x) = −
Table 2.36: Table p(x), q(x) singularites.
4x+1
p(x) = − 2x(x−1)
singularity
type
x=0
“regular”
x=1
“regular”
2
q(x) = x(x−1)
singularity
type
x=0
“regular”
x=1
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0, 1, ∞]
Irregular singular points : []
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
1
00
−x(x − 1) y +
+ 2x y 0 − 2y = 0
2
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
210
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
!
∞
X
−x(x − 1)
(n + r) (n + r − 1) an xn+r−2
(1)
n=0
+
1
+ 2x
2
X
∞
!
(n + r) an x
n+r−1
−2
n=0
∞
X
!
an xn+r
=0
n=0
Which simplifies to
∞
X
∞
X
−xn+r an (n + r) (n + r − 1) +
n =0
!
xn+r−1 an (n + r) (n + r − 1)
(2A)
n=0
∞
X
+
!
2xn+r an (n + r)
+
n=0
∞
X
(n + r) an xn+r−1
2
n=0
!
+
∞
X
−2an xn+r = 0
n =0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
∞
X
−xn+r an (n + r) (n + r − 1) =
−an−1 (n + r − 1) (n + r − 2) xn+r−1
n =0
n=1
∞
X
2x
n+r
an (n + r) =
n =0
∞
X
∞
X
2an−1 (n + r − 1) xn+r−1
n=1
−2an x
n+r
=
n =0
∞
X
−2an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
−an−1 (n + r − 1) (n + r − 2) xn+r−1
n =1
+
+
∞
X
n=0
∞
X
!
xn+r−1 an (n + r) (n + r − 1)
+
∞
X
!
2an−1 (n + r − 1) xn+r−1
n=1
(n + r) an x
2
n=0
n+r−1
!
+
∞
X
−2an−1 xn+r−1 = 0
n =1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) +
(n + r) an xn+r−1
=0
2
When n = 0 the above becomes
x−1+r a0 r(−1 + r) +
ra0 x−1+r
=0
2
(2B)
chapter 2 . book solved problems
Or
211
r x−1+r
−1+r
x
r(−1 + r) +
a0 = 0
2
Since a0 6= 0 then the above simplifies to
rx
−1+r
1
− +r =0
2
Since the above is true for all x then the indicial equation becomes
1
r2 − r = 0
2
Solving for r gives the roots of the indicial equation as
1
2
r2 = 0
r1 =
Since a0 6= 0 then the indicial equation becomes
1
−1+r
rx
− +r =0
2
Solving for r gives the roots of the indicial equation as 12 , 0 .
Since r1 − r2 = 12 is not an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
y2 (x) = xr2
n=0
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
1
an xn+ 2
n=0
y2 (x) =
∞
X
bn x n
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. For 1 ≤ n the recursive equation is
−an−1 (n + r − 1) (n + r − 2) + an (n + r) (n + r − 1)
an (n + r)
+ 2an−1 (n + r − 1) +
− 2an−1 = 0
2
(3)
Solving for an from recursive equation (4) gives
an =
2an−1 (n2 + 2nr + r2 − 5n − 5r + 6)
2n2 + 4nr + 2r2 − n − r
(4)
Which for the root r = 12 becomes
an =
an−1 (4n2 − 16n + 15)
4n2 + 2n
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 12
and after as more terms are found using the above recursive equation.
chapter 2 . book solved problems
n
a0
212
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 =
2r2 − 6r + 4
2r2 + 3r + 1
Which for the root r = 12 becomes
1
2
a1 =
And the table now becomes
n
a0
a1
an,r
1
an
1
2r2 −6r+4
2r2 +3r+1
1
2
For n = 2, using the above recursive equation gives
4(−1 + r)2 (−2 + r) r
a2 = 4
4r + 20r3 + 35r2 + 25r + 6
Which for the root r = 12 becomes
a2 = −
1
40
And the table now becomes
n
a0
a1
a2
an,r
1
an
1
2r2 −6r+4
2r2 +3r+1
4(−1+r)2 (−2+r)r
4
4r +20r3 +35r2 +25r+6
1
2
1
− 40
For n = 3, using the above recursive equation gives
8(−1 + r)2 (−2 + r) r2
a3 = 5
8r + 76r4 + 274r3 + 461r2 + 351r + 90
Which for the root r = 12 becomes
a3 = −
1
560
And the table now becomes
n
a0
a1
a2
a3
an,r
1
an
1
2r2 −6r+4
2r2 +3r+1
4(−1+r)2 (−2+r)r
4r4 +20r3 +35r2 +25r+6
8(−1+r)2 (−2+r)r2
5
8r +76r4 +274r3 +461r2 +351r+90
1
2
1
− 40
1
− 560
For n = 4, using the above recursive equation gives
16(−1 + r)2 (−2 + r) r2 (r + 1)
a4 =
16r6 + 240r5 + 1432r4 + 4296r3 + 6697r2 + 4959r + 1260
Which for the root r = 12 becomes
a4 = −
And the table now becomes
1
2688
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
213
an,r
1
an
1
2r2 −6r+4
2r2 +3r+1
4(−1+r)2 (−2+r)r
4r4 +20r3 +35r2 +25r+6
8(−1+r)2 (−2+r)r2
5
8r +76r4 +274r3 +461r2 +351r+90
16(−1+r)2 (−2+r)r2 (r+1)
16r6 +240r5 +1432r4 +4296r3 +6697r2 +4959r+1260
1
2
1
− 40
1
− 560
1
− 2688
For n = 5, using the above recursive equation gives
a5 =
32(−1 + r)2 r2 (r + 1) (r2 − 4)
32r7 + 688r6 + 6080r5 + 28360r4 + 74378r3 + 107347r2 + 76065r + 18900
Which for the root r = 12 becomes
a5 = −
1
8448
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
an
1
2r2 −6r+4
2r2 +3r+1
4(−1+r)2 (−2+r)r
4r4 +20r3 +35r2 +25r+6
8(−1+r)2 (−2+r)r2
8r5 +76r4 +274r3 +461r2 +351r+90
16(−1+r)2 (−2+r)r2 (r+1)
16r6 +240r5 +1432r4 +4296r3 +6697r2 +4959r+1260
32(−1+r)2 r2 (r+1) r2 −4
32r7 +688r6 +6080r5 +28360r4 +74378r3 +107347r2 +76065r+18900
1
2
1
− 40
1
− 560
1
− 2688
1
− 8448
Using the above table, then the solution y1 (x) is
√
y1 (x) = x a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
√
x x2
x3
x4
x5
6
= x 1+ −
−
−
−
+O x
2 40 560 2688 8448
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. For 1 ≤ n the recursive equation is
−bn−1 (n + r − 1) (n + r − 2) + bn (n + r) (n + r − 1)
(n + r) bn
+ 2bn−1 (n + r − 1) +
− 2bn−1 = 0
2
(3)
Solving for bn from recursive equation (4) gives
2bn−1 (n2 + 2nr + r2 − 5n − 5r + 6)
bn =
2n2 + 4nr + 2r2 − n − r
(4)
Which for the root r = 0 becomes
bn =
2bn−1 (n2 − 5n + 6)
n (2n − 1)
(5)
At this point, it is a good idea to keep track of bn in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
b0
bn,r
1
bn
1
For n = 1, using the above recursive equation gives
b1 =
2r2 − 6r + 4
2r2 + 3r + 1
chapter 2 . book solved problems
214
Which for the root r = 0 becomes
b1 = 4
And the table now becomes
n
b0
b1
bn,r
1
2r2 −6r+4
2r2 +3r+1
bn
1
4
For n = 2, using the above recursive equation gives
4(−1 + r)2 (−2 + r) r
b2 = 4
4r + 20r3 + 35r2 + 25r + 6
Which for the root r = 0 becomes
b2 = 0
And the table now becomes
n
b0
b1
b2
bn,r
1
2r2 −6r+4
2r2 +3r+1
4(−1+r)2 (−2+r)r
4r4 +20r3 +35r2 +25r+6
bn
1
4
0
For n = 3, using the above recursive equation gives
8(−1 + r)2 (−2 + r) r2
b3 = 5
8r + 76r4 + 274r3 + 461r2 + 351r + 90
Which for the root r = 0 becomes
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
2r2 −6r+4
2r2 +3r+1
4(−1+r)2 (−2+r)r
4r4 +20r3 +35r2 +25r+6
8(−1+r)2 (−2+r)r2
5
8r +76r4 +274r3 +461r2 +351r+90
bn
1
4
0
0
For n = 4, using the above recursive equation gives
16(−1 + r)2 (−2 + r) r2 (r + 1)
b4 =
16r6 + 240r5 + 1432r4 + 4296r3 + 6697r2 + 4959r + 1260
Which for the root r = 0 becomes
b4 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
2r2 −6r+4
2r2 +3r+1
4(−1+r)2 (−2+r)r
4r4 +20r3 +35r2 +25r+6
8(−1+r)2 (−2+r)r2
8r5 +76r4 +274r3 +461r2 +351r+90
16(−1+r)2 (−2+r)r2 (r+1)
6
5
16r +240r +1432r4 +4296r3 +6697r2 +4959r+1260
bn
1
4
0
0
0
chapter 2 . book solved problems
215
For n = 5, using the above recursive equation gives
32(−1 + r)2 r2 (r + 1) (r2 − 4)
b5 =
32r7 + 688r6 + 6080r5 + 28360r4 + 74378r3 + 107347r2 + 76065r + 18900
Which for the root r = 0 becomes
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
2r2 −6r+4
2r2 +3r+1
4(−1+r)2 (−2+r)r
4
4r +20r3 +35r2 +25r+6
8(−1+r)2 (−2+r)r2
8r5 +76r4 +274r3 +461r2 +351r+90
16(−1+r)2 (−2+r)r2 (r+1)
16r6 +240r5 +1432r4 +4296r3 +6697r2 +4959r+1260
32(−1+r)2 r2 (r+1) r2 −4
32r7 +688r6 +6080r5 +28360r4 +74378r3 +107347r2 +76065r+18900
bn
1
4
0
0
0
0
Using the above table, then the solution y2 (x) is
y2 (x) = b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
= 1 + 4x + O x6
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
√
x x2
x3
x4
x5
6
= c1 x 1 + −
−
−
−
+O x
+ c2 1 + 4x + O x6
2 40 560 2688 8448
Hence the final solution is
y = yh
√
x x2
x3
x4
x5
6
= c1 x 1 + −
−
−
−
+O x
+ c2 1 + 4x + O x6
2 40 560 2688 8448
3 Maple. Time used: 0.033 (sec). Leaf size: 36
Order:=6;
ode:=(1-x)*x*diff(diff(y(x),x),x)+(1/2+2*x)*diff(y(x),x)-2*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
y = c1
√
1
1 2
1 3
1 4
1 5
6
x 1+ x− x −
x −
x −
x +O x
+ c2 1 + 4x + O x6
2
40
560
2688
8448
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful
chapter 2 . book solved problems
216
Maple step by step
Let’s solve
d
d d
x(1 − x) dx
y(x) + 12 + 2x dx
y(x) − 2y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
(1+4x) dx
y(x)
2y(x)
d d
y(x)
=
−
+
dx dx
x(−1+x)
2x(−1+x)
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
◦
(1+4x)
d
y(x)
dx
2y(x)
+ x(−1+x)
=0
Check to see if x0 is a regular singular point
Define
functions
h
i
d d
y(x) −
dx dx
2x(−1+x)
1+4x
2
P2 (x) = − 2x(−1+x)
, P3 (x) = x(−1+x)
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
•
•
= 12
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
x=0
=0
x=0
x = 0is a regular singular point
Check to see if x0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
2x(−1 + x) dx
y(x)
+
(−1
−
4x)
y(x)
+ 4y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d
Convert xm · dx
y(x) to series expansion for m = 0..1
∞
P
d
xm · dx
y(x) =
ak (k + r) xk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
xm · dx
y(x) =
ak+1−m (k + 1 − m + r) xk+r
k=−1+m
d d
Convert xm · dx
y(x) to series expansion for m = 1..2
dx
∞
P
d d
xm · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−2+m
dx
k=0
◦
Shift index using k− >k + 2 − m
∞
P
d d
xm · dx
y(x)
=
ak+2−m (k + 2 − m + r) (k + 1 − m + r) xk+r
dx
k=−2+m
Rewrite ODE with series
expansions
∞
P
−1+r
k+r
−a0 r(−1 + 2r) x
+
(−ak+1 (k + 1 + r) (2k + 1 + 2r) + 2ak (k + r − 1) (k + r − 2)) x
k=0
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
−r(−1 + 2r) = 0
Values of r that satisfy the indicial equation
r ∈ 0, 12
Each term in the series must be 0, giving the recursion relation
−2 k + 12 + r (k + 1 + r) ak+1 + 2ak (k + r − 1) (k + r − 2) = 0
Recursion relation that defines series solution to ODE
k (k+r−1)(k+r−2)
ak+1 = 2a
(2k+1+2r)(k+1+r)
Recursion relation for r = 0 ; series terminates at k = 1
k (k−1)(k−2)
ak+1 = 2a(2k+1)(k+1)
Apply recursion relation for k = 0
a1 = 4a0
chapter 2 . book solved problems
•
•
217
Terminating series solution of the ODE for r = 0 . Use reduction of order to find the second linea
y(x) = a0 · (1 + 4x)
Recursion relation
for
r = 12
2ak k− 12 k− 32
(2k+2) k+ 32
Solution
for r = 12
∞
P
2ak k− 12 k− 32
k+ 12
y(x) =
ak x , ak+1 = (2k+2) k+ 3 2
k=0
ak+1 =
•
•
Combine
solutions and rename parameters
∞
P
2bk k− 12 k− 32
k+ 12
y(x) = a0 · (1 + 4x) +
bk x
, bk+1 = (2k+2) k+ 3 k=0
2
3 Mathematica. Time used: 0.004 (sec). Leaf size: 55
ode=x*(1-x)*D[y[x],{x,2}]+(1/2+2*x)*D[y[x],x]-2*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
√
x5
x4
x3
x2 x
y(x) → c1 x −
−
−
−
+ + 1 + c2 (4x + 1)
8448 2688 560 40 2
3 Sympy. Time used: 0.417 (sec). Leaf size: 73
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*(1 - x)*Derivative(y(x), (x, 2)) + (2*x + 1/2)*Derivative(y(x), x) 2*y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
128x5 32x4 32x3 8x2
y(x) = C2 −
+
−
+
− 4x + 1
14175
315
45
3
√ 32x4 32x3 8x2 4x
+ C1 x
−
+
−
+ 1 + O x6
2835
315
15
3
chapter 2 . book solved problems
2.2.15
218
Problem 17
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
Internal problem ID [8593]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 17
Date solved : Tuesday, November 11, 2025 at 12:03:56 PM
CAS classification : [[_Emden, _Fowler]]
4xy 00 + y 0 + 8y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
4xy 00 + y 0 + 8y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
4x
2
q(x) =
x
p(x) =
Table 2.38: Table p(x), q(x) singularites.
1
p(x) = 4x
singularity
type
x=0
“regular”
q(x) = x2
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
4xy 00 + y 0 + 8y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
219
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
4
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x +
n=0
!
(n + r) an xn+r−1
+8
n=0
∞
X
!
an xn+r
= 0 (1)
n=0
Which simplifies to
∞
X
!
4xn+r−1 an (n + r) (n + r − 1)
∞
X
+
n=0
!
(n + r) an xn+r−1
+
n=0
∞
X
!
8an xn+r
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
8an x
n+r
n =0
=
∞
X
8an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
4xn+r−1 an (n+r) (n+r −1) +
n=0
∞
X
!
(n+r) an xn+r−1 +
n=0
∞
X
!
8an−1 xn+r−1
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
4xn+r−1 an (n + r) (n + r − 1) + (n + r) an xn+r−1 = 0
When n = 0 the above becomes
4x−1+r a0 r(−1 + r) + ra0 x−1+r = 0
Or
4x−1+r r(−1 + r) + r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
r x−1+r (−3 + 4r) = 0
Since the above is true for all x then the indicial equation becomes
4r2 − 3r = 0
Solving for r gives the roots of the indicial equation as
3
4
r2 = 0
r1 =
Since a0 6= 0 then the indicial equation becomes
r x−1+r (−3 + 4r) = 0
= 0 (2B)
chapter 2 . book solved problems
220
3
Solving for r gives the roots of the indicial equation as
4
,0 .
Since r1 − r2 = 34 is not an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = xr2
∞
X
!
bn x n
n=0
Or
∞
X
y1 (x) =
n=0
∞
X
y2 (x) =
3
an xn+ 4
bn x n
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. For 1 ≤ n the recursive equation is
4an (n + r) (n + r − 1) + an (n + r) + 8an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an = −
8an−1
2
4n + 8nr + 4r2 − 3n − 3r
(4)
8an−1
n (4n + 3)
(5)
Which for the root r = 34 becomes
an = −
At this point, it is a good idea to keep track of an in a table both before substituting r = 34
and after as more terms are found using the above recursive equation.
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 = −
8
4r2 + 5r + 1
Which for the root r = 34 becomes
a1 = −
8
7
And the table now becomes
n
a0
a1
an,r
1
8
− 4r2 +5r+1
an
1
− 87
For n = 2, using the above recursive equation gives
a2 =
64
(4r2 + 5r + 1) (4r2 + 13r + 10)
Which for the root r = 34 becomes
a2 =
32
77
chapter 2 . book solved problems
221
And the table now becomes
n
a0
a1
a2
an,r
1
8
− 4r2 +5r+1
an
1
− 87
64
(4r2 +5r+1)(4r2 +13r+10)
32
77
For n = 3, using the above recursive equation gives
a3 = −
512
(4r2 + 5r + 1) (4r2 + 13r + 10) (4r2 + 21r + 27)
Which for the root r = 34 becomes
a3 = −
256
3465
And the table now becomes
n
a0
a1
a2
a3
an,r
1
8
− 4r2 +5r+1
an
1
− 87
64
(4r2 +5r+1)(4r2 +13r+10)
512
− (4r2 +5r+1)(4r2 +13r+10)(4r
2 +21r+27)
32
77
256
− 3465
For n = 4, using the above recursive equation gives
a4 =
4096
(4r2 + 5r + 1) (4r2 + 13r + 10) (4r2 + 21r + 27) (4r2 + 29r + 52)
Which for the root r = 34 becomes
a4 =
512
65835
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
8
− 4r2 +5r+1
an
1
− 87
64
(4r2 +5r+1)(4r2 +13r+10)
512
− (4r2 +5r+1)(4r2 +13r+10)(4r
2 +21r+27)
4096
(4r2 +5r+1)(4r2 +13r+10)(4r2 +21r+27)(4r2 +29r+52)
32
77
256
− 3465
512
65835
For n = 5, using the above recursive equation gives
a5 = −
32768
(4r2 + 5r + 1) (4r2 + 13r + 10) (4r2 + 21r + 27) (4r2 + 29r + 52) (4r2 + 37r + 85)
Which for the root r = 34 becomes
a5 = −
4096
7571025
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
8
− 4r2 +5r+1
an
1
− 87
64
(4r2 +5r+1)(4r2 +13r+10)
512
− (4r2 +5r+1)(4r2 +13r+10)(4r
2 +21r+27)
4096
(4r2 +5r+1)(4r2 +13r+10)(4r2 +21r+27)(4r2 +29r+52)
32768
− (4r2 +5r+1)(4r2 +13r+10)(4r2 +21r+27)(4r
2 +29r+52)(4r 2 +37r+85)
32
77
256
− 3465
512
65835
4096
− 7571025
chapter 2 . book solved problems
222
Using the above table, then the solution y1 (x) is
y1 (x) = x3/4 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
8x 32x2 256x3 512x4
4096x5
3/4
6
=x
1−
+
−
+
−
+O x
7
77
3465
65835 7571025
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. For 1 ≤ n the recursive equation is
4bn (n + r) (n + r − 1) + (n + r) bn + 8bn−1 = 0
(3)
Solving for bn from recursive equation (4) gives
bn = −
8bn−1
2
4n + 8nr + 4r2 − 3n − 3r
(4)
8bn−1
n (4n − 3)
(5)
Which for the root r = 0 becomes
bn = −
At this point, it is a good idea to keep track of bn in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
b0
bn,r
1
bn
1
For n = 1, using the above recursive equation gives
b1 = −
8
4r2 + 5r + 1
Which for the root r = 0 becomes
b1 = −8
And the table now becomes
n
b0
b1
bn,r
1
8
− 4r2 +5r+1
bn
1
−8
For n = 2, using the above recursive equation gives
b2 =
64
(4r2 + 5r + 1) (4r2 + 13r + 10)
Which for the root r = 0 becomes
b2 =
32
5
And the table now becomes
n
b0
b1
b2
bn,r
1
8
− 4r2 +5r+1
bn
1
−8
64
(4r2 +5r+1)(4r2 +13r+10)
32
5
For n = 3, using the above recursive equation gives
b3 = −
512
(4r2 + 5r + 1) (4r2 + 13r + 10) (4r2 + 21r + 27)
chapter 2 . book solved problems
223
Which for the root r = 0 becomes
b3 = −
256
135
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
8
− 4r2 +5r+1
bn
1
−8
64
(4r2 +5r+1)(4r2 +13r+10)
512
− (4r2 +5r+1)(4r2 +13r+10)(4r
2 +21r+27)
32
5
− 256
135
For n = 4, using the above recursive equation gives
b4 =
4096
(4r2 + 5r + 1) (4r2 + 13r + 10) (4r2 + 21r + 27) (4r2 + 29r + 52)
Which for the root r = 0 becomes
b4 =
512
1755
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
8
− 4r2 +5r+1
bn
1
−8
64
(4r2 +5r+1)(4r2 +13r+10)
512
− (4r2 +5r+1)(4r2 +13r+10)(4r
2 +21r+27)
4096
(4r2 +5r+1)(4r2 +13r+10)(4r2 +21r+27)(4r2 +29r+52)
32
5
− 256
135
512
1755
For n = 5, using the above recursive equation gives
b5 = −
32768
(4r2 + 5r + 1) (4r2 + 13r + 10) (4r2 + 21r + 27) (4r2 + 29r + 52) (4r2 + 37r + 85)
Which for the root r = 0 becomes
b5 = −
4096
149175
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
8
− 4r2 +5r+1
bn
1
−8
64
(4r2 +5r+1)(4r2 +13r+10)
512
− (4r2 +5r+1)(4r2 +13r+10)(4r
2 +21r+27)
4096
(4r2 +5r+1)(4r2 +13r+10)(4r2 +21r+27)(4r2 +29r+52)
32768
− (4r2 +5r+1)(4r2 +13r+10)(4r2 +21r+27)(4r
2 +29r+52)(4r 2 +37r+85)
32
5
− 256
135
512
1755
4096
− 149175
Using the above table, then the solution y2 (x) is
y2 (x) = b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
= 1 − 8x +
32x2 256x3 512x4 4096x5
−
+
−
+ O x6
5
135
1755
149175
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
8x 32x2 256x3 512x4
4096x5
3/4
6
= c1 x
1−
+
−
+
−
+O x
7
77
3465
65835 7571025
32x2 256x3 512x4 4096x5
6
+ c2 1 − 8x +
−
+
−
+O x
5
135
1755
149175
chapter 2 . book solved problems
224
Hence the final solution is
y = yh
8x 32x2 256x3 512x4
4096x5
6
= c1 x
1−
+
−
+
−
+O x
7
77
3465
65835 7571025
32x2 256x3 512x4 4096x5
6
+ c2 1 − 8x +
−
+
−
+O x
5
135
1755
149175
3/4
3 Maple. Time used: 0.033 (sec). Leaf size: 44
Order:=6;
ode:=4*diff(diff(y(x),x),x)*x+diff(y(x),x)+8*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
8
32 2
256 3
512 4
4096 5
6
y = c1 x
1− x+ x −
x +
x −
x +O x
7
77
3465
65835
7571025
32 2 256 3
512 4
4096 5
6
+ c2 1 − 8x + x −
x +
x −
x +O x
5
135
1755
149175
3/4
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
4x dx
y(x) + dx
y(x) + 8y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
d d
y(x) = − 2y(x)
− dx4x
dx dx
x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
d d
dx
y(x)
+
+ 2y(x)
=0
dx dx
4x
x
◦
◦
Check to see if x0 = 0 is a regular singular point
Define functions
1
P2 (x) = 4x
, P3 (x) = x2
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
= 14
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
x=0
=0
x=0
x = 0is a regular singular point
chapter 2 . book solved problems
225
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
4x dx
y(x) + dx
y(x) + 8y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
•
•
k=0
◦
Rewrite ODE with series expansions
d
Convert dx
y(x) to series expansion
∞
P
d
y(x) =
ak (k + r) xk+r−1
dx
k=0
◦
◦
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + 1 + r) xk+r
dx
k=−1
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series
∞expansions
P
−1+r
k+r
a0 r(−3 + 4r) x
+
(ak+1 (k + 1 + r) (4k + 1 + 4r) + 8ak ) x
=0
k=0
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r(−3 + 4r) = 0
Values of r that satisfy the indicial equation
r ∈ 0, 34
Each term in the series must be 0, giving the recursion relation
4(k + 1 + r) k + 14 + r ak+1 + 8ak = 0
Recursion relation that defines series solution to ODE
8ak
ak+1 = − (k+1+r)(4k+1+4r)
Recursion relation for r = 0
8ak
ak+1 = − (k+1)(4k+1)
Solution
for r = 0
∞
P
8ak
k
y(x) =
ak x , ak+1 = − (k+1)(4k+1)
k=0
•
Recursion relation for r = 34
k
ak+1 = − k+ 78a
(4k+4)
•
Solution
for r = 34
∞
P
8a
k+ 34
k
y(x) =
ak x , ak+1 = − k+ 7 (4k+4)
4
4
k=0
•
Combine
parameters
solutions
andrename
∞
∞
P
P
8ak
8b
k+ 34
k
k
y(x) =
ak x +
bk x
, ak+1 = − (k+1)(4k+1) , bk+1 = − k+ 7 (4k+4)
k=0
k=0
3 Mathematica. Time used: 0.002 (sec). Leaf size: 83
ode=4*x*D[y[x],{x,2}]+D[y[x],x]+8*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
4096x5 512x4 256x3 32x2
y(x) → c2 −
+
−
+
− 8x + 1
149175
1755
135
5
4096x5
512x4 256x3 32x2 8x
3/4
+ c1 x
−
+
−
+
−
+1
7571025 65835
3465
77
7
4
chapter 2 . book solved problems
3 Sympy. Time used: 0.266 (sec). Leaf size: 73
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(4*x*Derivative(y(x), (x, 2)) + 8*y(x) + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
4096x5 512x4 256x3 32x2
y(x) = C2 −
+
−
+
− 8x + 1
149175
1755
135
5
3
512x4 256x3 32x2 8x
4
+ C1 x
−
+
−
+ 1 + O x6
65835
3465
77
7
226
chapter 2 . book solved problems
2.2.16
227
Problem 18
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
Internal problem ID [8594]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 18
Date solved : Tuesday, November 11, 2025 at 12:03:57 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
4 t2 − 3t + 2 y 00 − 2y 0 + y = 0
Using series expansion around t = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.27)
(2.28)
(2)
(3)
chapter 2 . book solved problems
228
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
229
To find y(x) series solution around x = 0. Hence
2y 0 − y
4t2 − 12t + 8
dF0
F1 =
dt
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂t
∂y
∂y 0
(−2t2 − 2t + 10) y 0 + y(4t − 7)
=
8 (t2 − 3t + 2)2
dF1
F2 =
dt
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂t
∂y
∂y 0
(16t3 − 28t2 − 52t + 94) y 0 + (−23t2 + 81t − 73) y
=
16 (t2 − 3t + 2)3
dF2
F3 =
dt
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂t
∂y
∂y 0
845
1037
2
(−71t4 + 270t3 − 104t2 − 633t + 643) y 0 + 88y t3 − 467
t
+
t
−
88
88
176
=
16 (t2 − 3t + 2)4
dF3
F4 =
dt
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂t
∂y
∂y 0
10884 2
13391
6257
3
(1488t5 − 8466t4 + 13692t3 + 5550t2 − 33312t + 22938) y 0 − 1689 t4 − 3998
t
+
t
−
t
+
563
563
563
563
=
64 (t2 − 3t + 2)5
F0 =
And so on. Evaluating all the above at initial conditions t = 0 and y(0) = y(0) and y 0 (0) =
y 0 (0) gives
y 0 (0) y(0)
−
4
8
7y(0) 5y 0 (0)
F1 = −
+
32
16
73y(0) 47y 0 (0)
F2 = −
+
128
64
1037y(0) 643y 0 (0)
F3 = −
+
512
256
18771y(0) 11469y 0 (0)
F4 = −
+
2048
1024
F0 =
Substituting all the above in (7) and simplifying gives the solution as
1 2
7 3
73 4
1037 5
y = 1− t −
t −
t −
t y(0)
16
192
3072
61440
1 2
5 3
47 4
643 5 0
+ t+ t + t +
t +
t y (0) + O t6
8
96
1536
30720
Since the expansion point t = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
4t2 − 12t + 8 y 00 − 2y 0 + y = 0
Let the solution be represented as power series of the form
y=
∞
X
n=0
an tn
chapter 2 . book solved problems
230
Then
0
y =
∞
X
nan tn−1
n=1
y 00 =
∞
X
n(n − 1) an tn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
4t2 − 12t + 8
n(n − 1) an tn−2 − 2
nan tn−1 +
n=2
n=1
∞
X
!
an tn
=0
(1)
n=0
Which simplifies to
∞
X
!
n
4t an n(n − 1)
+
n=2
+
∞
X
−12n tn−1 an (n − 1)
(2)
n =2
∞
X
!
8n(n − 1) an tn−2
+
n=2
∞
X
−2nan tn−1 +
n =1
∞
X
!
an tn
=0
n=0
The next step is to make all powers of t be n in each summation term. Going over each
summation term above with power of t in it which is not already tn and adjusting the power
and the corresponding index gives
∞
X
−12n t
n−1
∞
X
an (n − 1) =
(−12(n + 1) an+1 n tn )
n =2
n=1
∞
X
n−2
8n(n − 1) an t
=
n =2
∞
X
8(n + 2) an+2 (n + 1) tn
n=0
∞
X
∞
X
−2nan tn−1 =
(−2(n + 1) an+1 tn )
n =1
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of t
are the same and equal to n.
∞
X
!
4tn an n(n − 1)
n=2
+
+
∞
X
(−12(n + 1) an+1 n tn )
(3)
n =1
∞
X
!
8(n + 2) an+2 (n + 1) tn
+
n=0
∞
X
(−2(n + 1) an+1 tn ) +
n =0
n = 0 gives
16a2 − 2a1 + a0 = 0
a2 = −
a0 a1
+
16
8
n = 1 gives
−28a2 + 48a3 + a1 = 0
Which after substituting earlier equations, simplifies to
a3 = −
7a0 5a1
+
192
96
∞
X
n=0
!
an tn
=0
chapter 2 . book solved problems
231
For 2 ≤ n, the recurrence equation is
4nan (n − 1) − 12(n + 1) an+1 n + 8(n + 2) an+2 (n + 1) − 2(n + 1) an+1 + an = 0
(4)
Solving for an+2 , gives
an+2 = −
=−
4n2 an − 12n2 an+1 − 4nan − 14nan+1 + an − 2an+1
8 (n + 2) (n + 1)
(4n2 − 4n + 1) an (−12n2 − 14n − 2) an+1
−
8 (n + 2) (n + 1)
8 (n + 2) (n + 1)
For n = 2 the recurrence equation gives
9a2 − 78a3 + 96a4 = 0
Which after substituting the earlier terms found becomes
a4 = −
73a0
47a1
+
3072 1536
For n = 3 the recurrence equation gives
25a3 − 152a4 + 160a5 = 0
Which after substituting the earlier terms found becomes
a5 = −
1037a0 643a1
+
61440
30720
For n = 4 the recurrence equation gives
49a4 − 250a5 + 240a6 = 0
Which after substituting the earlier terms found becomes
a6 = −
6257a0
3823a1
+
491520 245760
For n = 5 the recurrence equation gives
81a5 − 372a6 + 336a7 = 0
Which after substituting the earlier terms found becomes
a7 = −
137969a0
83791a1
+
13762560 6881280
And so on. Therefore the solution is
y=
∞
X
an tn
n=0
= a3 t3 + a2 t2 + a1 t + a0 + . . .
Substituting the values for an found above, the solution becomes
a
a1 2
7a0 5a1 3
0
y = a0 + a1 t + − +
t + −
+
t
16
8
192
96
73a0
47a1 4
1037a0 643a1 5
+ −
+
t + −
+
t + ...
3072 1536
61440
30720
(5)
chapter 2 . book solved problems
232
Collecting terms, the solution becomes
1 2
7 3
73 4
1037 5
y = 1− t −
t −
t −
t a0
16
192
3072
61440
1 2
5 3
47 4
643 5
+ t+ t + t +
t +
t a1 + O t 6
8
96
1536
30720
(3)
At t = 0 the solution above becomes
1 2
7 3
73 4
1037 5
y = 1− t −
t −
t −
t y(0)
16
192
3072
61440
1 2
5 3
47 4
643 5 0
+ t+ t + t +
t +
t y (0) + O t6
8
96
1536
30720
3 Maple. Time used: 0.007 (sec). Leaf size: 59
Order:=6;
ode:=4*(t^2-3*t+2)*diff(diff(y(t),t),t)-2*diff(y(t),t)+y(t) = 0;
dsolve(ode,y(t),type='series',t=0);
1 2
7 3
73 4
1037 5
y = 1− t −
t −
t −
t y(0)
16
192
3072
61440
1 2
5 3
47 4
643 5 0
+ t+ t + t +
t +
t y (0) + O t6
8
96
1536
30720
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful
Maple step by step
Let’s solve
4(t2 − 3t + 2) dtd dtd y(t) − 2 dtd y(t) + y(t) = 0
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Isolate 2nd derivative
•
•
d
y(t)
y(t)
d d
dt
y(t)
=
−
+
2
2
dt dt
4(t −3t+2)
2(t −3t+2)
•
Group terms with y(t) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(t)
d d
dt
y(t)
−
+ 4(t2y(t)
=0
2
dt dt
2(t −3t+2)
−3t+2)
◦
Check to see if t0 is a regular singular point
Define
functions
h
i
1
1
P2 (t) = − 2(t2 −3t+2)
, P3 (t) = 4(t2 −3t+2)
◦
(t − 1) · P2 (t) is analytic at t = 1
chapter 2 . book solved problems
((t − 1) · P2 (t))
t=1
233
= 12
◦
(t − 1)2 · P3 (t) is analytic at t = 1
(t − 1)2 · P3 (t)
=0
◦
t = 1is a regular singular point
Check to see if t0 is a regular singular point
t0 = 1
Multiply by denominators
(4t2 − 12t + 8) dtd dtd y(t) − 2 dtd y(t) + y(t) = 0
Change variables using t = u + 1 so that the regular singular point is at u = 0
d d
d
(4u2 − 4u) du
y(u) − 2 du
y(u) + y(u) = 0
du
Assume series solution for y(u)
∞
P
y(u) =
ak uk+r
t=1
•
•
•
k=0
◦
Rewrite ODE with series expansions
d
Convert du
y(u) to series expansion
∞
P
d
y(u) =
ak (k + r) uk+r−1
du
k=0
◦
◦
Shift index using k− >k + 1
∞
P
d
y(u)
=
ak+1 (k + 1 + r) uk+r
du
k=−1
d d
Convert um · du
y(u) to series expansion for m = 1..2
du
∞
P
d d
um · du
y(u)
=
ak (k + r) (k + r − 1) uk+r−2+m
du
k=0
◦
Shift index using k− >k + 2 − m
∞
P
d d
um · du
y(u)
=
ak+2−m (k + 2 − m + r) (k + 1 − m + r) uk+r
du
k=−2+m
Rewrite ODE with seriesexpansions
∞
P
2 k+r
−1+r
−2a0 r(−1 + 2r) u
+
−2ak+1 (k + 1 + r) (2k + 1 + 2r) + ak (2k + 2r − 1) u
=0
k=0
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
−2r(−1 + 2r) = 0
Values of r that satisfy the indicial equation
r ∈ 0, 12
Each term in the series must be 0, giving the recursion relation
ak (2k + 2r − 1)2 − 4 k + 12 + r ak+1 (k + 1 + r) = 0
Recursion relation that defines series solution to ODE
ak (2k+2r−1)2
ak+1 = 2(2k+1+2r)(k+1+r)
Recursion relation for r = 0
ak (2k−1)2
ak+1 = 2(2k+1)(k+1)
Solution
for r = 0
∞
P
ak (2k−1)2
k
y(u) =
ak u , ak+1 = 2(2k+1)(k+1)
k=0
•
Revert
the change of variables u = t − 1 ∞
P
ak (2k−1)2
y(t) =
ak (t − 1)k , ak+1 = 2(2k+1)(k+1)
•
Recursion relation for r = 12
2ak k2 ak+1 = (2k+2)
k+ 3
•
Solution
for r = 12
∞
P
2ak k2 k+ 12
y(u) =
ak u , ak+1 = (2k+2) k+ 3
k=0
2
k=0
2
•
Revert
the change of variables u = t − 1 ∞
1
P
2ak k2 y(t) =
ak (t − 1)k+ 2 , ak+1 = (2k+2)
k+ 3
•
Combine solutions and rename parameters
k=0
2
chapter 2 . book solved problems
y(t) =
∞
P
k=0
ak (t − 1)
k
+
∞
P
k=0
234
k+ 12
bk (t − 1)
ak (2k−1)2
2bk k2 , ak+1 = 2(2k+1)(k+1)
, bk+1 = (2k+2)
k+ 32
3 Mathematica. Time used: 0.001 (sec). Leaf size: 70
ode=4*(t^2-3*t+2)*D[y[t],{t,2}]-2*D[y[t],t]+y[t]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[t],{t,0,5}]
y(t) → c1
1037t5
73t4
7t3
t2
643t5
47t4
5t3 t2
−
−
−
−
+ 1 + c2
+
+
+ +t
61440
3072 192 16
30720 1536
96
8
3 Sympy. Time used: 0.344 (sec). Leaf size: 49
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq((4*t**2 - 12*t + 8)*Derivative(y(t), (t, 2)) + y(t) - 2*Derivative(y(t),
t),0)
ics = {}
dsolve(ode,func=y(t),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
73t4
7t3
t2
47t3
5t2
t
y(t) = C2 −
−
−
+ 1 + C1 t
+
+ + 1 + O t6
3072 192 16
1536
96
8
chapter 2 . book solved problems
2.2.17
235
Problem 19
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
Internal problem ID [8595]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 19
Date solved : Tuesday, November 11, 2025 at 12:03:58 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
2 t2 − 5t + 6 y 00 + (2t − 3) y 0 − 8y = 0
Using series expansion around t = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.30)
(2.31)
(2)
(3)
chapter 2 . book solved problems
236
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
237
To find y(x) series solution around x = 0. Hence
2y 0 t − 3y 0 − 8y
2 (t2 − 5t + 6)
dF0
F1 =
dt
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂t
∂y
∂y 0
(24t2 − 104t + 111) y 0 + (−48t + 104) y
=
4 (t2 − 5t + 6)2
dF1
F2 =
dt
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂t
∂y
∂y 0
37
13
23
0
30 t2 − 13
t
+
y
−
2y
t
−
−
+
t
3
8
6
10
=−
(t2 − 5t + 6)3
dF2
F3 =
dt
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂t
∂y
∂y 0
0
2 23
37
13
213
180 t2 − 13
t
+
y
−
2y
t
−
t
−
t
+
3
8
6
5
40
=
(t2 − 5t + 6)4
dF3
F4 =
dt
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂t
∂y
∂y 0
0
3 69 2 639
1260 t2 − 13
t + 37
y − 2y t − 13
t − 10 t + 40 t − 993
3
8
6
80
=−
(t2 − 5t + 6)5
F0 = −
And so on. Evaluating all the above at initial conditions t = 0 and y(0) = y(0) and y 0 (0) =
y 0 (0) gives
2y(0) y 0 (0)
+
3
4
13y(0) 37y 0 (0)
F1 =
+
18
48
299y(0) 851y 0 (0)
F2 =
+
216
576
923y(0) 2627y 0 (0)
F3 =
+
288
768
30121y(0) 85729y 0 (0)
F4 =
+
3456
9216
F0 =
Substituting all the above in (7) and simplifying gives the solution as
1 2
13 3
299 4
923 5
y = 1+ t +
t +
t +
t y(0)
3
108
5184
34560
1 2
37 3
851 4
2627 5 0
+ t+ t +
t +
t +
t y (0) + O t6
8
288
13824
92160
Since the expansion point t = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
2t2 − 10t + 12 y 00 + (2t − 3) y 0 − 8y = 0
Let the solution be represented as power series of the form
y=
∞
X
n=0
an tn
chapter 2 . book solved problems
238
Then
0
y =
∞
X
nan tn−1
n=1
y 00 =
∞
X
n(n − 1) an tn−2
n=2
Substituting the above back into the ode gives
!
!
!
∞
∞
∞
X
X
X
2t2 − 10t + 12
n(n − 1) an tn−2 + (2t − 3)
nan tn−1 − 8
an tn = 0 (1)
n=2
n=1
n=0
Which simplifies to
∞
X
!
n
2t an n(n − 1)
+
n=2
+
∞
X
!
2nan tn
+
n=1
∞
X
n =2
∞
X
−10n t
n−1
an (n − 1) +
∞
X
!
n−2
12n(n − 1) an t
(2)
n=2
∞
X
−3nan tn−1 +
(−8an tn ) = 0
n =1
n =0
The next step is to make all powers of t be n in each summation term. Going over each
summation term above with power of t in it which is not already tn and adjusting the power
and the corresponding index gives
∞
X
−10n t
n−1
∞
X
an (n − 1) =
(−10(n + 1) an+1 n tn )
n =2
n=1
∞
X
12n(n − 1) an t
n−2
n =2
=
∞
X
12(n + 2) an+2 (n + 1) tn
n=0
∞
X
∞
X
−3nan tn−1 =
(−3(n + 1) an+1 tn )
n =1
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of t
are the same and equal to n.
∞
X
!
2tn an n(n − 1) +
n=2
+
∞
X
n=1
!
2nan tn
+
∞
X
n =1
∞
X
∞
X
(−10(n + 1) an+1 n tn ) +
!
12(n + 2) an+2 (n + 1) tn
n=0
(−3(n + 1) an+1 tn ) +
n =0
∞
X
(−8an tn ) = 0
n =0
n = 0 gives
24a2 − 3a1 − 8a0 = 0
a2 =
a0 a1
+
3
8
n = 1 gives
−26a2 + 72a3 − 6a1 = 0
Which after substituting earlier equations, simplifies to
a3 =
13a0 37a1
+
108
288
(3)
chapter 2 . book solved problems
239
For 2 ≤ n, the recurrence equation is
2nan (n − 1) − 10(n + 1) an+1 n + 12(n + 2) an+2 (n + 1) + 2nan − 3(n + 1) an+1 − 8an = 0 (4)
Solving for an+2 , gives
an+2 = −
=−
2n2 an − 10n2 an+1 − 13nan+1 − 8an − 3an+1
12 (n + 2) (n + 1)
(2n2 − 8) an
(−10n2 − 13n − 3) an+1
−
12 (n + 2) (n + 1)
12 (n + 2) (n + 1)
For n = 2 the recurrence equation gives
−69a3 + 144a4 = 0
Which after substituting the earlier terms found becomes
a4 =
299a0 851a1
+
5184
13824
For n = 3 the recurrence equation gives
10a3 − 132a4 + 240a5 = 0
Which after substituting the earlier terms found becomes
a5 =
923a0 2627a1
+
34560
92160
For n = 4 the recurrence equation gives
24a4 − 215a5 + 360a6 = 0
Which after substituting the earlier terms found becomes
a6 =
30121a0
85729a1
+
2488320 6635520
For n = 5 the recurrence equation gives
42a5 − 318a6 + 504a7 = 0
Which after substituting the earlier terms found becomes
a7 =
161603a0
459947a1
+
29859840 79626240
And so on. Therefore the solution is
y=
∞
X
an tn
n=0
= a3 t3 + a2 t2 + a1 t + a0 + . . .
Substituting the values for an found above, the solution becomes
a1 2
13a0 37a1 3
y = a0 + a1 t +
+
t +
+
t
3
8
108
288
299a0 851a1 4
923a0 2627a1 5
+
+
t +
+
t + ...
5184
13824
34560
92160
a
0
(5)
chapter 2 . book solved problems
240
Collecting terms, the solution becomes
1 2
13 3
299 4
923 5
y = 1+ t +
t +
t +
t a0
3
108
5184
34560
1 2
37 3
851 4
2627 5
+ t+ t +
t +
t +
t a1 + O t6
8
288
13824
92160
(3)
At t = 0 the solution above becomes
1 2
13 3
299 4
923 5
y = 1+ t +
t +
t +
t y(0)
3
108
5184
34560
1 2
37 3
851 4
2627 5 0
+ t+ t +
t +
t +
t y (0) + O t6
8
288
13824
92160
3 Maple. Time used: 0.007 (sec). Leaf size: 59
Order:=6;
ode:=2*(t^2-5*t+6)*diff(diff(y(t),t),t)+(2*t-3)*diff(y(t),t)-8*y(t) = 0;
dsolve(ode,y(t),type='series',t=0);
1 2
13 3
299 4
923 5
y = 1+ t +
t +
t +
t y(0)
3
108
5184
34560
1 2
37 3
851 4
2627 5 0
+ t+ t +
t +
t +
t y (0) + O t6
8
288
13824
92160
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Reducible group (found another exponential solution)
<- Kovacics algorithm successful
Maple step by step
Let’s solve
2(t2 − 5t + 6) dtd dtd y(t) + (2t − 3) dtd y(t) − 8y(t) = 0
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Isolate 2nd derivative
•
•
d
(2t−3) dt
y(t)
4y(t)
d d
y(t)
=
−
dt dt
t2 −5t+6
2(t2 −5t+6)
•
Group terms with
y(t)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
− t24y(t)
=0
−5t+6
Check to see if t0 is a regular singular point
Define
functions
h
i
d d
y(t) +
dt dt
◦
(2t−3)
d
y(t)
dt
2(t2 −5t+6)
4
P2 (t) = 2(t22t−3
, P3 (t) = − t2 −5t+6
−5t+6)
chapter 2 . book solved problems
◦
241
(t − 2) · P2 (t) is analytic at t = 2
((t − 2) · P2 (t))
t=2
= − 12
◦
(t − 2)2 · P3 (t) is analytic at t = 2
(t − 2)2 · P3 (t)
=0
◦
t = 2is a regular singular point
Check to see if t0 is a regular singular point
t0 = 2
Multiply by denominators
(2t2 − 10t + 12) dtd dtd y(t) + (2t − 3) dtd y(t) − 8y(t) = 0
Change variables using t = u + 2 so that the regular singular point is at u = 0
d d
d
(2u2 − 2u) du
y(u) + (2u + 1) du
y(u) − 8y(u) = 0
du
Assume series solution for y(u)
∞
P
y(u) =
ak uk+r
t=2
•
•
•
k=0
◦
Rewrite ODE with series expansions
d
Convert um · du
y(u) to series expansion for m = 0..1
∞
P
d
um · du
y(u) =
ak (k + r) uk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
um · du
y(u) =
ak+1−m (k + 1 − m + r) uk+r
k=−1+m
d d
Convert um · du
y(u) to series expansion for m = 1..2
du
∞
P
d d
um · du
y(u) =
ak (k + r) (k + r − 1) uk+r−2+m
du
k=0
◦
Shift index using k− >k + 2 − m
∞
P
d d
um · du
y(u)
=
ak+2−m (k + 2 − m + r) (k + 1 − m + r) uk+r
du
k=−2+m
Rewrite ODE with series
expansions
∞
P
−a0 r(−3 + 2r) u−1+r +
(−ak+1 (k + 1 + r) (2k − 1 + 2r) + 2ak (k + r + 2) (k + r − 2)) uk+r
k=0
•
•
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
−r(−3 + 2r) = 0
Values of r that satisfy the indicial equation
r ∈ 0, 32
Each term in the series must be 0, giving the recursion relation
−2(k + 1 + r) k − 12 + r ak+1 + 2ak (k + r + 2) (k + r − 2) = 0
Recursion relation that defines series solution to ODE
k (k+r+2)(k+r−2)
ak+1 = 2a
(k+1+r)(2k−1+2r)
Recursion relation for r = 0 ; series terminates at k = 2
k (k+2)(k−2)
ak+1 = 2a(k+1)(2k−1)
Apply recursion relation for k = 0
a1 = 8a0
Apply recursion relation for k = 1
a2 = −3a1
Express in terms of a0
a2 = −24a0
Terminating series solution of the ODE for r = 0 . Use reduction of order to find the second linea
y(u) = a0 · (−24u2 + 8u + 1)
Revert the change of variables u = t − 2
[y(t) = a0 (−24t2 + 104t − 111)]
Recursion relation
for
r = 32
2ak k+ 72 k− 12
k+ 52 (2k+2)
Solution for r = 32
ak+1 =
•
chapter 2 . book solved problems
∞
P
y(u) =
k=0
k+ 32
ak u
242
2ak k+ 72 k− 12
, ak+1 = k+ 5 (2k+2)
2
•
Revert
the change of variables u = t − 2
∞
P
2ak k+ 72 k− 12
k+ 32
y(t) =
ak (t − 2)
, ak+1 = k+ 5 (2k+2)
•
Combine
solutions and rename parameters
∞
P
2bk k+ 72 k− 12
k+ 32
2
y(t) = a0 (−24t + 104t − 111) +
bk (t − 2)
, bk+1 = k+ 5 (2k+2)
k=0
2
k=0
2
3 Mathematica. Time used: 0.002 (sec). Leaf size: 70
ode=2*(t^2-5*t+6)*D[y[t],{t,2}]+(2*t-3)*D[y[t],t]-8*y[t]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[t],{t,0,5}]
y(t) → c1
923t5
299t4 13t3 t2
2627t5
851t4
37t3 t2
+
+
+ + 1 + c2
+
+
+ +t
34560
5184
108
3
92160
13824
288
8
3 Sympy. Time used: 0.367 (sec). Leaf size: 49
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq((2*t - 3)*Derivative(y(t), t) + (2*t**2 - 10*t + 12)*Derivative(y(t), (t,
2)) - 8*y(t),0)
ics = {}
dsolve(ode,func=y(t),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
y(t) = C2
299t4 13t3 t2
851t3
37t2
t
+
+ + 1 + C1 t
+
+ + 1 + O t6
5184
108
3
13824
288
8
chapter 2 . book solved problems
2.2.18
243
Problem 20
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
Internal problem ID [8596]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number : 20
Date solved : Tuesday, November 11, 2025 at 12:03:58 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
3t(1 + t) y 00 + y 0 t − y = 0
Using series expansion around t = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
3t2 + 3t y 00 + y 0 t − y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(t)y 0 + q(t)y = 0
Where
1
3 + 3t
1
q(t) = −
3t (1 + t)
p(t) =
Table 2.42: Table p(t), q(t) singularites.
1
q(t) = − 3t(1+t)
singularity
type
t = −1
“regular”
t=0
“regular”
1
p(t) = 3+3t
singularity
type
t = −1
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [−1, 0, ∞]
Irregular singular points : []
Since t = 0 is regular singular point, then Frobenius power series is used. The ode is normalized
to be
3t(1 + t) y 00 + y 0 t − y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an tn+r
chapter 2 . book solved problems
244
Then
0
y =
∞
X
(n + r) an tn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an tn+r−2
n=0
Substituting the above back into the ode gives
3t(1 + t)
∞
X
!
(n + r) (n + r − 1) an tn+r−2
+
n=0
∞
X
!
(n + r) an tn+r−1 t −
n=0
∞
X
!
an tn+r
=0
n=0
(1)
Which simplifies to
∞
X
!
3tn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
3tn+r−1 an (n + r) (n + r − 1)
(2A)
n=0
∞
X
!
tn+r an (n + r)
+
n=0
∞
X
−an tn+r = 0
n =0
The next step is to make all powers of t be n + r − 1 in each summation term. Going over
each summation term above with power of t in it which is not already tn+r−1 and adjusting
the power and the corresponding index gives
∞
X
n+r
3t
an (n + r) (n + r − 1) =
n =0
∞
X
3an−1 (n + r − 1) (n + r − 2) tn+r−1
n=1
∞
X
n+r
t
an (n + r) =
n =0
∞
X
∞
X
an−1 (n + r − 1) tn+r−1
n=1
n+r
−an t
n =0
=
∞
X
−an−1 tn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of t
are the same and equal to n + r − 1.
∞
X
!
3an−1 (n + r − 1) (n + r − 2) tn+r−1
n=1
+
+
∞
X
!
3tn+r−1 an (n + r) (n + r − 1)
n=0
∞
X
!
an−1 (n + r − 1) tn+r−1
n=1
+
∞
X
−an−1 tn+r−1 = 0
n =1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
3tn+r−1 an (n + r) (n + r − 1) = 0
When n = 0 the above becomes
3t−1+r a0 r(−1 + r) = 0
Or
3t−1+r a0 r(−1 + r) = 0
(2B)
chapter 2 . book solved problems
245
Since a0 6= 0 then the above simplifies to
3t−1+r r(−1 + r) = 0
Since the above is true for all t then the indicial equation becomes
3r(−1 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 1
r2 = 0
Since a0 6= 0 then the indicial equation becomes
3t−1+r r(−1 + r) = 0
Solving for r gives the roots of the indicial equation as [1, 0].
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (t) = tr1
an t n
n=0
y2 (t) = Cy1 (t) ln (t) + tr2
∞
X
!
bn t n
n=0
Or
y1 (t) = t
∞
X
!
an tn
n=0
y2 (t) = Cy1 (t) ln (t) +
∞
X
!
bn t n
n=0
Or
y1 (t) =
∞
X
an tn+1
n=0
y2 (t) = Cy1 (t) ln (t) +
∞
X
!
bn t n
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. For 1 ≤ n the recursive equation is
3an−1 (n + r − 1) (n + r − 2) + 3an (n + r) (n + r − 1) + an−1 (n + r − 1) − an−1 = 0 (3)
Solving for an from recursive equation (4) gives
an = −
an−1 (3n2 + 6nr + 3r2 − 8n − 8r + 4)
3 (n + r) (n + r − 1)
(4)
Which for the root r = 1 becomes
an = −
an−1 (3n2 − 2n − 1)
3 (n + 1) n
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 1
and after as more terms are found using the above recursive equation.
n
a0
an,r
1
an
1
chapter 2 . book solved problems
246
For n = 1, using the above recursive equation gives
−3r2 + 2r + 1
a1 =
3 (1 + r) r
Which for the root r = 1 becomes
a1 = 0
And the table now becomes
n
a0
a1
an,r
1
−3r2 +2r+1
3(1+r)r
an
1
0
For n = 2, using the above recursive equation gives
a2 =
9r3 + 6r2 − 11r − 4
9 (1 + r)2 (2 + r)
Which for the root r = 1 becomes
a2 = 0
And the table now becomes
n
a0
a1
a2
an,r
1
−3r2 +2r+1
3(1+r)r
9r3 +6r2 −11r−4
9(1+r)2 (2+r)
an
1
0
0
For n = 3, using the above recursive equation gives
a3 =
−27r4 − 81r3 − 9r2 + 89r + 28
27 (3 + r) (2 + r)2 (1 + r)
Which for the root r = 1 becomes
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
−3r2 +2r+1
3(1+r)r
9r3 +6r2 −11r−4
9(1+r)2 (2+r)
−27r4 −81r3 −9r2 +89r+28
27(3+r)(2+r)2 (1+r)
an
1
0
0
0
For n = 4, using the above recursive equation gives
a4 =
81r5 + 513r4 + 837r3 − 177r2 − 974r − 280
81 (4 + r) (1 + r) (2 + r) (3 + r)2
Which for the root r = 1 becomes
a4 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
−3r2 +2r+1
3(1+r)r
9r3 +6r2 −11r−4
9(1+r)2 (2+r)
−27r4 −81r3 −9r2 +89r+28
27(3+r)(2+r)2 (1+r)
81r5 +513r4 +837r3 −177r2 −974r−280
81(4+r)(1+r)(2+r)(3+r)2
an
1
0
0
0
0
chapter 2 . book solved problems
247
For n = 5, using the above recursive equation gives
a5 =
−243r6 − 2592r5 − 9180r4 − 10350r3 + 5223r2 + 13502r + 3640
243 (5 + r) (3 + r) (2 + r) (1 + r) (4 + r)2
Which for the root r = 1 becomes
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
−3r2 +2r+1
3(1+r)r
9r3 +6r2 −11r−4
9(1+r)2 (2+r)
−27r4 −81r3 −9r2 +89r+28
27(3+r)(2+r)2 (1+r)
81r5 +513r4 +837r3 −177r2 −974r−280
81(4+r)(1+r)(2+r)(3+r)2
−243r6 −2592r5 −9180r4 −10350r3 +5223r2 +13502r+3640
243(5+r)(3+r)(2+r)(1+r)(4+r)2
an
1
0
0
0
0
0
Using the above table, then the solution y1 (t) is
y1 (t) = t a0 + a1 t + a2 t2 + a3 t3 + a4 t4 + a5 t5 + a6 t6 . . .
= t 1 + O t6
Now the second solution y2 (t) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=
−3r2 + 2r + 1
3 (1 + r) r
Therefore
−3r2 + 2r + 1
−3r2 + 2r + 1
= lim
r→r2
r→0
3 (1 + r) r
3 (1 + r) r
lim
= undefined
Since the limit does not exist then the log term is needed. Therefore the second solution has
the form
!
∞
X
y2 (t) = Cy1 (t) ln (t) +
bn tn+r2
n=0
Therefore
d
Cy1 (t)
y2 (t) = Cy10 (t) ln (t) +
+
dt
t
Cy1 (t)
= Cy10 (t) ln (t) +
+
t
∞
X
bn tn+r2 (n + r2 )
t
n=0
∞
X
!
!
−1+n+r2
t
bn (n + r2 )
n=0
∞
d2
2Cy10 (t) Cy1 (t) X
00
y2 (t) = Cy1 (t) ln (t) +
−
+
dt2
t
t2
n=0
2Cy10 (t) Cy1 (t)
= Cy100 (t) ln (t) +
−
+
t
t2
∞
X
n=0
bn tn+r2 (n + r2 )2 bn tn+r2 (n + r2 )
−
t2
t2
!
t−2+n+r2 bn (n + r2 ) (−1 + n + r2 )
!
chapter 2 . book solved problems
248
Substituting these back into the given ode (3t2 + 3t) y 00 + y 0 t − y = 0 gives
∞
2Cy10 (t) Cy1 (t) X
00
3 Cy1 (t) ln (t) +
−
+
t
t2
n=0
+ 3 Cy100 (t) ln (t) +
+
2Cy10 (t)
t
−
Cy1 (t)
Cy10 (t) ln (t) +
+
t
− Cy1 (t) ln (t) −
∞
X
bn tn+r2
bn tn+r2 (n + r2 )2 bn tn+r2 (n + r2 )
−
t2
t2
∞
X
Cy1 (t)
+
t2
n=0
bn t
∞
X
bn tn+r2 (n + r2 )
n=0
!
t
n+r2
2
n+r2
(n + r2 )
bn t
−
2
t
!!
(n + r2 )
t2
t2
!!
t
!!
t
=0
n=0
Which can be written as
3y100 (t) t2 + 3y100 (t) t + y10 (t) t − y1 (t) ln (t)
0
0
2y1 (t) y1 (t) 2
2y1 (t) y1 (t)
+3
− 2
t +3
− 2
t + y1 (t) C
t
t
t
t
!!
∞
X
bn tn+r2 (n + r2 )2 bn tn+r2 (n + r2 )
+3
−
t2
2
2
t
t
n=0
!!
∞
X
bn tn+r2 (n + r2 )2 bn tn+r2 (n + r2 )
+3
−
t
t2
t2
n=0
!
!
∞
∞
X
X
bn tn+r2 (n + r2 )
+
t−
bn tn+r2 = 0
t
n=0
n=0
(7)
But since y1 (t) is a solution to the ode, then
3t2 + 3t y100 (t) + y10 (t) t − y1 (t) = 0
Eq (7) simplifes to
3y100 (t) t2 + 3y100 (t) t + y10 (t) t − y1 (t) ln (t)
0
0
2y1 (t) y1 (t) 2
2y1 (t) y1 (t)
+3
− 2
t +3
− 2
t + y1 (t) C
t
t
t
t
!!
∞
X
bn tn+r2 (n + r2 )2 bn tn+r2 (n + r2 )
+3
−
t2
2
2
t
t
n=0
!!
∞
X
bn tn+r2 (n + r2 )2 bn tn+r2 (n + r2 )
+3
−
t
t2
t2
n=0
!
!
∞
∞
X
X
bn tn+r2 (n + r2 )
n+r2
+
t−
bn t
=0
t
n=0
n=0
Substituting y1 =
∞
P
(8)
an tn+r1 into the above gives
n=0
∞
∞
P −2+n+r1
P −1+n+r1
2
3t ln (t) (1 + t)
t
an (n + r1 ) (−1 + n + r1 ) + t(t ln (t) + 6t + 6)
t
an (n + r1 ) −
n=0
n=0
(9)
t
∞
∞
∞
P −2+n+r2
P −1+n+r2
P
3
2
2
n+r2
(3t + 3t )
t
bn (n + r2 ) (−1 + n + r2 ) +
t
bn (n + r2 ) t −
bn t
t
n=0
n=0
n=0
+
t
=0
chapter 2 . book solved problems
249
Since r1 = 1 and r2 = 0 then the above becomes
∞
∞
∞
P n−1
P n
P
2
3t ln (t) (1 + t)
t an (n + 1) n + t(t ln (t) + 6t + 6)
t an (n + 1) − (t ln (t) + 2t + 3)
an
n=0
n=0
n=0
(10)
t
∞
∞
∞
P −2+n
P n−1
P
(3t3 + 3t2 )
t
bn n(n − 1) +
t bn n t2 −
bn tn t
n=0
n=0
n=0
+
=0
t
Which simplifies to
∞
X
!
3tn an n(n + 1) C ln (t)
n=0
∞
X
+
+
+
3Cn tn+1 an ln (t) (n + 1)
n=0
!
6tn an (n + 1) C
n=0
∞
X
−2C t
+
+
!
! n=0 ∞
!
X
C tn+1 an ln (t) (n + 1) +
6C tn+1 an (n + 1)
n=0
∞
X
∞
X
n =0
∞
X
n+1
∞
X
+
an +
n =0
∞
X
n
(−3an t C) +
n =0
!
3n tn−1 bn (n − 1)
+
n=0
(2A)
−C tn+1 an ln (t)
∞
X
tn bn n
∞
X
!
n
3t bn n(n − 1)
! n=0∞
X
+
n=0
(−bn tn ) = 0
n =0
The next step is to make all powers of t be n − 1 in each summation term. Going over each
summation term above with power of t in it which is not already tn−1 and adjusting the
power and the corresponding index gives
∞
X
3tn an n(n + 1) C ln (t) =
n =0
∞
X
∞
X
3C(n − 1) an−1 n ln (t) tn−1
n=1
3Cn t
n+1
an ln (t) (n + 1) =
n =0
∞
X
3C(−2 + n) a−2+n (n − 1) ln (t) tn−1
n=2
∞
X
Ct
n+1
an ln (t) (n + 1) =
n =0
∞
X
Ca−2+n (n − 1) ln (t) tn−1
n=2
∞
X
6C t
n+1
an (n + 1) =
n =0
6Ca−2+n (n − 1) tn−1
n=2
∞
X
n
6t an (n + 1) C =
n =0
∞
X
∞
X
∞
X
6Can−1 n tn−1
n=1
−C t
n+1
∞
X
an ln (t) =
−Ca−2+n ln (t) tn−1
n =0
n=2
∞
X
−2C t
n+1
an =
n =0
n
(−3an t C) =
n =0
n =0
−2Ca−2+n tn−1
n=2
∞
X
∞
X
∞
X
n
3t bn n(n − 1) =
∞
X
−3Can−1 tn−1
n=1
∞
X
n=1
3bn−1 (n − 1) (−2 + n) tn−1
chapter 2 . book solved problems
∞
X
∞
X
n
t bn n =
n =0
∞
X
250
bn−1 (n − 1) tn−1
n=1
∞
X
n
(−bn t ) =
n =0
−bn−1 tn−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of t
are the same and equal to n − 1.
∞
X
!
3C(n−1) an−1 n ln (t) tn−1 +
n=1
+
+
∞
X
n=2
∞
X
!
3C(−2+n) a−2+n (n−1) ln (t) tn−1
! n=2 ∞
!
X
n−1
n−1
Ca−2+n (n − 1) ln (t) t
+
6Ca−2+n (n − 1) t
n=2
!
6Can−1 n tn−1
n=1
∞
X
+
+
∞
X
n =1
∞
X
+
∞
X
∞
X
−Ca−2+n ln (t) tn−1 +
−2Ca−2+n tn−1
n =2
−3Can−1 tn−1 +
∞
X
(2B)
n =2
!
3bn−1 (n − 1) (−2 + n) tn−1
! n=1 ∞
! ∞
X
X
3n tn−1 bn (n − 1) +
bn−1 (n − 1) tn−1 +
−bn−1 tn−1 = 0
n=0
n=1
n =1
For n = 0 in Eq. (2B), we choose arbitray value for b0 as b0 = 1. For n = N , where N = 1
which is the difference between the two roots, we are free to choose b1 = 0. Hence for n = 1,
Eq (2B) gives
3C − 1 = 0
Which is solved for C. Solving for C gives
C=
1
3
For n = 2, Eq (2B) gives
6Ca1 ln (t) + (4a0 + 9a1 ) C + 6b2 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
4
+ 6b2 = 0
3
Solving the above for b2 gives
2
b2 = −
9
For n = 3, Eq (2B) gives
18a2
7C a1 +
ln (t) + (10a1 + 15a2 ) C + 7b2 + 18b3 = 0
7
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
14
− + 18b3 = 0
9
Solving the above for b3 gives
7
b3 =
81
For n = 4, Eq (2B) gives
9a3
20C a2 +
ln (t) + (16a2 + 21a3 ) C + 20b3 + 36b4 = 0
5
chapter 2 . book solved problems
251
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
140
+ 36b4 = 0
81
Solving the above for b4 gives
35
b4 = −
729
For n = 5, Eq (2B) gives
20a4
39 a3 +
C ln (t) + (22a3 + 27a4 ) C + 39b4 + 60b5 = 0
13
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
455
−
+ 60b5 = 0
243
Solving the above for b5 gives
91
b5 =
2916
Now that we found all bn and C, we can calculate the second solution from
!
∞
X
y2 (t) = Cy1 (t) ln (t) +
bn tn+r2
n=0
Using the above value found for C = 13 and all bn , then the second solution becomes
1
2t2 7t3 35t4
91t5
t 1 + O t6
ln (t) + 1 −
+
−
+
+ O t6
3
9
81
729
2916
y2 (t) =
Therefore the homogeneous solution is
yh (t) = c1 y1 (t) + c2 y2 (t)
6
= c1 t 1 + O t
1
2t2 7t3 35t4
91t5
6
6
t 1+O t
ln (t) + 1 −
+
−
+
+O t
3
9
81
729
2916
t(1 + O(t6 )) ln (t)
2t2 7t3 35t4
91t5
6
+1−
+
−
+
+O t
3
9
81
729
2916
+ c2
Hence the final solution is
y = yh
6
= c1 t 1 + O t
+ c2
3 Maple. Time used: 0.024 (sec). Leaf size: 42
Order:=6;
ode:=3*t*(1+t)*diff(diff(y(t),t),t)+diff(y(t),t)*t-y(t) = 0;
dsolve(ode,y(t),type='series',t=0);
6
y = c1 t 1 + O t
+ ln (t)
1
1
2
7
35 4
91 5
6
t+O t
c2 + 1 − t − t2 + t3 −
t +
t
3
3
9
81
729
2916
6
+O t
c2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
chapter 2 . book solved problems
252
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
Solution has integrals. Trying a special function solution free of integrals\
...
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Whittaker
-> hyper3: Equivalence to 1F1 under a power @ Moebius
-> hypergeometric
-> heuristic approach
<- heuristic approach successful
<- hypergeometric successful
<- special function solution successful
-> Trying to convert hypergeometric functions to elementary form...
<- elementary form for at least one hypergeometric solution is achieved -\
returning with no uncomputed integrals
<- Kovacics algorithm successful
Maple step by step
Let’s solve
3t(1 + t) dtd dtd y(t) + t dtd y(t) − y(t) = 0
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Isolate 2nd derivative
•
•
d
y(t)
y(t)
d d
dt
y(t)
=
−
dt dt
3t(1+t)
3(1+t)
•
Group terms with y(t) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(t)
y(t)
d d
dt
y(t)
+
− 3t(1+t)
=0
dt dt
3(1+t)
◦
Check to see if t0 is a regular singular point
Define
functions
h
i
1
1
P2 (t) = 3(1+t)
, P3 (t) = − 3t(1+t)
◦
(1 + t) · P2 (t) is analytic at t = −1
((1 + t) · P2 (t))
◦
t=−1
= 13
(1 + t)2 · P3 (t) is analytic at t = −1
(1 + t)2 · P3 (t)
=0
t=−1
◦
•
•
•
t = −1is a regular singular point
Check to see if t0 is a regular singular point
t0 = −1
Multiply by denominators
3t(1 + t) dtd dtd y(t) + t dtd y(t) − y(t) = 0
Change variables using t = u − 1 so that the regular singular point is at u = 0
d d
d
(3u2 − 3u) du
y(u) + (u − 1) du
y(u) − y(u) = 0
du
Assume series solution for y(u)
∞
P
y(u) =
ak uk+r
k=0
chapter 2 . book solved problems
◦
253
Rewrite ODE with series expansions
d
Convert um · du
y(u) to series expansion for m = 0..1
∞
P
d
um · du
y(u) =
ak (k + r) uk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
um · du
y(u) =
ak+1−m (k + 1 − m + r) uk+r
k=−1+m
d d
Convert um · du
y(u) to series expansion for m = 1..2
du
∞
P
d d
um · du
y(u) =
ak (k + r) (k + r − 1) uk+r−2+m
du
k=0
◦
Shift index using k− >k + 2 − m
∞
P
d d
um · du
y(u)
=
ak+2−m (k + 2 − m + r) (k + 1 − m + r) uk+r
du
k=−2+m
Rewrite ODE with series
expansions
∞
P
−a0 r(−2 + 3r) u−1+r +
(−ak+1 (k + 1 + r) (3k + 3r + 1) + ak (3k + 3r + 1) (k + r − 1)) uk+r
k=0
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
−r(−2 + 3r) = 0
Values of r that satisfy the indicial equation
r ∈ 0, 23
Each term in the series must be 0, giving the recursion relation
3((−k − r − 1) ak+1 + ak (k + r − 1)) k + r + 13 = 0
Recursion relation that defines series solution to ODE
(k+r−1)
ak+1 = akk+1+r
Recursion relation for r = 0 ; series terminates at k = 1
(k−1)
ak+1 = akk+1
Apply recursion relation for k = 0
a1 = −a0
Terminating series solution of the ODE for r = 0 . Use reduction of order to find the second linea
y(u) = a0 · (1 − u)
Revert the change of variables u = 1 + t
[y(t) = −a0 t]
Recursion relation
for r = 23
ak+1 =
•
ak k− 13
k+ 53
Solution
for r = 23
∞
P
ak k− 13
k+ 23
y(u) =
ak u , ak+1 = k+ 5
3
k=0
•
Revert
the change of variables u = 1 + t
∞
P
ak k− 13
k+ 23
y(t) =
ak (1 + t)
, ak+1 = k+ 5
•
Combine
solutions
∞and rename parameters
P
bk k− 13
k+ 23
y(t) = −a0 t +
bk (1 + t)
, bk+1 = k+ 5
3
k=0
k=0
3
3 Mathematica. Time used: 0.032 (sec). Leaf size: 43
ode=3*t*(1+t)*D[y[t],{t,2}]+t*D[y[t],t]-y[t]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[t],{t,0,5}]
y(t) → c1
1
1
4
3
2
−35t + 63t − 162t + 243t + 729 + t log(t) + c2 t
729
3
chapter 2 . book solved problems
254
3 Sympy. Time used: 0.309 (sec). Leaf size: 29
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(3*t*(t + 1)*Derivative(y(t), (t, 2)) + t*Derivative(y(t), t) - y(t),0)
ics = {}
dsolve(ode,func=y(t),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
4
t
t3
t2
t
y(t) = C1 t
+
+
+ + 1 + O t6
2880 144 12 2
chapter 2 . book solved problems
2.3
255
Chapter 5. Series Solutions of ODEs. Special
Functions. Problem set 5.4. Bessels Equation page
195
Local contents
2.3.1
2.3.2
2.3.3
2.3.4
2.3.5
2.3.6
Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
256
264
271
279
288
297
chapter 2 . book solved problems
2.3.1
256
Problem 2
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
Internal problem ID [8597]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.4. Bessels
Equation page 195
Problem number : 2
Date solved : Tuesday, November 11, 2025 at 12:03:59 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
4
2
x y +yx+ x −
y=0
49
2 00
0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
4
2 00
0
2
x y +yx+ x −
y=0
49
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
49x2 − 4
q(x) =
49x2
p(x) =
Table 2.44: Table p(x), q(x) singularites.
2
−4
q(x) = 49x
49x2
singularity
type
x=0
“regular”
p(x) = x1
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
4
2 00
0
2
x y +yx+ x −
y=0
49
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
257
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
x
∞
X
2
+
!
(n + r) (n + r − 1) an x
n=0
∞
X
!
(n + r) an xn+r−1
n=0
n+r−2
(1)
!
X
∞
4
x + x2 −
an xn+r = 0
49
n=0
Which simplifies to
∞
X
∞
X
!
x
n+r
an (n + r) (n + r − 1)
+
n=0
+
!
x
n+r
an (n + r)
(2A)
n=0
∞
X
!
xn+r+2 an
n=0
∞ X
4an xn+r
+
−
=0
49
n =0
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives
∞
X
xn+r+2 an =
n =0
∞
X
an−2 xn+r
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r.
∞
X
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
+
!
xn+r an (n + r)
n=0
∞
X
!
an−2 xn+r
n=2
+
∞ X
n =0
n+r
−
4an x
49
=0
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r an (n + r) (n + r − 1) + xn+r an (n + r) −
4an xn+r
=0
49
When n = 0 the above becomes
xr a0 r(−1 + r) + xr a0 r −
Or
4a0 xr
=0
49
4xr
r
r
x r(−1 + r) + x r −
a0 = 0
49
Since a0 6= 0 then the above simplifies to
(49r2 − 4) xr
=0
49
(2B)
chapter 2 . book solved problems
258
Since the above is true for all x then the indicial equation becomes
r2 −
4
=0
49
Solving for r gives the roots of the indicial equation as
r1 =
2
7
r2 = −
2
7
Since a0 6= 0 then the indicial equation becomes
(49r2 − 4) xr
=0
49
Solving for r gives the roots of the indicial equation as
2
7
, − 27 .
Since r1 − r2 = 47 is not an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
y2 (x) = xr2
n=0
∞
X
!
bn x n
n=0
Or
∞
X
y1 (x) =
n=0
∞
X
y2 (x) =
2
an xn+ 7
2
bn xn− 7
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + an (n + r) + an−2 −
4an
=0
49
(3)
Solving for an from recursive equation (4) gives
an = −
49an−2
2
49n + 98nr + 49r2 − 4
(4)
7an−2
n (7n + 4)
(5)
Which for the root r = 27 becomes
an = −
At this point, it is a good idea to keep track of an in a table both before substituting r = 27
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
a2 = −
49
49r2 + 196r + 192
chapter 2 . book solved problems
259
Which for the root r = 27 becomes
a2 = −
7
36
And the table now becomes
n
a0
a1
a2
an,r
1
0
49
− 49r2 +196r+192
an
1
0
7
− 36
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
49
− 49r2 +196r+192
0
an
1
0
7
− 36
0
For n = 4, using the above recursive equation gives
a4 =
2401
(49r2 + 196r + 192) (49r2 + 392r + 780)
Which for the root r = 27 becomes
a4 =
49
4608
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
49
− 49r2 +196r+192
0
an
1
0
7
− 36
0
2401
(49r2 +196r+192)(49r2 +392r+780)
49
4608
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
49
− 49r2 +196r+192
0
an
1
0
7
− 36
0
2401
(49r2 +196r+192)(49r2 +392r+780)
49
4608
0
0
Using the above table, then the solution y1 (x) is
y1 (x) = x2/7 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
7x2 49x4
2/7
6
=x
1−
+
+O x
36
4608
chapter 2 . book solved problems
260
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. Substituting n = 1 in Eq. (2B) gives
b1 = 0
For 2 ≤ n the recursive equation is
bn (n + r) (n + r − 1) + bn (n + r) + bn−2 −
4bn
=0
49
(3)
Solving for bn from recursive equation (4) gives
bn = −
49bn−2
49n2 + 98nr + 49r2 − 4
(4)
7bn−2
n (7n − 4)
(5)
Which for the root r = − 27 becomes
bn = −
At this point, it is a good idea to keep track of bn in a table both before substituting r = − 27
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
0
bn
1
0
For n = 2, using the above recursive equation gives
b2 = −
49
49r2 + 196r + 192
Which for the root r = − 27 becomes
b2 = −
7
20
And the table now becomes
n
b0
b1
b2
bn,r
1
0
49
− 49r2 +196r+192
bn
1
0
7
− 20
For n = 3, using the above recursive equation gives
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
0
49
− 49r2 +196r+192
0
bn
1
0
7
− 20
0
For n = 4, using the above recursive equation gives
b4 =
2401
(49r2 + 196r + 192) (49r2 + 392r + 780)
chapter 2 . book solved problems
Which for the root r = − 27 becomes
b4 =
261
49
1920
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
49
− 49r2 +196r+192
0
bn
1
0
7
− 20
0
2401
(49r2 +196r+192)(49r2 +392r+780)
49
1920
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
0
49
− 49r2 +196r+192
0
bn
1
0
7
− 20
0
2401
(49r2 +196r+192)(49r2 +392r+780)
49
1920
0
0
Using the above table, then the solution y2 (x) is
y2 (x) = x2/7 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
2
4
1 − 7x
+ 49x
+ O(x6 )
20
1920
=
x2/7
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
c 1 − 7x2 + 49x4 + O(x6 )
2
4
2
20
1920
7x
49x
= c1 x2/7 1 −
+
+ O x6 +
36
4608
x2/7
Hence the final solution is
y = yh
c 1 − 7x2 + 49x4 + O(x6 )
2
4
2
20
1920
7x
49x
= c1 x2/7 1 −
+
+ O x6 +
36
4608
x2/7
3 Maple. Time used: 0.024 (sec). Leaf size: 36
Order:=6;
ode:=x^2*diff(diff(y(x),x),x)+diff(y(x),x)*x+(x^2-4/49)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
7 2
49
7 2
49
c2 x4/7 1 − 36
x + 4608
x4 + O (x6 ) + c1 1 − 20
x + 1920
x4 + O (x6 )
y=
x2/7
Maple trace
chapter 2 . book solved problems
262
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
4
x2 dx
y(x) + x dx
y(x) + x2 − 49
y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
49x2 −4 y(x)
d d
dx
y(x)
=
−
−
2
dx dx
49x
x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
49x2 −4 y(x)
d d
dx
y(x)
+
+
=0
dx dx
x
49x2
◦
Check to see if x0 = 0 is a regular singular point
Define
functions
h
i
2
−4
P2 (x) = x1 , P3 (x) = 49x
49x2
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=1
x=0
x=0
4
= − 49
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
49x2 dx
y(x)
+
49x
y(x)
+ (49x2 − 4) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..2
∞
P
xm · y(x) =
ak xk+r+m
k=0
◦
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert x · dx
y(x) to series expansion
∞
P
d
x · dx
y(x) =
ak (k + r) xk+r
k=0
d d
Convert x2 · dx
y(x) to series expansion
dx
∞
P
d d
x2 · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r
dx
k=0
Rewrite ODE with series expansions
chapter 2 . book solved problems
263
r
a0 (2 + 7r) (−2 + 7r) x + a1 (9 + 7r) (5 + 7r) x
1+r
+
∞
P
(ak (7k + 7r + 2) (7k + 7r − 2) + 49ak−2 )
k=2
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
(2 + 7r) (−2 + 7r) = 0
Values of r that satisfy the indicial equation
r ∈ − 27 , 27
Each term must be 0
a1 (9 + 7r) (5 + 7r) = 0
Solve for the dependent coefficient(s)
a1 = 0
Each term in the series must be 0, giving the recursion relation
ak (7k + 7r + 2) (7k + 7r − 2) + 49ak−2 = 0
Shift index using k− >k + 2
ak+2 (7k + 16 + 7r) (7k + 12 + 7r) + 49ak = 0
Recursion relation that defines series solution to ODE
49ak
ak+2 = − (7k+16+7r)(7k+12+7r)
Recursion relation for r = − 27
49ak
ak+2 = − (7k+14)(7k+10)
Solution
for r = − 27
∞
P
49ak
k− 27
y(x) =
ak x , ak+2 = − (7k+14)(7k+10) , a1 = 0
k=0
•
•
Recursion relation for r = 27
49ak
ak+2 = − (7k+18)(7k+14)
Solution
for r = 27
∞
P
49ak
k+ 27
y(x) =
ak x , ak+2 = − (7k+18)(7k+14) , a1 = 0
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
49ak
49bk
k− 27
k+ 27
y(x) =
ak x
+
bk x
, ak+2 = − (7k+14)(7k+10)
, a1 = 0, bk+2 = − (7k+18)(7k+14)
, b1 =
k=0
k=0
3 Mathematica. Time used: 0.003 (sec). Leaf size: 52
ode=x^2*D[y[x],{x,2}]+x*D[y[x],x]+(x^2-4/49)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1 x2/7
c
2
49x
7x
−
+1 +
4608
36
4
2
2
49x4
− 7x
+1
1920
20
x2/7
3 Sympy. Time used: 0.316 (sec). Leaf size: 49
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x**2*Derivative(y(x), (x, 2)) + x*Derivative(y(x), x) + (x**2 - 4/49)*y(
x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C2 x
2
7
C
1
49x4 7x2
−
+1 +
4608
36
2
49x4
− 7x
+1
1920
20
x
2
7
+ O x6
chapter 2 . book solved problems
2.3.2
264
Problem 3
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
Internal problem ID [8598]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.4. Bessels
Equation page 195
Problem number : 3
Date solved : Tuesday, November 11, 2025 at 12:04:00 PM
CAS classification : [[_Emden, _Fowler]]
xy 00 + y 0 +
y
=0
4
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
y
xy 00 + y 0 + = 0
4
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
1
q(x) =
4x
p(x) =
Table 2.46: Table p(x), q(x) singularites.
p(x) = x1
singularity
type
x=0
“regular”
1
q(x) = 4x
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
y
xy 00 + y 0 + = 0
4
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
265
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
∞
X
!
(n + r) (n + r − 1) an x
n+r−2
x+
n=0
∞
P
!
(n + r) an x
n+r−1
+
an x
n+r
n=0
= 0 (1)
4
n=0
Which simplifies to
∞
X
∞
X
!
xn+r−1 an (n + r) (n + r − 1)
+
n=0
!
(n + r) an xn+r−1
+
n=0
∞
X
an xn+r
!
= 0 (2A)
4
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
an xn+r
n =0
4
=
∞
X
an−1 xn+r−1
n=1
4
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
!
(n + r) an xn+r−1 +
n=0
∞
X
an−1 xn+r−1
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + (n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
x−1+r r2 = 0
Since the above is true for all x then the indicial equation becomes
r2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r2 = 0
4
!
= 0 (2B)
chapter 2 . book solved problems
266
Solving for r gives the roots of the indicial equation as [0, 0].
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
y1 (x) =
∞
X
an xn+r
(1A)
n=0
Now the second solution y2 is found using
∞
X
y2 (x) = y1 (x) ln (x) +
!
bn xn+r
(1B)
n=1
Then the general solution will be
y = c1 y1 (x) + c2 y2 (x)
In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), a0 is never zero, and is arbitrary
and is typically taken as a0 = 1, and {c1 , c2 } are two arbitray constants of integration which
can be found from initial conditions. We start by finding the first solution y1 (x). Eq (2B)
derived above is now used to find all an coefficients. The case n = 0 is skipped since it was
used to find the roots of the indicial equation. a0 is arbitrary and taken as a0 = 1. For 1 ≤ n
the recursive equation is
an (n + r) (n + r − 1) + an (n + r) +
an−1
=0
4
(3)
Solving for an from recursive equation (4) gives
an = −
an−1
4 (n2 + 2nr + r2 )
Which for the root r = 0 becomes
(4)
an−1
(5)
4n2
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
an = −
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 = −
1
4 (r + 1)2
Which for the root r = 0 becomes
a1 = −
1
4
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
For n = 2, using the above recursive equation gives
a2 =
1
16 (r + 1)2 (r + 2)2
Which for the root r = 0 becomes
a2 =
1
64
chapter 2 . book solved problems
267
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
a2
1
16(r+1)2 (r+2)2
1
64
For n = 3, using the above recursive equation gives
a3 = −
1
64 (r + 1) (r + 2)2 (r + 3)2
2
Which for the root r = 0 becomes
a3 = −
1
2304
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
a2
1
16(r+1)2 (r+2)2
1
− 64(r+1)2 (r+2)
2
(r+3)2
1
64
a3
1
− 2304
For n = 4, using the above recursive equation gives
a4 =
1
256 (r + 1) (r + 2)2 (r + 3)2 (r + 4)2
2
Which for the root r = 0 becomes
a4 =
1
147456
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
a2
1
16(r+1)2 (r+2)2
1
− 64(r+1)2 (r+2)
2
(r+3)2
1
256(r+1)2 (r+2)2 (r+3)2 (r+4)2
1
64
a3
a4
1
− 2304
1
147456
For n = 5, using the above recursive equation gives
a5 = −
1
1024 (r + 1) (r + 2) (r + 3)2 (r + 4)2 (r + 5)2
2
2
Which for the root r = 0 becomes
a5 = −
1
14745600
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
a2
1
16(r+1)2 (r+2)2
1
− 64(r+1)2 (r+2)
2
(r+3)2
1
256(r+1)2 (r+2)2 (r+3)2 (r+4)2
1
− 1024(r+1)2 (r+2)2 (r+3)
2
(r+4)2 (r+5)2
1
64
a3
a4
a5
1
− 2304
1
147456
1
− 14745600
chapter 2 . book solved problems
268
Using the above table, then the first solution y1 (x) becomes
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
=1−
x x2
x3
x4
x5
+
−
+
−
+ O x6
4 64 2304 147456 14745600
Now the second solution is found. The second solution is given by
!
∞
X
y2 (x) = y1 (x) ln (x) +
bn xn+r
n=1
Where bn is found using
d
an,r
dr
And the above is then evaluated at r = 0. The above table for an,r is used for this purpose.
Computing the derivatives gives the following table
bn =
n
b0
b1
bn,r
1
1
− 4(r+1)
2
an
1
− 14
b2
1
16(r+1)2 (r+2)2
1
− 64(r+1)2 (r+2)
2
(r+3)2
1
256(r+1)2 (r+2)2 (r+3)2 (r+4)2
1
− 1024(r+1)2 (r+2)2 (r+3)
2
(r+4)2 (r+5)2
1
64
b3
b4
b5
1
− 2304
1
147456
1
− 14745600
d
bn,r = dr
an,r
N/A since bn starts from 1
bn (r = 0)
N/A
1
2(r+1)3
−2r−3
8(r+1)3 (r+2)3
3r2 +12r+11
32(r+1)3 (r+2)3 (r+3)3
−2r3 −15r2 −35r−25
64(r+1)3 (r+2)3 (r+3)3 (r+4)3
5r4 +60r3 +255r2 +450r+274
512(r+1)3 (r+2)3 (r+3)3 (r+4)3 (r+5)3
1
2
3
− 64
11
6912
25
− 884736
137
442368000
The above table gives all values of bn needed. Hence the second solution is
y2 (x) = y1 (x) ln (x) + b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x x2
x3
x4
x5
6
= 1− +
−
+
−
+O x
ln (x)
4 64 2304 147456 14745600
x 3x2 11x3
25x4
137x5
+ −
+
−
+
+ O x6
2
64
6912 884736 442368000
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x x2
x3
x4
x5
6
= c1 1 − +
−
+
−
+O x
4 64 2304 147456 14745600
x x2
x3
x4
x5
x 3x2
6
+ c2
1− +
−
+
−
+O x
ln (x) + −
4 64 2304 147456 14745600
2
64
3
4
5
11x
25x
137x
6
+
−
+
+O x
6912 884736 442368000
Hence the final solution is
y = yh
x x2
x3
x4
x5
6
= c1 1 − +
−
+
−
+O x
4 64 2304 147456 14745600
x x2
x3
x4
x5
x 3x2 11x3
6
+ c2
1− +
−
+
−
+O x
ln (x) + −
+
4 64 2304 147456 14745600
2
64
6912
4
5
25x
137x
6
−
+
+O x
884736 442368000
chapter 2 . book solved problems
3 Maple. Time used: 0.031 (sec). Leaf size: 44
Order:=6;
ode:=diff(diff(y(x),x),x)*x+diff(y(x),x)+1/4*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
269
1
1 2
1 3
1
1
4
5
6
y = (c2 ln (x) + c1 ) 1 − x + x −
x +
x −
x +O x
4
64
2304
147456
14745600
1
3
11 3
25
137
+
x − x2 +
x −
x4 +
x 5 + O x 6 c2
2
64
6912
884736
442368000
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
x dx
y(x)
+ dx
y(x) + y(x)
=0
dx
4
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
d d
y(x) = − y(x)
− dx x
dx dx
4x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
d d
dx
y(x)
+
+ y(x)
=0
dx dx
x
4x
◦
◦
Check to see if x0 = 0 is a regular singular point
Define functions
1
P2 (x) = x1 , P3 (x) = 4x
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=1
x=0
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
4x dx
y(x) + y(x) + 4 dx
y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d
Convert dx
y(x) to series expansion
chapter 2 . book solved problems
d
y(x) =
dx
◦
◦
∞
P
270
ak (k + r) xk+r−1
k=0
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + 1 + r) xk+r
dx
k=−1
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODEwith series expansions
∞
k+r
P
2
2 −1+r
4a0 r x
+
4ak+1 (k + 1 + r) + ak x
=0
k=0
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
4r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term in the series must be 0, giving the recursion relation
4ak+1 (k + 1)2 + ak = 0
Recursion relation that defines series solution to ODE
ak
ak+1 = − 4(k+1)
2
•
Recursion relation for r = 0
ak
ak+1 = − 4(k+1)
2
•
Solution
for r = 0
∞
P
a
k
y(x) =
ak xk , ak+1 = − 4(k+1)
2
k=0
3 Mathematica. Time used: 0.003 (sec). Leaf size: 117
ode=x*D[y[x],{x,2}]+D[y[x],x]+1/4*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
x5
x4
x3
x2 x
137x5
25x4
11x3
−
+
−
+
− + 1 + c2
−
+
14745600 147456 2304 64 4
442368000 884736 6912
2
5
4
3
3x
x
x
x
x2 x
x
−
+ −
+
−
+
− + 1 log(x) +
64
14745600 147456 2304 64 4
2
3 Sympy. Time used: 0.200 (sec). Leaf size: 32
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*Derivative(y(x), (x, 2)) + y(x)/4 + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C1 −
x5
x4
x3
x2 x
+
−
+
− + 1 + O x6
14745600 147456 2304 64 4
chapter 2 . book solved problems
2.3.3
271
Problem 4
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
Internal problem ID [8599]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.4. Bessels
Equation page 195
Problem number : 4
Date solved : Tuesday, November 11, 2025 at 12:04:01 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
00
−2x
y + e
1
−
9
y=0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.33)
(2.34)
(2)
(3)
chapter 2 . book solved problems
272
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
273
To find y(x) series solution around x = 0. Hence
(9 e−2x − 1) y
9
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
y0
= (2y − y 0 ) e−2x +
9
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
2
(−9 e−2x + 1) y
=
− 4 e−2x (y − y 0 )
81
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
10 e−2x (8y − 11y 0 )
y0
=
+
+ e−4x (−8y + y 0 )
9
81
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
(133y − 36y 0 ) e−4x (−517y + 900y 0 ) e−2x
y
=
+
− y e−6x +
3
27
729
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = −
F0 = −
8y(0)
9
8y 0 (0)
9
260y(0)
F2 = −
+ 4y 0 (0)
81
8y(0) 908y 0 (0)
F3 =
−
9
81
17632y(0) 64y 0 (0)
F4 =
+
729
3
Substituting all the above in (7) and simplifying gives the solution as
4 2 1 3
65 4
1 5
4 3 1 4
227 5 0
y = 1− x + x −
x +
x y(0) + x − x + x −
x y (0) + O x6
9
3
486
135
27
6
2430
F1 = 2y(0) −
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
9 e2x y 00 + 9 − e2x y = 0
Let the solution be represented as power series of the form
∞
X
y=
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n=2
n(n − 1) an xn−2
chapter 2 . book solved problems
274
Substituting the above back into the ode gives
!
!
∞
∞
X
X
9 e2x
n(n − 1) an xn−2 + 9 − e2x
an x n = 0
(1)
n=2
n=0
Hence the ODE in Eq (1) becomes
!
X
∞
12
4
9 + 18x + 18x2 + 12x3 + 6x4 + x5 + x6
n(n − 1) an xn−2
5
5
n=2
!
X
∞
4
2
4
4
+ 8 − 2x − 2x2 − x3 − x4 − x5 − x6
an x n = 0
3
3
15
45
n=0
Expanding the first term in (1) gives
9·
·
∞
X
n=2
∞
X
!
n(n − 1) an x
n−2
∞
X
+ 18x ·
!
n(n − 1) an xn−2
+ 12x3 ·
n(n − 1) an x
n=2
∞
X
n=2
!
!
n−2
+ 18x2
!
n(n − 1) an xn−2
+ 6x4
n=2
∞
X
!
12x
4x6
·
n(n − 1) an xn−2 +
·
n(n − 1) an xn−2 +
5
5
n=2
n=2
!
!
X
∞
∞
X
4
2
4
4
·
n(n−1) an xn−2 + 8−2x−2x2 − x3 − x4 − x5 − x6
an x n = 0
3
3
15
45
n=2
n=0
∞
X
5
Expanding the second term in (1) gives
Expression too large to display
Which simplifies to
∞
X
4n xn+4 an (n − 1)
5
n=2
+
+
∞
X
n=2
∞
X
+
!
6n xn+2 an (n − 1) +
n=2
∞
X
!
18n xn−1 an (n − 1)
n=2
∞
X
∞
X
12n xn+3 an (n − 1)
!
+
5
!
12n x1+n an (n − 1) +
n=2
∞
X
!
9n(n − 1) an xn−2
n=2
+
∞
X
!
18nan xn (n − 1)
n=2
∞
X
!
8an xn
(2)
n=0
∞ X
4xn+3 an
1+n
n+2
+
−2x an +
−2x an +
−
3
n =0
n =0
n =0
∞
∞
∞
X
X
X
2xn+4 an
4xn+5 an
4xn+6 an
+
−
+
−
+
−
=0
3
15
45
n =0
n =0
n =0
∞
X
!
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
chapter 2 . book solved problems
275
and the corresponding index gives
∞
X
4n xn+4 an (n − 1)
5
n =2
5
∞
X
6n x
n+2
=
an (n − 1) =
5
∞
X
6(n − 2) an−2 (n − 3) xn
n=4
12n x
1+n
an (n − 1) =
n =2
∞
X
∞
X
12(n − 3) an−3 (n − 4) xn
n=5
n =2
∞
X
5
n=6
∞
X
12n xn+3 an (n − 1)
n =2
=
∞
X
4(n − 4) an−4 (n − 5) xn
∞
X
12(n − 1) an−1 (n − 2) xn
n=3
18n xn−1 an (n − 1) =
n =2
∞
X
18(1 + n) a1+n n xn
n=1
∞
X
9n(n − 1) an x
n−2
=
n =2
∞
X
9(n + 2) an+2 (1 + n) xn
n=0
∞
X
−2x
1+n
an =
∞
X
(−2an−1 xn )
n =0
n=1
∞
X
∞
X
−2xn+2 an =
(−2an−2 xn )
n =0
n=2
X
∞ ∞ X
4xn+3 an
4an−3 xn
−
=
−
3
3
n =0
n=3
X
∞ ∞ X
2xn+4 an
2an−4 xn
−
=
−
3
3
n =0
n=4
X
∞ ∞ X
4xn+5 an
4an−5 xn
−
=
−
15
15
n =0
n=5
X
∞ ∞ X
4xn+6 an
4an−6 xn
−
=
−
45
45
n =0
n=6
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
4(n − 4) an−4 (n − 5) xn
5
n=6
+
∞
X
∞
X
12(n − 3) an−3 (n − 4) xn
!
+
6(n − 2) an−2 (n − 3) xn
+
n=4
+
+
∞
X
n=2
∞
X
n=0
∞
X
5
n=5
!
∞
X
!
!
12(n − 1) an−1 (n − 2) xn
n=3
!
18nan xn (n − 1)
+
∞
X
!
18(1 + n) a1+n n xn
n=1
!
9(n + 2) an+2 (1 + n) x
∞
X
n
+
∞
X
!
8an xn
n=0
∞ X
4an−3 xn
+
(−2an−1 x ) +
(−2an−2 x ) +
−
3
n =1
n =2
n =3
X
X
∞ ∞ ∞ X
2an−4 xn
4an−5 xn
4an−6 xn
+
−
+
−
+
−
=0
3
15
45
n =4
n =5
n =6
n
n
(3)
chapter 2 . book solved problems
276
n = 0 gives
18a2 + 8a0 = 0
a2 = −
4a0
9
n = 1 gives
36a2 + 54a3 + 8a1 − 2a0 = 0
Which after substituting earlier equations, simplifies to
a3 =
a0 4a1
−
3
27
n = 2 gives
44a2 + 108a3 + 108a4 − 2a1 − 2a0 = 0
Which after substituting earlier equations, simplifies to
a4 = −
65a0 a1
+
486
6
n = 3 gives
22a2 + 116a3 + 216a4 + 180a5 − 2a1 −
4a0
=0
3
Which after substituting earlier equations, simplifies to
a0
227a1
−
135
2430
a5 =
n = 4 gives
10a2 + 70a3 + 224a4 + 360a5 + 270a6 −
4a1 2a0
−
=0
3
3
Which after substituting earlier equations, simplifies to
a6 =
1102a0 4a1
+
32805
135
n = 5 gives
52a2
2a1 4a0
+ 34a3 + 142a4 + 368a5 + 540a6 + 378a7 −
−
=0
15
3
15
Which after substituting earlier equations, simplifies to
a7 = −
19a0
503a1
+
630
459270
For 6 ≤ n, the recurrence equation is
4(n − 4) an−4 (n − 5) 12(n − 3) an−3 (n − 4)
+
+ 6(n − 2) an−2 (n − 3)
5
5
(4)
+ 12(n − 1) an−1 (n − 2) + 18nan (n − 1) + 18(1 + n) a1+n n + 9(n + 2) an+2 (1 + n)
4an−3 2an−4 4an−5 4an−6
+ 8an − 2an−1 − 2an−2 −
−
−
−
=0
3
3
15
45
chapter 2 . book solved problems
277
Solving for an+2 , gives
an+2 =
(5)
2(405n2 an + 405n2 a1+n + 18n2 an−4 + 54n2 an−3 + 135n2 an−2 + 270n2 an−1 − 405nan + 405na1+n − 162
−
405 (
2(405n2 − 405n + 180) an 2(405n2 + 405n) a1+n
4an−6
−
+
405 (n + 2) (1 + n)
405 (n + 2) (1 + n)
405 (n + 2) (1 + n)
2
4an−5
2(18n − 162n + 345) an−4 2(54n2 − 378n + 618) an−3
+
−
−
135 (n + 2) (1 + n)
405 (n + 2) (1 + n)
405 (n + 2) (1 + n)
2
2
2(135n − 675n + 765) an−2 2(270n − 810n + 495) an−1
−
−
405 (n + 2) (1 + n)
405 (n + 2) (1 + n)
=−
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
4a0 x2
y = a0 + a1 x −
+
9
a0 4a1
−
3
27
65a0 a1
a0
227a1
4
x + −
+
x +
−
x5 + . . .
486
6
135
2430
3
Collecting terms, the solution becomes
y=
4 2 1 3
65 4
1 5
4 3 1 4
227 5
1− x + x −
x +
x a0 + x − x + x −
x a1 + O x6 (3)
9
3
486
135
27
6
2430
At x = 0 the solution above becomes
y=
4 2 1 3
65 4
1 5
4 3 1 4
227 5 0
1− x + x −
x +
x y(0) + x − x + x −
x y (0) + O x6
9
3
486
135
27
6
2430
3 Maple. Time used: 0.007 (sec). Leaf size: 54
Order:=6;
ode:=diff(diff(y(x),x),x)+(exp(-2*x)-1/9)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
y=
4 2 1 3
65 4
1 5
4 3 1 4
227 5 0
1− x + x −
x +
x y(0) + x − x + x −
x y (0) + O x6
9
3
486
135
27
6
2430
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \
@ Moebius
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x),\
dx)) * 2F1([a1, a2], [b1], f)
-> Trying changes of variables to rationalize or make the ODE simpler
trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
chapter 2 . book solved problems
278
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Change of variables used:
[x = -1/2*ln(t)]
Linear ODE actually solved:
(9*t-1)*u(t)+36*t*diff(u(t),t)+36*t^2*diff(diff(u(t),t),t) = 0
<- change of variables successful
3 Mathematica. Time used: 0.001 (sec). Leaf size: 63
ode=D[y[x],{x,2}]+(Exp[-2*x]-1/9)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
5
227x5 x4 4x3
x
65x4 x3 4x2
−
+
−
+ x + c1
−
+
−
+1
2430
6
27
135
486
3
9
3 Sympy. Time used: 0.335 (sec). Leaf size: 63
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((-1/9 + exp(-2*x))*y(x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
2
y(x) = C2
!
x4 (e2x − 9) e−4x x2 x2 e−2x
+
−
+1
1944
18
2
2
x
x2 e−2x
+ C1 x
−
+ 1 + O x6
54
6
chapter 2 . book solved problems
2.3.4
279
Problem 6
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
Internal problem ID [8600]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.4. Bessels
Equation page 195
Problem number : 6
Date solved : Tuesday, November 11, 2025 at 12:04:01 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
x + 34 y
xy +
=0
4
2 00
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
x
3
2 00
xy +
+
y=0
4 16
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
p(x) = 0
4x + 3
q(x) =
16x2
Table 2.48: Table p(x), q(x) singularites.
q(x) = 4x+3
16x2
singularity
type
x=0
“regular”
p(x) = 0
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
x
3
2 00
xy +
+
y=0
4 16
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
280
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
x2
∞
X
!
(n + r) (n + r − 1) an xn+r−2
+
n=0
X
∞
3
x
+
4 16
!
an xn+r
=0
∞
X
3an xn+r
!
(1)
n=0
Which simplifies to
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
∞
X
x1+n+r an
!
+
4
n=0
n=0
=0
16
(2A)
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives
∞
X
x1+n+r an
4
n =0
=
∞
X
an−1 xn+r
4
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r.
∞
X
!
xn+r an (n + r) (n + r − 1)
n=0
+
∞
X
an−1 xn+r
!
4
n=1
+
∞
X
3an xn+r
n=0
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r an (n + r) (n + r − 1) +
3an xn+r
=0
16
When n = 0 the above becomes
xr a0 r(−1 + r) +
Or
3a0 xr
=0
16
3xr
x r(−1 + r) +
16
r
a0 = 0
Since a0 6= 0 then the above simplifies to
(16r2 − 16r + 3) xr
=0
16
Since the above is true for all x then the indicial equation becomes
r2 − r +
3
=0
16
Solving for r gives the roots of the indicial equation as
3
4
1
r2 =
4
r1 =
16
!
=0
(2B)
chapter 2 . book solved problems
281
Since a0 6= 0 then the indicial equation becomes
(16r2 − 16r + 3) xr
=0
16
Solving for r gives the roots of the indicial equation as 34 , 14 .
Since r1 − r2 = 12 is not an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = xr2
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
3
an xn+ 4
n=0
y2 (x) =
∞
X
1
bn xn+ 4
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. For 1 ≤ n the recursive equation is
an (n + r) (n + r − 1) +
an−1 3an
+
=0
4
16
(3)
Solving for an from recursive equation (4) gives
an = −
4an−1
2
16n + 32nr + 16r2 − 16n − 16r + 3
(4)
Which for the root r = 34 becomes
an = −
an−1
4n2 + 2n
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 34
and after as more terms are found using the above recursive equation.
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 = −
4
16r2 + 16r + 3
Which for the root r = 34 becomes
a1 = −
1
6
And the table now becomes
n
a0
a1
an,r
1
4
− 16r2 +16r+3
an
1
− 16
For n = 2, using the above recursive equation gives
a2 =
16
(16r2 + 16r + 3) (16r2 + 48r + 35)
chapter 2 . book solved problems
282
Which for the root r = 34 becomes
a2 =
1
120
And the table now becomes
n
a0
a1
a2
an,r
1
4
− 16r2 +16r+3
an
1
− 16
16
(16r2 +16r+3)(16r2 +48r+35)
1
120
For n = 3, using the above recursive equation gives
a3 = −
64
(16r2 + 16r + 3) (16r2 + 48r + 35) (16r2 + 80r + 99)
Which for the root r = 34 becomes
a3 = −
1
5040
And the table now becomes
n
a0
a1
a2
a3
an,r
1
4
− 16r2 +16r+3
an
1
− 16
16
(16r2 +16r+3)(16r2 +48r+35)
64
− (16r2 +16r+3)(16r2 +48r+35)(16r
2 +80r+99)
1
120
1
− 5040
For n = 4, using the above recursive equation gives
a4 =
256
(16r2 + 16r + 3) (16r2 + 48r + 35) (16r2 + 80r + 99) (16r2 + 112r + 195)
Which for the root r = 34 becomes
a4 =
1
362880
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
4
− 16r2 +16r+3
an
1
− 16
16
(16r2 +16r+3)(16r2 +48r+35)
64
− (16r2 +16r+3)(16r2 +48r+35)(16r
2 +80r+99)
256
(16r2 +16r+3)(16r2 +48r+35)(16r2 +80r+99)(16r2 +112r+195)
1
120
1
− 5040
1
362880
For n = 5, using the above recursive equation gives
a5 = −
1024
(16r2 + 16r + 3) (16r2 + 48r + 35) (16r2 + 80r + 99) (16r2 + 112r + 195) (16r2 + 144r + 323)
Which for the root r = 34 becomes
a5 = −
And the table now becomes
1
39916800
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
283
an,r
1
4
− 16r2 +16r+3
an
1
− 16
16
(16r2 +16r+3)(16r2 +48r+35)
64
− (16r2 +16r+3)(16r2 +48r+35)(16r
2 +80r+99)
256
(16r2 +16r+3)(16r2 +48r+35)(16r2 +80r+99)(16r2 +112r+195)
1024
− (16r2 +16r+3)(16r2 +48r+35)(16r2 +80r+99)(16r
2 +112r+195)(16r 2 +144r+323)
1
120
1
− 5040
1
362880
1
− 39916800
Using the above table, then the solution y1 (x) is
y1 (x) = x3/4 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x
x2
x3
x4
x5
3/4
6
=x
1− +
−
+
−
+O x
6 120 5040 362880 39916800
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. For 1 ≤ n the recursive equation is
bn (n + r) (n + r − 1) +
bn−1 3bn
+
=0
4
16
(3)
Solving for bn from recursive equation (4) gives
bn = −
4bn−1
2
16n + 32nr + 16r2 − 16n − 16r + 3
(4)
Which for the root r = 14 becomes
bn = −
bn−1
4n2 − 2n
(5)
At this point, it is a good idea to keep track of bn in a table both before substituting r = 14
and after as more terms are found using the above recursive equation.
n
b0
bn,r
1
bn
1
For n = 1, using the above recursive equation gives
b1 = −
4
16r2 + 16r + 3
Which for the root r = 14 becomes
b1 = −
1
2
And the table now becomes
n
b0
b1
bn,r
1
4
− 16r2 +16r+3
bn
1
− 12
For n = 2, using the above recursive equation gives
b2 =
16
(16r2 + 16r + 3) (16r2 + 48r + 35)
Which for the root r = 14 becomes
b2 =
1
24
chapter 2 . book solved problems
284
And the table now becomes
n
b0
b1
b2
bn,r
1
4
− 16r2 +16r+3
bn
1
− 12
16
(16r2 +16r+3)(16r2 +48r+35)
1
24
For n = 3, using the above recursive equation gives
b3 = −
64
(16r2 + 16r + 3) (16r2 + 48r + 35) (16r2 + 80r + 99)
Which for the root r = 14 becomes
b3 = −
1
720
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
4
− 16r2 +16r+3
bn
1
− 12
16
(16r2 +16r+3)(16r2 +48r+35)
64
− (16r2 +16r+3)(16r2 +48r+35)(16r
2 +80r+99)
1
24
1
− 720
For n = 4, using the above recursive equation gives
b4 =
256
(16r2 + 16r + 3) (16r2 + 48r + 35) (16r2 + 80r + 99) (16r2 + 112r + 195)
Which for the root r = 14 becomes
b4 =
1
40320
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
4
− 16r2 +16r+3
bn
1
− 12
16
(16r2 +16r+3)(16r2 +48r+35)
64
− (16r2 +16r+3)(16r2 +48r+35)(16r
2 +80r+99)
256
(16r2 +16r+3)(16r2 +48r+35)(16r2 +80r+99)(16r2 +112r+195)
1
24
1
− 720
1
40320
For n = 5, using the above recursive equation gives
b5 = −
1024
(16r2 + 16r + 3) (16r2 + 48r + 35) (16r2 + 80r + 99) (16r2 + 112r + 195) (16r2 + 144r + 323)
Which for the root r = 14 becomes
b5 = −
1
3628800
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
4
− 16r2 +16r+3
bn
1
− 12
16
(16r2 +16r+3)(16r2 +48r+35)
64
− (16r2 +16r+3)(16r2 +48r+35)(16r
2 +80r+99)
256
(16r2 +16r+3)(16r2 +48r+35)(16r2 +80r+99)(16r2 +112r+195)
1024
− (16r2 +16r+3)(16r2 +48r+35)(16r2 +80r+99)(16r
2 +112r+195)(16r 2 +144r+323)
1
24
1
− 720
1
40320
1
− 3628800
chapter 2 . book solved problems
285
Using the above table, then the solution y2 (x) is
y2 (x) = x3/4 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x x2
x3
x4
x5
1/4
6
=x
1− +
−
+
−
+O x
2 24 720 40320 3628800
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x
x2
x3
x4
x5
3/4
6
= c1 x
1− +
−
+
−
+O x
6 120 5040 362880 39916800
x x2
x3
x4
x5
1/4
6
+ c2 x
1− +
−
+
−
+O x
2 24 720 40320 3628800
Hence the final solution is
y = yh
x
x2
x3
x4
x5
6
= c1 x
1− +
−
+
−
+O x
6 120 5040 362880 39916800
x x2
x3
x4
x5
1/4
6
+ c2 x
1− +
−
+
−
+O x
2 24 720 40320 3628800
3/4
3 Maple. Time used: 0.023 (sec). Leaf size: 47
Order:=6;
ode:=x^2*diff(diff(y(x),x),x)+1/4*(x+3/4)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1
1 2
1 3
1
1
4
5
6
y = c1 x
1− x+ x −
x +
x −
x +O x
2
24
720
40320
3628800
1
1 2
1 3
1
1
3/4
4
5
6
+ c2 x
1− x+
x −
x +
x −
x +O x
6
120
5040
362880
39916800
1/4
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Group is reducible or imprimitive
<- Kovacics algorithm successful
Maple step by step
Let’s solve
2
•
•
d d
y(x)
dx dx
x+ 34 y(x)
=0
4
x
+
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
d d
y(x) = − (4x+3)y(x)
dx dx
16x2
chapter 2 . book solved problems
•
◦
◦
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(x) + (4x+3)y(x)
=0
dx dx
16x2
Check to see if x0 = 0 is a regular singular point
Define functions
P2 (x) = 0, P3 (x) = 4x+3
16x2
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
=0
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
286
x=0
3
= 16
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
16x2 dx
y(x) + (4x + 3) y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..1
∞
P
xm · y(x) =
ak xk+r+m
k=0
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d d
Convert x2 · dx
y(x) to series expansion
dx
∞
P
d d
x2 · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r
dx
k=0
Rewrite ODE with series expansions
∞
P
r
k+r
a0 (−1 + 4r) (−3 + 4r) x +
(ak (4k + 4r − 1) (4k + 4r − 3) + 4ak−1 ) x
=0
k=1
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
(−1 + 4r) (−3 + 4r) = 0
Values of r that satisfy the indicial equation
r ∈ 14 , 34
Each term in the series must be 0, giving the recursion relation
16 k + r − 14 k + r − 34 ak + 4ak−1 = 0
Shift index using k− >k + 1
16 k + 34 + r k + 14 + r ak+1 + 4ak = 0
Recursion relation that defines series solution to ODE
4ak
ak+1 = − (4k+3+4r)(4k+1+4r)
Recursion relation for r = 14
4ak
ak+1 = − (4k+4)(4k+2)
Solution
for r = 14
∞
P
4ak
k+ 14
y(x) =
ak x , ak+1 = − (4k+4)(4k+2)
k=0
•
•
Recursion relation for r = 34
4ak
ak+1 = − (4k+6)(4k+4)
Solution
for r = 34
∞
P
4ak
k+ 34
y(x) =
ak x , ak+1 = − (4k+6)(4k+4)
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
4ak
4bk
k+ 14
k+ 34
y(x) =
ak x
+
bk x
, ak+1 = − (4k+4)(4k+2) , bk+1 = − (4k+6)(4k+4)
k=0
k=0
chapter 2 . book solved problems
287
3 Mathematica. Time used: 0.003 (sec). Leaf size: 90
ode=x^2*D[y[x],{x,2}]+1/4*(x+3/4)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
√
4
x5
x4
x3
x2 x
y(x) → c2 x −
+
−
+
− +1
3628800 40320 720 24 2
x5
x4
x3
x2
x
3/4
+ c1 x
−
+
−
+
− +1
39916800 362880 5040 120 6
3 Sympy. Time used: 0.313 (sec). Leaf size: 60
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x**2*Derivative(y(x), (x, 2)) + (x + 3/4)*y(x)/4,0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C2 x
3
4
4
√
x4
x3
x2
x
x
x3
x2 x
4
−
+
− + 1 + C1 x
−
+ − + 1 + O x6
362880 5040 120 6
40320 720 24 2
chapter 2 . book solved problems
2.3.5
288
Problem 7
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
Internal problem ID [8601]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.4. Bessels
Equation page 195
Problem number : 7
Date solved : Tuesday, November 11, 2025 at 12:04:02 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
x2 y 00 + y 0 x +
(x2 − 1) y
=0
4
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
2
x
1
2 00
0
x y +yx+
−
y=0
4
4
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
x2 − 1
q(x) =
4x2
p(x) =
Table 2.50: Table p(x), q(x) singularites.
2
q(x) = x4x−1
2
singularity
type
x=0
“regular”
p(x) = x1
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
2
x
1
2 00
0
x y +yx+
−
y=0
4
4
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
289
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
x
∞
X
2
+
!
(n + r) (n + r − 1) an x
n=0
∞
X
!
(n + r) an xn+r−1
n=0
n+r−2
(1)
!
2
X
∞
x
1
x+
−
an xn+r = 0
4
4
n=0
Which simplifies to
∞
X
!
x
n+r
an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
x
n+r
an (n + r)
(2A)
n=0
∞
X
xn+r+2 an
n=0
!
4
∞ X
an xn+r
+
−
=0
4
n =0
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives
∞
X
xn+r+2 an
4
n =0
=
∞
X
an−2 xn+r
4
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r.
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
xn+r an (n + r)
n=0
∞
X
an−2 xn+r
n=2
4
!
+
∞ X
n+r
−
n =0
an x
4
=0
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r an (n + r) (n + r − 1) + xn+r an (n + r) −
an xn+r
=0
4
When n = 0 the above becomes
xr a0 r(−1 + r) + xr a0 r −
Or
xr
x r(−1 + r) + x r −
4
r
r
Since a0 6= 0 then the above simplifies to
(4r2 − 1) xr
=0
4
a0 x r
=0
4
a0 = 0
(2B)
chapter 2 . book solved problems
290
Since the above is true for all x then the indicial equation becomes
1
=0
4
r2 −
Solving for r gives the roots of the indicial equation as
r1 =
1
2
r2 = −
1
2
Since a0 6= 0 then the indicial equation becomes
(4r2 − 1) xr
=0
4
Solving for r gives the roots of the indicial equation as
1
2
, − 12 .
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = Cy1 (x) ln (x) + x
∞
X
r2
!
bn x
n
n=0
Or
∞
X
√
y1 (x) = x
an x n
!
n=0
∞
P
bn x n
y2 (x) = Cy1 (x) ln (x) + n=0√
x
Or
y1 (x) =
∞
X
1
an xn+ 2
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x
n− 12
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + an (n + r) +
an−2 an
−
=0
4
4
(3)
Solving for an from recursive equation (4) gives
an = −
an−2
2
4n + 8nr + 4r2 − 1
(4)
an−2
4n (n + 1)
(5)
Which for the root r = 12 becomes
an = −
At this point, it is a good idea to keep track of an in a table both before substituting r = 12
and after as more terms are found using the above recursive equation.
chapter 2 . book solved problems
n
a0
a1
291
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
a2 = −
1
4r2 + 16r + 15
Which for the root r = 12 becomes
a2 = −
1
24
And the table now becomes
n
a0
a1
a2
an,r
1
0
1
− 4r2 +16r+15
an
1
0
1
− 24
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
1
− 4r2 +16r+15
0
an
1
0
1
− 24
0
For n = 4, using the above recursive equation gives
a4 =
1
16r4 + 192r3 + 824r2 + 1488r + 945
Which for the root r = 12 becomes
a4 =
1
1920
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
1
− 4r2 +16r+15
0
an
1
0
1
− 24
0
1
16r4 +192r3 +824r2 +1488r+945
1
1920
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
292
an,r
1
0
1
− 4r2 +16r+15
0
an
1
0
1
− 24
0
1
16r4 +192r3 +824r2 +1488r+945
1
1920
0
0
Using the above table, then the solution y1 (x) is
√
y1 (x) = x a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
√
x2
x4
6
= x 1−
+
+O x
24 1920
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=0
Therefore
lim 0 = lim1 0
r→r2
r→− 2
=0
The limit is 0. Since the limit exists then the log term is not needed and we can set C = 0.
Therefore the second solution has the form
y2 (x) =
∞
X
bn xn+r
n=0
=
∞
X
1
bn xn− 2
n=0
Eq (3) derived above is used to find all bn coefficients. The case n = 0 is skipped since
it was used to find the roots of the indicial equation. b0 is arbitrary and taken as b0 = 1.
Substituting n = 1 in Eq(3) gives
b1 = 0
For 2 ≤ n the recursive equation is
bn−2 bn
−
=0
4
4
(4)
1
3
1
bn−2 bn
bn n −
n−
+ bn n −
+
−
=0
2
2
2
4
4
(4A)
bn (n + r) (n + r − 1) + bn (n + r) +
Which for for the root r = − 12 becomes
Solving for bn from the recursive equation (4) gives
bn = −
bn−2
2
4n + 8nr + 4r2 − 1
(5)
chapter 2 . book solved problems
293
Which for the root r = − 12 becomes
bn = −
bn−2
4n2 − 4n
(6)
At this point, it is a good idea to keep track of bn in a table both before substituting r = − 12
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
0
bn
1
0
For n = 2, using the above recursive equation gives
b2 = −
1
4r2 + 16r + 15
Which for the root r = − 12 becomes
b2 = −
1
8
And the table now becomes
n
b0
b1
b2
bn,r
1
0
1
− 4r2 +16r+15
bn
1
0
− 18
For n = 3, using the above recursive equation gives
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
0
1
− 4r2 +16r+15
0
bn
1
0
− 18
0
For n = 4, using the above recursive equation gives
b4 =
1
(4r2 + 16r + 15) (4r2 + 32r + 63)
Which for the root r = − 12 becomes
b4 =
1
384
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
1
− 4r2 +16r+15
0
bn
1
0
− 18
0
1
16r4 +192r3 +824r2 +1488r+945
1
384
For n = 5, using the above recursive equation gives
b5 = 0
chapter 2 . book solved problems
294
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
0
1
− 4r2 +16r+15
0
bn
1
0
− 18
0
1
16r4 +192r3 +824r2 +1488r+945
1
384
0
0
Using the above table, then the solution y2 (x) is
y2 (x) =
√
x b 0 + b1 x + b2 x 2 + b3 x 3 + b 4 x 4 + b5 x 5 + b6 x 6 . . .
2
4
x
1 − x8 + 384
+ O(x6 )
√
=
x
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
c 1 − x2 + x4 + O(x6 )
2
4
2
√
8
384
x
x
√
= c1 x 1 −
+
+ O x6 +
24 1920
x
Hence the final solution is
y = yh
√
x
x
= c1 x 1 −
+
+O x
24 1920
2
4
6
2
x4
c2 1 − x8 + 384
+ O(x6 )
√
+
x
3 Maple. Time used: 0.021 (sec). Leaf size: 34
Order:=6;
ode:=x^2*diff(diff(y(x),x),x)+diff(y(x),x)*x+1/4*(x^2-1)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2
1
1
c1 x 1 − 24
x + 1920
x4 + O (x6 ) + c2 1 − 18 x2 + 384
x4 + O (x6 )
√
y=
x
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Group is reducible or imprimitive
<- Kovacics algorithm successful
Maple step by step
chapter 2 . book solved problems
295
Let’s solve
2
d d
y(x)
dx dx
d
y(x)
dx
x2 −1 y(x)
=0
4
x
+x
+
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
x2 −1 y(x)
d d
y(x)
=
−
− dx x
dx dx
4x2
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
x2 −1 y(x)
d d
dx
y(x)
+
+
=0
dx dx
x
4x2
◦
Check to see if x0 = 0 is a regular singular point
Define
functions
h
i
2
P2 (x) = x1 , P3 (x) = x4x−1
2
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=1
x=0
x=0
= − 14
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
4x2 dx
y(x)
+
4x
y(x)
+ (x2 − 1) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..2
∞
P
xm · y(x) =
ak xk+r+m
k=0
◦
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert x · dx
y(x) to series expansion
∞
P
d
x · dx
y(x) =
ak (k + r) xk+r
k=0
d d
Convert x2 · dx
y(x) to series expansion
dx
∞
P
d d
x2 · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r
dx
k=0
Rewrite ODE with series expansions
r
a0 (1 + 2r) (−1 + 2r) x + a1 (3 + 2r) (1 + 2r) x
1+r
+
∞
P
(ak (2k + 2r + 1) (2k + 2r − 1) + ak−2 ) xk
k=2
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
(1 + 2r) (−1 + 2r) = 0
Values of r that satisfy the indicial equation
r ∈ − 12 , 12
Each term must be 0
a1 (3 + 2r) (1 + 2r) = 0
Solve for the dependent coefficient(s)
a1 = 0
Each term in the series must be 0, giving the recursion relation
ak (4k 2 + 8kr + 4r2 − 1) + ak−2 = 0
Shift index using k− >k + 2
ak+2 4(k + 2)2 + 8(k + 2) r + 4r2 − 1 + ak = 0
Recursion relation that defines series solution to ODE
k
ak+2 = − 4k2 +8kr+4r2a+16k+16r+15
Recursion relation for r = − 12
chapter 2 . book solved problems
•
296
ak
ak+2 = − 4k2 +12k+8
Solution
for r = − 12
∞
P
ak
k− 12
y(x) =
ak x , ak+2 = − 4k2 +12k+8 , a1 = 0
k=0
•
•
Recursion relation for r = 12
ak
ak+2 = − 4k2 +20k+24
Solution
for r = 12
∞
P
ak
k+ 12
y(x) =
ak x , ak+2 = − 4k2 +20k+24 , a1 = 0
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
1
ak
bk
k− 2
k+ 12
y(x) =
ak x
+
bk x
, ak+2 = − 4k2 +12k+8 , a1 = 0, bk+2 = − 4k2 +20k+24 , b1 = 0
k=0
k=0
3 Mathematica. Time used: 0.007 (sec). Leaf size: 58
ode=x^2*D[y[x],{x,2}]+x*D[y[x],x]+1/4*(x^2-1)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
7/2
9/2
x
x3/2
1
x
x5/2 √
y(x) → c1
−
+√
+ c2
−
+ x
384
8
1920
24
x
3 Sympy. Time used: 0.317 (sec). Leaf size: 42
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x**2*Derivative(y(x), (x, 2)) + x*Derivative(y(x), x) + (x**2 - 1)*y(x)
/4,0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
√
y(x) = C2 x
4
2
x
x
−
+1 +
1920 24
C1
4
2
x
− x8 + 1
384
√
x
+ O x6
chapter 2 . book solved problems
2.3.6
297
Problem 8
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
Internal problem ID [8602]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.4. Bessels
Equation page 195
Problem number : 8
Date solved : Tuesday, November 11, 2025 at 12:04:03 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
(2x + 1)2 y 00 + 2(2x + 1) y 0 + 16x(1 + x) y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X
x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.36)
(2.37)
(2)
(3)
chapter 2 . book solved problems
298
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
299
To find y(x) series solution around x = 0. Hence
2(8yx2 + 2y 0 x + 8yx + y 0 )
F0 = −
(2x + 1)2
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
32 x2 + x − 12
x + 12 y 0 − y
=−
(2x + 1)3
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
(128x3 + 192x2 − 96x − 80) y 0 + 256 x4 + 2x3 + 14 x2 − 34 x + 12 y
=
(2x + 1)4
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
(512x5 + 1280x4 + 128x3 − 1088x2 + 1216x + 832) y 0 − 1024y x4 + 2x3 − 32 x2 − 52 x + 19
16
=
(2x + 1)5
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
(−3072x5 − 7680x4 + 5376x3 + 15744x2 − 14208x − 9984) y 0 − 4096y x6 + 3x5 + 34 x4 − 72 x3 + 21
x2 +
4
=
(2x + 1)6
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = −2y 0 (0)
F1 = −16y(0) + 8y 0 (0)
F2 = 128y(0) − 80y 0 (0)
F3 = −1216y(0) + 832y 0 (0)
F4 = 14720y(0) − 9984y 0 (0)
Substituting all the above in (7) and simplifying gives the solution as
y=
8 3 16 4 152 5
4 3 10 4 104 5 0
2
1− x + x −
x y(0) + x − x + x − x +
x y (0) + O x6
3
3
15
3
3
15
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
4x2 + 4x + 1 y 00 + (4x + 2) y 0 + 16x2 + 16x y = 0
Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n=2
n(n − 1) an xn−2
chapter 2 . book solved problems
300
Substituting the above back into the ode gives
!
!
!
∞
∞
∞
X
X
X
4x2 + 4x + 1
n(n − 1) an xn−2 + (4x + 2)
nan xn−1 + 16x2 + 16x
an x n = 0
n=2
n=1
n=0
(1)
Which simplifies to
∞
X
∞
X
!
4xn an n(n − 1)
+
n=2
4n xn−1 an (n − 1)
+
n=2
∞
X
+
∞
X
!
!
4nan xn
∞
X
+
n=1
!
n(n − 1) an xn−2
(2)
n=2
!
2nan xn−1
∞
X
+
n=1
!
16xn+2 an
+
n=0
∞
X
!
16x1+n an
=0
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
4n x
n−1
an (n − 1) =
n =2
∞
X
4(1 + n) a1+n n xn
n=1
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (1 + n) xn
n=0
∞
X
2nan xn−1 =
n =1
∞
X
2(1 + n) a1+n xn
n=0
16x
n+2
an =
n =0
∞
X
∞
X
∞
X
16an−2 xn
n=2
16x
1+n
an =
n =0
∞
X
16an−1 xn
n=1
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
4xn an n(n − 1)
+
n=2
+
∞
X
n=1
∞
X
!
4(1 + n) a1+n n xn
+
n=1
!
4nan x
n
+
∞
X
∞
X
!
(n + 2) an+2 (1 + n) xn
(3)
n=0
!
2(1+n) a1+n x
n
n=0
+
∞
X
!
16an−2 x
n=2
n = 0 gives
2a2 + 2a1 = 0
a2 = −a1
n = 1 gives
12a2 + 6a3 + 4a1 + 16a0 = 0
Which after substituting earlier equations, simplifies to
a3 = −
8a0 4a1
+
3
3
n
+
∞
X
n=1
!
16an−1 xn
=0
chapter 2 . book solved problems
301
For 2 ≤ n, the recurrence equation is
4nan (n − 1) + 4(1 + n) a1+n n + (n + 2) an+2 (1 + n)
+ 4nan + 2(1 + n) a1+n + 16an−2 + 16an−1 = 0
(4)
Solving for an+2 , gives
2(2n2 an + 2n2 a1+n + 3na1+n + a1+n + 8an−2 + 8an−1 )
an+2 = −
(n + 2) (1 + n)
=−
(5)
4n2 an
2(2n2 + 3n + 1) a1+n
16an−2
16an−1
−
−
−
(n + 2) (1 + n)
(n + 2) (1 + n)
(n + 2) (1 + n) (n + 2) (1 + n)
For n = 2 the recurrence equation gives
16a2 + 30a3 + 12a4 + 16a0 + 16a1 = 0
Which after substituting the earlier terms found becomes
a4 = −
10a1 16a0
+
3
3
For n = 3 the recurrence equation gives
36a3 + 56a4 + 20a5 + 16a1 + 16a2 = 0
Which after substituting the earlier terms found becomes
a5 = −
152a0 104a1
+
15
15
For n = 4 the recurrence equation gives
64a4 + 90a5 + 30a6 + 16a2 + 16a3 = 0
Which after substituting the earlier terms found becomes
a6 = −
208a1 184a0
+
15
9
For n = 5 the recurrence equation gives
100a5 + 132a6 + 42a7 + 16a3 + 16a4 = 0
Which after substituting the earlier terms found becomes
a7 = −
288a0 8768a1
+
7
315
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
8a0 4a1
10a1 16a0
152a0 104a1
3
4
y = a0 + a1 x − a1 x + −
+
x + −
+
x + −
+
x5 + . . .
3
3
3
3
15
15
2
chapter 2 . book solved problems
302
Collecting terms, the solution becomes
y=
8 3 16 4 152 5
4 3 10 4 104 5
2
1− x + x −
x a0 + x − x + x − x +
x a1 + O x6 (3)
3
3
15
3
3
15
At x = 0 the solution above becomes
y=
8 3 16 4 152 5
4 3 10 4 104 5 0
2
1− x + x −
x y(0) + x − x + x − x +
x y (0) + O x6
3
3
15
3
3
15
3 Maple. Time used: 0.007 (sec). Leaf size: 54
Order:=6;
ode:=(2*x+1)^2*diff(diff(y(x),x),x)+2*(2*x+1)*diff(y(x),x)+16*(x+1)*x*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
y=
8 3 16 4 152 5
4 3 10 4 104 5 0
2
1− x + x −
x y(0) + x − x + x − x +
x y (0) + O x6
3
3
15
3
3
15
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
-> elliptic
-> Legendre
-> Whittaker
-> hyper3: Equivalence to 1F1 under a power @ Moebius
<- hyper3 successful: received ODE is equivalent to the 1F1 ODE
<- Whittaker successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
(2x + 1)2 dx
y(x) + 2(2x + 1) dx
y(x) + 16x(x + 1) y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d d
y(x) = − 16x(x+1)y(x)
−
dx dx
(2x+1)2
•
2
d
y(x)
dx
2x+1
Group termswith y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(x) +
dx dx
◦
◦
2
d
y(x)
dx
2x+1
+ 16x(x+1)y(x)
=0
(2x+1)2
Check to see if x0 = − 12 is a regular singular point
Define
functions
h
i
2
P2 (x) = 2x+1
, P3 (x) = 16x(x+1)
(2x+1)2
x + 12 · P2 (x) is analytic at x = − 12
chapter 2 . book solved problems
◦
◦
•
303
x + 12 · P2 (x)
=1
x=− 12
2
x + 12 · P3 (x) is analytic at x = − 12
2
x + 12 · P3 (x)
= −1
x=− 12
1
x = − 2 is a regular singular point
Check to see if x0 = − 12 is a regular singular point
x0 = − 12
Multiply by denominators
d d
d
(2x + 1)2 dx
y(x) + (4x + 2) dx
y(x) + 16x(x + 1) y(x) = 0
dx
Change variables using x = u − 12 so that the regular singular point is at u = 0
d d
d
4u2 du
y(u) + 4u du
y(u) + (16u2 − 4) y(u) = 0
du
Assume series solution for y(u)
∞
P
y(u) =
ak uk+r
•
•
k=0
◦
Rewrite ODE with series expansions
Convert um · y(u) to series expansion for m = 0..2
∞
P
um · y(u) =
ak uk+r+m
k=0
◦
◦
◦
Shift index using k− >k − m
∞
P
um · y(u) =
ak−m uk+r
k=m
d
Convert u · du
y(u) to series expansion
∞
P
d
u · du
y(u) =
ak (k + r) uk+r
k=0
d d
Convert u2 · du
y(u)
to series expansion
du
∞
P
d d
u2 · du
y(u) =
ak (k + r) (k + r − 1) uk+r
du
k=0
Rewrite ODE with series expansions
r
4a0 (1 + r) (−1 + r) u + 4a1 (2 + r) r u
1+r
+
∞
P
(4ak (k + r + 1) (k + r − 1) + 16ak−2 ) u
k=2
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
4(1 + r) (−1 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {−1, 1}
Each term must be 0
4a1 (2 + r) r = 0
Solve for the dependent coefficient(s)
a1 = 0
Each term in the series must be 0, giving the recursion relation
4ak (k + r + 1) (k + r − 1) + 16ak−2 = 0
Shift index using k− >k + 2
4ak+2 (k + 3 + r) (k + r + 1) + 16ak = 0
Recursion relation that defines series solution to ODE
4ak
ak+2 = − (k+3+r)(k+r+1)
Recursion relation for r = −1
4ak
ak+2 = − (k+2)k
Solution
for r = −1
∞
P
4ak
k−1
y(u) =
ak u , ak+2 = − (k+2)k , a1 = 0
k=0
•
Revert
the change of variables u = x + 12
∞
P
4ak
1 k−1
y(x) =
ak x + 2
, ak+2 = − (k+2)k , a1 = 0
k=0
•
•
Recursion relation for r = 1
4ak
ak+2 = − (k+4)(k+2)
Solution for r = 1
k+r
=0
chapter 2 . book solved problems
y(u) =
∞
P
k+1
ak u
k=0
4ak
, ak+2 = − (k+4)(k+2)
, a1 = 0
304
•
Revert
the change of variables u = x + 12
∞
P
4ak
1 k+1
y(x) =
ak x + 2
, ak+2 = − (k+4)(k+2) , a1 = 0
•
Combine
and rename
parameters
solutions
∞
∞
P
P
4ak
4bk
1 k−1
1 k+1
y(x) =
ak x + 2
+
bk x + 2
, ak+2 = − (k+2)k
, a1 = 0, bk+2 = − (k+4)(k+2)
, b1
k=0
k=0
k=0
3 Mathematica. Time used: 0.002 (sec). Leaf size: 61
ode=(2*x+1)^2*D[y[x],{x,2}]+2*(2*x+1)*D[y[x],x]+16*x*(x+1)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
152x5 16x4 8x3
104x5 10x4 4x3
2
−
+
−
+ 1 + c2
−
+
−x +x
15
3
3
15
3
3
3 Sympy. Time used: 0.323 (sec). Leaf size: 51
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(16*x*(x + 1)*y(x) + (2*x + 1)**2*Derivative(y(x), (x, 2)) + (4*x + 2)*
Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
y(x) = −
5x4 r(3) 26x5 r(3)
+
+ C2 −8x5 + 4x4 − 4x2 + 1 + C1 x(1 − x) + O x6
2
5
chapter 2 . book solved problems
2.4
305
Chapter 5. Series Solutions of ODEs. Special
Functions. Problem set 5.5. Bessel Functions Y(x).
General Solution page 200
Local contents
2.4.1
2.4.2
2.4.3
2.4.4
2.4.5
2.4.6
2.4.7
2.4.8
2.4.9
Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
306
315
324
332
338
345
352
358
364
chapter 2 . book solved problems
2.4.1
306
Problem 1
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314
Internal problem ID [8603]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 1
Date solved : Tuesday, November 11, 2025 at 12:04:03 PM
CAS classification : [_Bessel]
x2 y 00 + y 0 x + x2 − 6 y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
x2 y 00 + y 0 x + x2 − 6 y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
x2 − 6
q(x) =
x2
p(x) =
Table 2.53: Table p(x), q(x) singularites.
2
q(x) = x x−6
2
singularity
type
x=0
“regular”
p(x) = x1
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
x2 y 00 + y 0 x + x2 − 6 y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
307
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
x2
∞
X
!
(n + r) (n + r − 1) an xn+r−2 +
n=0
∞
X
!
(n + r) an xn+r−1
∞
X
2
x+ x −6
an xn+r
n=0
!
=0
n=0
(1)
Which simplifies to
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
xn+r an (n + r)
(2A)
n=0
∞
X
!
xn+r+2 an
+
n=0
∞
X
−6an xn+r = 0
n =0
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives
∞
X
x
n+r+2
an =
n =0
∞
X
an−2 xn+r
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r.
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
xn+r an (n + r)
n=0
∞
X
!
an−2 x
n=2
n+r
+
∞
X
−6an xn+r = 0
n =0
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r an (n + r) (n + r − 1) + xn+r an (n + r) − 6an xn+r = 0
When n = 0 the above becomes
xr a0 r(−1 + r) + xr a0 r − 6a0 xr = 0
Or
(xr r(−1 + r) + xr r − 6xr ) a0 = 0
Since a0 6= 0 then the above simplifies to
r 2 − 6 xr = 0
Since the above is true for all x then the indicial equation becomes
r2 − 6 = 0
(2B)
chapter 2 . book solved problems
308
Solving for r gives the roots of the indicial equation as
√
r1 = 6
√
r2 = − 6
Since a0 6= 0 then the indicial equation becomes
r 2 − 6 xr = 0
√ √
Solving for r gives the roots of the indicial equation as
6, − 6 .
√
Since r1 − r2 = 2 6 is not an integer, then we can construct two linearly independent
solutions
!
∞
X
y1 (x) = xr1
an x n
y2 (x) = xr2
n=0
∞
X
!
bn x n
n=0
Or
y1 (x) =
y2 (x) =
∞
X
n=0
∞
X
√
an xn+ 6
√
bn xn− 6
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + an (n + r) + an−2 − 6an = 0
(3)
Solving for an from recursive equation (4) gives
an−2
an = − 2
n + 2nr + r2 − 6
Which for the root r =
√
6 becomes
an = −
an−2
√
n 2 6+n
At this point, it is a good idea to keep track of an in a table both before substituting r =
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
1
a2 = − 2
r + 4r − 2
Which for the root r =
(4)
√
6 becomes
a2 = −
1
√
4+4 6
(5)
√
6
chapter 2 . book solved problems
309
And the table now becomes
n
a0
a1
a2
an,r
1
0
1
− r2 +4r−2
an
1
0
− 4+41√6
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
1
− r2 +4r−2
0
an
1
0
− 4+41√6
0
For n = 4, using the above recursive equation gives
a4 =
Which for the root r =
1
(r2 + 4r − 2) (r2 + 8r + 10)
√
6 becomes
a4 =
1
√ √ 32 1 + 6 2 + 6
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
1
− r2 +4r−2
0
an
1
0
− 4+41√6
0
1
(r2 +4r−2)(r2 +8r+10)
√ 1 √ 32 1+ 6 2+ 6
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
1
− r2 +4r−2
0
an
1
0
− 4+41√6
0
1
(r2 +4r−2)(r2 +8r+10)
√ 1 √ 32 1+ 6 2+ 6
a5
0
0
Using the above table, then the solution y1 (x) is
√
y1 (x) = x 6 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x2
x4
√ +
√ √ + O x6
1−
4 + 4 6 32 1 + 6 2 + 6
!
√
=x 6
chapter 2 . book solved problems
310
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. Substituting n = 1 in Eq. (2B) gives
b1 = 0
For 2 ≤ n the recursive equation is
bn (n + r) (n + r − 1) + bn (n + r) + bn−2 − 6bn = 0
(3)
Solving for bn from recursive equation (4) gives
bn−2
bn = − 2
n + 2nr + r2 − 6
(4)
√
Which for the root r = − 6 becomes
bn = −
bn−2
√
n −2 6 + n
(5)
√
At this point, it is a good idea to keep track of bn in a table both before substituting r = − 6
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
0
bn
1
0
For n = 2, using the above recursive equation gives
1
b2 = − 2
r + 4r − 2
√
Which for the root r = − 6 becomes
b2 =
1
√
−4 + 4 6
And the table now becomes
n
b0
b1
b2
bn,r
1
0
1
− r2 +4r−2
bn
1
0
1√
−4+4 6
For n = 3, using the above recursive equation gives
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
0
1
− r2 +4r−2
0
bn
1
0
1√
−4+4 6
0
For n = 4, using the above recursive equation gives
b4 =
1
(r2 + 4r − 2) (r2 + 8r + 10)
chapter 2 . book solved problems
311
√
Which for the root r = − 6 becomes
b4 =
1
√ √ 32 −1 + 6 −2 + 6
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
1
− r2 +4r−2
0
bn
1
0
1
(r2 +4r−2)(r2 +8r+10)
1
√ √ 32 −1+ 6 −2+ 6
1√
−4+4 6
0
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
1
− r2 +4r−2
0
bn
1
0
1
(r2 +4r−2)(r2 +8r+10)
1
√ √ 32 −1+ 6 −2+ 6
b5
0
0
1√
−4+4 6
0
Using the above table, then the solution y2 (x) is
√
y2 (x) = x 6 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
=x
x2
x4
√ +
√ √ + O x6
1+
−4 + 4 6 32 −1 + 6 −2 + 6
√
− 6
!
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
√
= c1 x 6
+ c2 x
!
x2
x4
√ +
√ √ + O x6
1−
4 + 4 6 32 1 + 6 2 + 6
√
− 6
!
x2
x4
6
√ +
√ √ +O x
1+
−4 + 4 6 32 −1 + 6 −2 + 6
Hence the final solution is
y = yh
√
= c1 x 6
+ c2 x
!
x2
x4
√ +
√ √ + O x6
1−
4 + 4 6 32 1 + 6 2 + 6
√
− 6
x2
x4
√ +
√ √ + O x6
1+
−4 + 4 6 32 −1 + 6 −2 + 6
!
chapter 2 . book solved problems
3 Maple. Time used: 0.026 (sec). Leaf size: 89
Order:=6;
ode:=x^2*diff(diff(y(x),x),x)+diff(y(x),x)*x+(x^2-6)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
312
!
1
1
1
√ x2 +
√ √ x4 + O x6
y = c1 x − 6 1 +
32
−4 + 4 6
−2 + 6 −1 + 6
!
√
1
1
1
√ x2 +
√ √ x 4 + O x6
+ c2 x 6 1 −
32
4+4 6
2+ 6 1+ 6
√
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
x2 dx
y(x) + x dx
y(x) + (x2 − 6) y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
x2 −6 y(x)
d d
dx
y(x)
=
−
−
2
dx dx
x
x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
x2 −6 y(x)
d d
dx
y(x)
+
+
=0
dx dx
x
x2
◦
Check to see if x0 = 0 is a regular singular point
Define
functions
h
i
2
P2 (x) = x1 , P3 (x) = x x−6
2
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=1
x=0
= −6
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x2 dx
y(x)
+
x
y(x)
+ (x2 − 6) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
chapter 2 . book solved problems
◦
313
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..2
∞
P
xm · y(x) =
ak xk+r+m
k=0
◦
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert x · dx
y(x) to series expansion
∞
P
d
x · dx
y(x) =
ak (k + r) xk+r
k=0
d d
Convert x2 · dx
y(x)
to series expansion
dx
∞
P
d d
x2 · dx
y(x) =
ak (k + r) (k + r − 1) xk+r
dx
k=0
Rewrite ODE with series expansions r
2
2
a0 (r − 6) x + a1 (r + 2r − 5) x
1+r
+
∞
P
2
2
(ak (k + 2kr + r − 6) + ak−2 ) x
k+r
=0
k=2
•
•
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r2 − 6 = 0
Values of r that satisfy the indicial equation
√
√
r∈
6, − 6
Each term must be 0
a1 (r2 + 2r − 5) = 0
Solve for the dependent coefficient(s)
a1 = 0
Each term in the series must be 0, giving the recursion relation
ak (k 2 + 2kr + r2 − 6) + ak−2 = 0
Shift index using k− >k + 2
ak+2 (k + 2)2 + 2(k + 2) r + r2 − 6 + ak = 0
Recursion relation that defines series solution to ODE
ak+2 = − k2 +2kr+ra2k+4k+4r−2
√
Recursion relation for r = 6
ak
√
ak+2 = − k2 +2k√6+4+4k+4
6
√
Solution
for
r
=
6
∞
√
P
a
k+ 6
k
√ , a1 = 0
y(x) =
ak x
, ak+2 = − k2 +2k√6+4+4k+4
6
k=0
√
Recursion relation for r = − 6
ak
√
ak+2 = − k2 −2k√6+4+4k−4
6
√
Solution
for
r
=
−
6
∞
√
P
a
k− 6
k
y(x) =
ak x
, ak+2 = − k2 −2k√6+4+4k−4√6 , a1 = 0
k=0
•
Combine
and rename
solutions
∞ parameters
∞
√
√
P
P
ak
k+ 6
k− 6
√ , a1 = 0, bk+2 = −
√ bk
y(x) =
ak x
+
bk x
, ak+2 = − k2 +2k√6+4+4k+4
6
k2 −2k 6+4+4
k=0
k=0
chapter 2 . book solved problems
314
3 Mathematica. Time used: 0.003 (sec). Leaf size: 210
ode=x^2*D[y[x],{x,2}]+x*D[y[x],x]+(x^2-6)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
x4
√
√ √ √
√ √ −4 − 6 + 1 − 6 2 − 6
−2 − 6 + 3 − 6 4 − 6
!
√
x2
√
√ √ + 1 x− 6
−
−4 − 6 + 1 − 6 2 − 6
+ c1
−4 +
√
x4
√ √ √
√ √ 6+ 1+ 6 2+ 6
−2 + 6 + 3 + 6 4 + 6
2
−
−4 +
√
!
√
x
√ √ +1 x 6
6+ 1+ 6 2+ 6
7 Sympy
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x**2*Derivative(y(x), (x, 2)) + x*Derivative(y(x), x) + (x**2 - 6)*y(x)
,0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
NotImplementedError : Not sure of sign of 6 - x0
chapter 2 . book solved problems
2.4.2
315
Problem 2
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323
Internal problem ID [8604]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 2
Date solved : Tuesday, November 11, 2025 at 12:04:04 PM
CAS classification : [_Lienard]
xy 00 + 5y 0 + yx = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + 5y 0 + yx = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
5
x
q(x) = 1
p(x) =
Table 2.55: Table p(x), q(x) singularites.
p(x) = x5
singularity
type
x=0
“regular”
q(x) = 1
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + 5y 0 + yx = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
316
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x + 5
n=0
!
(n + r) an xn+r−1
∞
X
+
n=0
!
an xn+r
x = 0 (1)
n=0
Which simplifies to
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
!
5(n + r) an xn+r−1
+
n=0
∞
X
!
x1+n+r an
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
x
1+n+r
an =
n =0
∞
X
an−2 xn+r−1
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
!
5(n + r) an xn+r−1 +
n=0
∞
X
!
an−2 xn+r−1
n=2
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + 5(n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + 5ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + 5r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
r x−1+r (4 + r) = 0
Since the above is true for all x then the indicial equation becomes
r(4 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = −4
Since a0 6= 0 then the indicial equation becomes
r x−1+r (4 + r) = 0
= 0 (2B)
chapter 2 . book solved problems
317
Solving for r gives the roots of the indicial equation as [0, −4].
Since r1 − r2 = 4 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) + xr2
!
bn x n
n=0
Or
y1 (x) =
∞
X
an x n
n=0
∞
P
y2 (x) = Cy1 (x) ln (x) +
bn x n
n=0
x4
Or
y1 (x) =
∞
X
an xn
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn xn−4
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + 5an (n + r) + an−2 = 0
(3)
Solving for an from recursive equation (4) gives
an−2
an = − 2
n + 2nr + r2 + 4n + 4r
(4)
Which for the root r = 0 becomes
an = −
an−2
n (n + 4)
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
1
a2 = − 2
r + 8r + 12
Which for the root r = 0 becomes
a2 = −
And the table now becomes
1
12
chapter 2 . book solved problems
n
a0
a1
a2
318
an,r
1
0
1
− r2 +8r+12
an
1
0
1
− 12
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
1
− r2 +8r+12
0
an
1
0
1
− 12
0
For n = 4, using the above recursive equation gives
a4 =
1
(r + 6) (r + 2) (r + 8) (4 + r)
Which for the root r = 0 becomes
a4 =
1
384
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
1
− r2 +8r+12
0
an
1
0
1
− 12
0
1
(r+6)(r+2)(r+8)(4+r)
1
384
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
1
− r2 +8r+12
0
an
1
0
1
− 12
0
1
(r+6)(r+2)(r+8)(4+r)
1
384
0
0
Using the above table, then the solution y1 (x) is
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x2
x4
=1−
+
+ O x6
12 384
Now the second solution y2 (x) is found. Let
r1 − r2 = N
chapter 2 . book solved problems
319
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 4. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a4 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a4
=
1
(r + 6) (r + 2) (r + 8) (4 + r)
Therefore
1
1
= lim
r→r2 (r + 6) (r + 2) (r + 8) (4 + r)
r→−4 (r + 6) (r + 2) (r + 8) (4 + r)
lim
= undefined
Since the limit does not exist then the log term is needed. Therefore the second solution has
the form
!
∞
X
y2 (x) = Cy1 (x) ln (x) +
bn xn+r2
n=0
Therefore
d
Cy1 (x)
y2 (x) = Cy10 (x) ln (x) +
+
dx
x
Cy1 (x)
= Cy10 (x) ln (x) +
+
x
∞
X
bn xn+r2 (n + r2 )
n=0
∞
X
!
x
!
x−1+n+r2 bn (n + r2 )
n=0
∞
d2
2Cy10 (x) Cy1 (x) X
00
y
(x)
=
Cy
(x)
ln
(x)
+
−
+
2
1
dx2
x
x2
n=0
2Cy10 (x) Cy1 (x)
= Cy100 (x) ln (x) +
−
+
x
x2
∞
X
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!
!
x−2+n+r2 bn (n + r2 ) (−1 + n + r2 )
n=0
Substituting these back into the given ode xy 00 + 5y 0 + yx = 0 gives
∞
2Cy10 (x) Cy1 (x) X
00
Cy1 (x) ln (x) +
−
+
x
x2
n=0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!
∞
X
5Cy1 (x)
bn xn+r2 (n + r2 )
0
+ 5Cy1 (x) ln (x) +
+5
x
x
n=0
!!
∞
X
+ Cy1 (x) ln (x) +
bn xn+r2
x=0
!!
x
n=0
Which can be written as
0
2y1 (x) y1 (x)
5y1 (x)
00
0
(y1 (x) x + y1 (x) x + 5y1 (x)) ln (x) + x
− 2
+
C
x
x
x
!!
∞
X
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
+
−
x
x2
x2
n=0
!
!
∞
∞
X
X
bn xn+r2 (n + r2 )
n+r2
+
bn x
x+5
=0
x
n=0
n=0
But since y1 (x) is a solution to the ode, then
y1 (x) x + y100 (x) x + 5y10 (x) = 0
(7)
chapter 2 . book solved problems
320
Eq (7) simplifes to
0
2y1 (x) y1 (x)
5y1 (x)
x
− 2
+
C
x
x
x
!!
∞
2
n+r2
n+r2
X
bn x
(n + r2 )
bn x
(n + r2 )
+
−
x
2
x
x2
n=0
!
!
∞
∞
n+r2
X
X
b
x
(n
+
r
)
n
2
+
bn xn+r2 x + 5
=0
x
n=0
n=0
Substituting y1 =
∞
P
(8)
an xn+r1 into the above gives
n=0
∞
∞
P −1+n+r1
P
n+r1
2
x
an (n + r1 ) x + 4
an x
C
n=0
n=0
(9)
∞
∞
P
P −1+n+r2
−2+n+r2
2
n+r2
2
x
bn (n + r2 ) (−1 + n + r2 ) x +
bn x
x +5
x
bn (n + r2 ) x
x
∞
P
n=0
+
n=0
n=0
x
=0
Since r1 = 0 and r2 = −4 then the above becomes
∞
∞
P −1+n
P
n
2
x
an n x + 4
an x
C
n=0
n=0
x
∞
P
+
x
−6+n
∞
∞
P
P −5+n
2
n−4
2
bn (n − 4) (−5 + n) x +
bn x
x +5
x
bn (n − 4) x
n=0
n=0
n=0
=0
x
(10)
Which simplifies to
∞
X
2x
+
an nC
n=0
+
∞
X
!
−1+n
∞
X
!
x−3+n bn
+
n=0
!
4C x
−1+n
an
n=0
∞
X
+
∞
X
!
x
−5+n
bn (−5 + n) (n − 4)
n=0
(2A)
!
5x−5+n bn (n − 4)
=0
n=0
The next step is to make all powers of x be −5 + n in each summation term. Going over each
summation term above with power of x in it which is not already x−5+n and adjusting the
power and the corresponding index gives
∞
X
2x−1+n an nC =
n =0
∞
X
2C(n − 4) an−4 x−5+n
n=4
∞
X
4C x
−1+n
an =
n =0
∞
X
4Can−4 x−5+n
n=4
∞
X
n =0
x
−3+n
bn =
∞
X
n=2
bn−2 x−5+n
chapter 2 . book solved problems
321
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to −5 + n.
∞
X
!
2C(n − 4) an−4 x−5+n
n=4
+
∞
X
+
∞
X
!n=4
x−5+n bn (−5+n) (n−4) +
n=0
!
4Can−4 x−5+n
∞
X
(2B)
!
bn−2 x−5+n +
n=2
∞
X
!
5x−5+n bn (n−4) = 0
n=0
For n = 0 in Eq. (2B), we choose arbitray value for b0 as b0 = 1. For n = 1, Eq (2B) gives
−3b1 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
−3b1 = 0
Solving the above for b1 gives
b1 = 0
For n = 2, Eq (2B) gives
−4b2 + b0 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
−4b2 + 1 = 0
Solving the above for b2 gives
b2 =
1
4
For n = 3, Eq (2B) gives
−3b3 + b1 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
−3b3 = 0
Solving the above for b3 gives
b3 = 0
For n = N , where N = 4 which is the difference between the two roots, we are free to choose
b4 = 0. Hence for n = 4, Eq (2B) gives
4C +
1
=0
4
Which is solved for C. Solving for C gives
C=−
1
16
For n = 5, Eq (2B) gives
6Ca1 + b3 + 5b5 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
5b5 = 0
Solving the above for b5 gives
b5 = 0
Now that we found all bn and C, we can calculate the second solution from
!
∞
X
y2 (x) = Cy1 (x) ln (x) +
bn xn+r2
n=0
chapter 2 . book solved problems
322
1
Using the above value found for C = − 16
and all bn , then the second solution becomes
2
1 + x4 + O(x6 )
1
x2
x4
6
y2 (x) = −
1−
+
+O x
ln (x) +
16
12 384
x4
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
= c1
!
2
1 + x4 + O(x6 )
x2 x4
1
x2 x4
6
6
1− +
+O x
+c2 −
1− +
+O x
ln (x)+
12 384
16
12 384
x4
Hence the final solution is
y = yh
!
x2
6
2
4
6
1
+
+
O(x
)
x2
x4
1
x
x
O(x
)
4
= c1 1 − +
+ O x 6 + c2
− +
−
−
ln (x) +
12 384
16 192 6144
16
x4
3 Maple. Time used: 0.025 (sec). Leaf size: 44
Order:=6;
ode:=diff(diff(y(x),x),x)*x+5*diff(y(x),x)+x*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2
1
c1 1 − 12
x + 384
x4 + O (x6 ) x4 + c2 (ln (x) (9x4 + O (x6 )) + (−144 − 36x2 + O (x6 )))
y=
x4
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
•
Let’s solve
d d
d
x dx
y(x) + 5 dx
y(x) + xy(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative d d
y(x) = −y(x) −
dx dx
•
5
d
y(x)
dx
x
Group termswith y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(x) +
dx dx
5
d
y(x)
dx
x
+ y(x) = 0
chapter 2 . book solved problems
•
•
•
•
•
•
•
•
323
Simplify ODE
d d
d
x2 dx
y(x) + 5 dx
y(x) x + x2 y(x) = 0
dx
Make a change of variables
y(x) = u(x)
x2
d
Compute dx
y(x)
d
u(x)
2u(x)
d
dx
y(x)
=
−
+
3
dx
x
x2
d d
Compute dx
y(x) dx
d
d d
4 dx
u(x)
u(x)
6u(x)
d d
dx dx
y(x)
=
−
+
4
3
dx dx
x
x
x2
Apply change of variables to the ODE
2
d d
d
u(x) x2 + dx
u(x)
x
+
u(x)
x − 4u(x) = 0
dx
dx
ODE is now of the Bessel form
Solution to Bessel ODE
u(x) = C 1 BesselJ (2, x) + C 2 BesselY (2, x)
Make the change from y(x) back to y(x)
2 BesselY (2,x)
y(x) = C 1 BesselJ (2,x)+C
x2
3 Mathematica. Time used: 0.006 (sec). Leaf size: 47
ode=x*D[y[x],{x,2}]+5*D[y[x],x]+x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
!
2
x4
x2
(x2 + 8)
log(x)
−
+ 1 + c1
−
384 12
64x4
16
3 Sympy. Time used: 0.230 (sec). Leaf size: 19
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*y(x) + x*Derivative(y(x), (x, 2)) + 5*Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C1
x4
x2
−
+ 1 + O x6
384 12
chapter 2 . book solved problems
2.4.3
324
Problem 3
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331
Internal problem ID [8605]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 3
Date solved : Tuesday, November 11, 2025 at 12:04:05 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
9x2 y 00 + 9y 0 x + 36x4 − 16 y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
9x2 y 00 + 9y 0 x + 36x4 − 16 y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
4x4 − 16
9
q(x) =
2
x
p(x) =
Table 2.57: Table p(x), q(x) singularites.
4x4 − 16
q(x) = x2 9
singularity
type
x=0
“regular”
x=∞
“regular”
x = −∞ “regular”
p(x) = x1
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0, ∞, −∞]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
9x2 y 00 + 9y 0 x + 36x4 − 16 y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
325
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
9x2
+9
!
(n + r) (n + r − 1) an xn+r−2
n=0
∞
X
(1)
!
(n + r) an x
n+r−1
4
x + 36x − 16
∞
X
n=0
!
an xn+r
=0
n=0
Which simplifies to
∞
X
!
9xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
9xn+r an (n + r)
n=0
∞
X
!
36xn+r+4 an
+
n=0
∞
X
(2A)
−16an xn+r = 0
n =0
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives
∞
X
36xn+r+4 an =
n =0
∞
X
36an−4 xn+r
n=4
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r.
∞
X
!
9xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
9xn+r an (n + r)
n=0
∞
X
!
36an−4 x
n+r
+
n=4
∞
X
−16an xn+r = 0
n =0
The indicial equation is obtained from n = 0. From Eq (2B) this gives
9xn+r an (n + r) (n + r − 1) + 9xn+r an (n + r) − 16an xn+r = 0
When n = 0 the above becomes
9xr a0 r(−1 + r) + 9xr a0 r − 16a0 xr = 0
Or
(9xr r(−1 + r) + 9xr r − 16xr ) a0 = 0
Since a0 6= 0 then the above simplifies to
9r2 − 16 xr = 0
Since the above is true for all x then the indicial equation becomes
9r2 − 16 = 0
(2B)
chapter 2 . book solved problems
326
Solving for r gives the roots of the indicial equation as
r1 =
4
3
r2 = −
4
3
Since a0 6= 0 then the indicial equation becomes
9r2 − 16 xr = 0
Solving for r gives the roots of the indicial equation as
4
4
,
−
.
3
3
Since r1 − r2 = 83 is not an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = xr2
∞
X
!
bn x n
n=0
Or
∞
X
y1 (x) =
4
an xn+ 3
n=0
∞
X
y2 (x) =
4
bn xn− 3
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
Substituting n = 2 in Eq. (2B) gives
a2 = 0
Substituting n = 3 in Eq. (2B) gives
a3 = 0
For 4 ≤ n the recursive equation is
9an (n + r) (n + r − 1) + 9an (n + r) + 36an−4 − 16an = 0
(3)
Solving for an from recursive equation (4) gives
an = −
36an−4
2
9n + 18nr + 9r2 − 16
(4)
12an−4
n (3n + 8)
(5)
Which for the root r = 43 becomes
an = −
At this point, it is a good idea to keep track of an in a table both before substituting r = 43
and after as more terms are found using the above recursive equation.
n
a0
a1
a2
a3
an,r
1
0
0
0
an
1
0
0
0
chapter 2 . book solved problems
327
For n = 4, using the above recursive equation gives
a4 = −
36
9r2 + 72r + 128
Which for the root r = 43 becomes
a4 = −
3
20
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
0
0
36
− 9r2 +72r+128
an
1
0
0
0
3
− 20
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
0
0
36
− 9r2 +72r+128
0
an
1
0
0
0
3
− 20
0
Using the above table, then the solution y1 (x) is
y1 (x) = x4/3 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
3x4
4/3
6
=x
1−
+O x
20
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. Substituting n = 1 in Eq. (2B) gives
b1 = 0
Substituting n = 2 in Eq. (2B) gives
b2 = 0
Substituting n = 3 in Eq. (2B) gives
b3 = 0
For 4 ≤ n the recursive equation is
9bn (n + r) (n + r − 1) + 9bn (n + r) + 36bn−4 − 16bn = 0
(3)
Solving for bn from recursive equation (4) gives
bn = −
36bn−4
2
9n + 18nr + 9r2 − 16
(4)
12bn−4
n (3n − 8)
(5)
Which for the root r = − 43 becomes
bn = −
chapter 2 . book solved problems
328
At this point, it is a good idea to keep track of bn in a table both before substituting r = − 43
and after as more terms are found using the above recursive equation.
n
b0
b1
b2
b3
bn,r
1
0
0
0
bn
1
0
0
0
For n = 4, using the above recursive equation gives
b4 = −
36
9r2 + 72r + 128
Which for the root r = − 43 becomes
b4 = −
3
4
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
0
0
36
− 9r2 +72r+128
bn
1
0
0
0
− 34
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
0
0
0
36
− 9r2 +72r+128
0
bn
1
0
0
0
− 34
0
Using the above table, then the solution y2 (x) is
y2 (x) = x4/3 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
4
1 − 3x4 + O(x6 )
=
x4/3
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
c 1 − 3x4 + O(x6 )
4
2
4
3x
= c1 x4/3 1 −
+ O x6 +
20
x4/3
Hence the final solution is
y = yh
3x4
6
c2 1 − 4 + O(x )
3x4
= c1 x4/3 1 −
+ O x6 +
20
x4/3
chapter 2 . book solved problems
3 Maple. Time used: 0.025 (sec). Leaf size: 32
Order:=6;
ode:=9*x^2*diff(diff(y(x),x),x)+9*diff(y(x),x)*x+(36*x^4-16)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
329
3 4
c2 x8/3 1 − 20
x + O (x6 ) + c1 1 − 34 x4 + O (x6 )
y=
x4/3
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
9x2 dx
y(x) + 9x dx
y(x) + (36x4 − 16) y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
4 9x4 −4 y(x)
d d
dx
y(x)
=
−
−
2
dx dx
9x
x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
4 9x4 −4 y(x)
d d
dx
y(x) + x +
=0
dx dx
9x2
◦
Check to see if x0 = 0 is a regular singular point
Define
functions
h
i
P2 (x) = x1 , P3 (x) =
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
•
•
=1
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
4 9x4 −4
9x2
x=0
= − 16
9
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
9x2 dx
y(x) + 9x dx
y(x) + (36x4 − 16) y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..4
∞
P
xm · y(x) =
ak xk+r+m
k=0
chapter 2 . book solved problems
◦
◦
◦
330
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert x · dx
y(x) to series expansion
∞
P
d
x · dx
y(x) =
ak (k + r) xk+r
k=0
d d
Convert x2 · dx
y(x) to series expansion
dx
∞
P
d d
x2 · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r
dx
k=0
Rewrite ODE with series expansions
a0 (4 + 3r) (−4 + 3r) xr + a1 (7 + 3r) (−1 + 3r) x1+r + a2 (10 + 3r) (2 + 3r) x2+r + a3 (13 + 3r) (5 +
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
(4 + 3r) (−4 + 3r) = 0
Values of r that satisfy the indicial equation
r ∈ − 43 , 43
The coefficients of each power of x must be 0
[a1 (7 + 3r) (−1 + 3r) = 0, a2 (10 + 3r) (2 + 3r) = 0, a3 (13 + 3r) (5 + 3r) = 0]
Solve for the dependent coefficient(s)
{a1 = 0, a2 = 0, a3 = 0}
Each term in the series must be 0, giving the recursion relation
ak (3k + 3r + 4) (3k + 3r − 4) + 36ak−4 = 0
Shift index using k− >k + 4
ak+4 (3k + 16 + 3r) (3k + 8 + 3r) + 36ak = 0
Recursion relation that defines series solution to ODE
36ak
ak+4 = − (3k+16+3r)(3k+8+3r)
Recursion relation for r = − 43
36ak
ak+4 = − (3k+12)(3k+4)
Solution
for r = − 43
∞
P
36ak
k− 43
y(x) =
ak x , ak+4 = − (3k+12)(3k+4) , a1 = 0, a2 = 0, a3 = 0
k=0
•
•
Recursion relation for r = 43
36ak
ak+4 = − (3k+20)(3k+12)
Solution
for r = 43
∞
P
36ak
k+ 43
y(x) =
ak x , ak+4 = − (3k+20)(3k+12) , a1 = 0, a2 = 0, a3 = 0
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
4
4
36ak
y(x) =
ak xk− 3 +
bk xk+ 3 , ak+4 = − (3k+12)(3k+4)
, a1 = 0, a2 = 0, a3 = 0, bk+4 = − (3k+
k=0
k=0
3 Mathematica. Time used: 0.003 (sec). Leaf size: 38
ode=9*x^2*D[y[x],{x,2}]+9*x*D[y[x],x]+(36*x^4-16)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
3x4
c
1
−
4
2
4
3x
y(x) → c1 1 −
x4/3 +
4/3
20
x
chapter 2 . book solved problems
331
3 Sympy. Time used: 0.323 (sec). Leaf size: 27
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(9*x**2*Derivative(y(x), (x, 2)) + 9*x*Derivative(y(x), x) + (36*x**4 16)*y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) =
4
C2 1 − 3x4
x
4
3
4
+ C 1 x 3 + O x6
chapter 2 . book solved problems
2.4.4
332
Problem 4
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
Internal problem ID [8606]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 4
Date solved : Tuesday, November 11, 2025 at 12:04:05 PM
CAS classification : [[_Emden, _Fowler]]
y 00 + yx = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.39)
(2.40)
(2)
(3)
chapter 2 . book solved problems
333
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = −yx
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −y 0 x − y
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= yx2 − 2y 0
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= x(y 0 x + 4y)
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= −yx3 + 6y 0 x + 4y
(7)
chapter 2 . book solved problems
334
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = 0
F1 = −y(0)
F2 = −2y 0 (0)
F3 = 0
F4 = 4y(0)
Substituting all the above in (7) and simplifying gives the solution as
x3
1 4 0
y = 1−
y(0) + x − x y (0) + O x6
6
12
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
y 00 =
∞
X
n=1
∞
X
nan xn−1
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
∞
X
n(n − 1) an xn−2 +
n=2
∞
X
!
an xn x = 0
(1)
n=0
Which simplifies to
∞
X
!
n(n − 1) an xn−2
+
n=2
∞
X
!
x1+n an
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an xn−2 =
n =2
∞
X
(n + 2) an+2 (1 + n) xn
n=0
∞
X
x
1+n
n =0
an =
∞
X
an−1 xn
n=1
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
(n + 2) an+2 (1 + n) xn
n=0
+
∞
X
!
an−1 xn
=0
(3)
n=1
For 1 ≤ n, the recurrence equation is
(n + 2) an+2 (1 + n) + an−1 = 0
(4)
chapter 2 . book solved problems
335
Solving for an+2 , gives
an+2 = −
an−1
(n + 2) (1 + n)
For n = 1 the recurrence equation gives
6a3 + a0 = 0
Which after substituting the earlier terms found becomes
a3 = −
a0
6
For n = 2 the recurrence equation gives
12a4 + a1 = 0
Which after substituting the earlier terms found becomes
a4 = −
a1
12
For n = 3 the recurrence equation gives
20a5 + a2 = 0
Which after substituting the earlier terms found becomes
a5 = 0
For n = 4 the recurrence equation gives
30a6 + a3 = 0
Which after substituting the earlier terms found becomes
a6 =
a0
180
For n = 5 the recurrence equation gives
42a7 + a4 = 0
Which after substituting the earlier terms found becomes
a7 =
a1
504
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
1
1
y = a0 + a1 x − a0 x 3 − a1 x 4 + . . .
6
12
(5)
chapter 2 . book solved problems
336
Collecting terms, the solution becomes
x3
1 4
y = 1−
a0 + x − x a1 + O x 6
6
12
(3)
At x = 0 the solution above becomes
x3
1 4 0
y = 1−
y(0) + x − x y (0) + O x6
6
12
3 Maple. Time used: 0.006 (sec). Leaf size: 29
Order:=6;
ode:=x*y(x)+diff(diff(y(x),x),x) = 0;
dsolve(ode,y(x),type='series',x=0);
x3
1 4 0
y = 1−
y(0) + x − x y (0) + O x6
6
12
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
•
Let’s solve
d d
y(x) + xy(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Assume series solution for y(x)
∞
P
y(x) =
ak x k
k=0
Rewrite ODE with series expansions
◦ Convert x · y(x) to series expansion
∞
P
x · y(x) =
ak xk+1
k=0
◦ Shift index using k− >k − 1
∞
P
x · y(x) =
ak−1 xk
k=1
d d
◦ Convert dx
y(x) to series expansion
dx
∞
P
d d
y(x) =
ak k(k − 1) xk−2
dx dx
k=2
◦ Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=0
chapter 2 . book solved problems
337
Rewrite
ODE with series expansions
∞
P
2a2 +
(ak+2 (k + 2) (k + 1) + ak−1 ) xk = 0
k=1
•
•
•
•
Each term must be 0
2a2 = 0
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 + ak−1 = 0
Shift index using k− >k + 1
(k + 1)2 + 3k + 5 ak+3 + ak = 0
Recursion
relation that defines the series solution
to the ODE
∞
P
ak
y(x) =
ak xk , ak+3 = − k2 +5k+6
, 2a2 = 0
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 28
ode=D[y[x],{x,2}]+x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
x4
x−
12
3 Sympy. Time used: 0.158 (sec). Leaf size: 24
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*y(x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
y(x) = C2
x3
1−
6
x3
+ c1 1 −
6
x3
+ C1 x 1 −
+ O x6
12
chapter 2 . book solved problems
2.4.5
338
Problem 5
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
Internal problem ID [8607]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 5
Date solved : Tuesday, November 11, 2025 at 12:04:06 PM
CAS classification : [[_Emden, _Fowler]]
4xy 00 + 4y 0 + y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
4xy 00 + 4y 0 + y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
1
q(x) =
4x
p(x) =
Table 2.60: Table p(x), q(x) singularites.
p(x) = x1
singularity
type
x=0
“regular”
1
q(x) = 4x
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
4xy 00 + 4y 0 + y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
339
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
4
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x + 4
n=0
!
(n + r) an xn+r−1
+
n=0
∞
X
!
an xn+r
= 0 (1)
n=0
Which simplifies to
∞
X
!
4xn+r−1 an (n + r) (n + r − 1)
∞
X
+
n=0
!
4(n + r) an xn+r−1
+
n=0
∞
X
!
an xn+r
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
an x
n+r
n =0
=
∞
X
an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
4xn+r−1 an (n+r) (n+r −1) +
n=0
∞
X
!
4(n+r) an xn+r−1 +
n=0
∞
X
!
an−1 xn+r−1
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
4xn+r−1 an (n + r) (n + r − 1) + 4(n + r) an xn+r−1 = 0
When n = 0 the above becomes
4x−1+r a0 r(−1 + r) + 4ra0 x−1+r = 0
Or
4x−1+r r(−1 + r) + 4r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
4x−1+r r2 = 0
Since the above is true for all x then the indicial equation becomes
4r2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = 0
Since a0 6= 0 then the indicial equation becomes
4x−1+r r2 = 0
= 0 (2B)
chapter 2 . book solved problems
340
Solving for r gives the roots of the indicial equation as [0, 0].
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
y1 (x) =
∞
X
an xn+r
(1A)
n=0
Now the second solution y2 is found using
∞
X
y2 (x) = y1 (x) ln (x) +
!
bn xn+r
(1B)
n=1
Then the general solution will be
y = c1 y1 (x) + c2 y2 (x)
In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), a0 is never zero, and is arbitrary
and is typically taken as a0 = 1, and {c1 , c2 } are two arbitray constants of integration which
can be found from initial conditions. We start by finding the first solution y1 (x). Eq (2B)
derived above is now used to find all an coefficients. The case n = 0 is skipped since it was
used to find the roots of the indicial equation. a0 is arbitrary and taken as a0 = 1. For 1 ≤ n
the recursive equation is
4an (n + r) (n + r − 1) + 4an (n + r) + an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an = −
an−1
2
4 (n + 2nr + r2 )
Which for the root r = 0 becomes
(4)
an−1
(5)
4n2
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
an = −
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 = −
1
4 (r + 1)2
Which for the root r = 0 becomes
a1 = −
1
4
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
For n = 2, using the above recursive equation gives
a2 =
1
16 (r + 1)2 (r + 2)2
Which for the root r = 0 becomes
a2 =
1
64
chapter 2 . book solved problems
341
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
a2
1
16(r+1)2 (r+2)2
1
64
For n = 3, using the above recursive equation gives
a3 = −
1
64 (r + 1) (r + 2)2 (r + 3)2
2
Which for the root r = 0 becomes
a3 = −
1
2304
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
a2
1
16(r+1)2 (r+2)2
1
− 64(r+1)2 (r+2)
2
(r+3)2
1
64
a3
1
− 2304
For n = 4, using the above recursive equation gives
a4 =
1
256 (r + 1) (r + 2)2 (r + 3)2 (r + 4)2
2
Which for the root r = 0 becomes
a4 =
1
147456
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
a2
1
16(r+1)2 (r+2)2
1
− 64(r+1)2 (r+2)
2
(r+3)2
1
256(r+1)2 (r+2)2 (r+3)2 (r+4)2
1
64
a3
a4
1
− 2304
1
147456
For n = 5, using the above recursive equation gives
a5 = −
1
1024 (r + 1) (r + 2) (r + 3)2 (r + 4)2 (r + 5)2
2
2
Which for the root r = 0 becomes
a5 = −
1
14745600
And the table now becomes
n
a0
a1
an,r
1
1
− 4(r+1)
2
an
1
− 14
a2
1
16(r+1)2 (r+2)2
1
− 64(r+1)2 (r+2)
2
(r+3)2
1
256(r+1)2 (r+2)2 (r+3)2 (r+4)2
1
− 1024(r+1)2 (r+2)2 (r+3)
2
(r+4)2 (r+5)2
1
64
a3
a4
a5
1
− 2304
1
147456
1
− 14745600
chapter 2 . book solved problems
342
Using the above table, then the first solution y1 (x) becomes
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
=1−
x x2
x3
x4
x5
+
−
+
−
+ O x6
4 64 2304 147456 14745600
Now the second solution is found. The second solution is given by
!
∞
X
y2 (x) = y1 (x) ln (x) +
bn xn+r
n=1
Where bn is found using
d
an,r
dr
And the above is then evaluated at r = 0. The above table for an,r is used for this purpose.
Computing the derivatives gives the following table
bn =
n
b0
b1
bn,r
1
1
− 4(r+1)
2
an
1
− 14
b2
1
16(r+1)2 (r+2)2
1
− 64(r+1)2 (r+2)
2
(r+3)2
1
256(r+1)2 (r+2)2 (r+3)2 (r+4)2
1
− 1024(r+1)2 (r+2)2 (r+3)
2
(r+4)2 (r+5)2
1
64
b3
b4
b5
1
− 2304
1
147456
1
− 14745600
d
bn,r = dr
an,r
N/A since bn starts from 1
bn (r = 0)
N/A
1
2(r+1)3
−2r−3
8(r+1)3 (r+2)3
3r2 +12r+11
32(r+1)3 (r+2)3 (r+3)3
−2r3 −15r2 −35r−25
64(r+1)3 (r+2)3 (r+3)3 (r+4)3
5r4 +60r3 +255r2 +450r+274
512(r+1)3 (r+2)3 (r+3)3 (r+4)3 (r+5)3
1
2
3
− 64
11
6912
25
− 884736
137
442368000
The above table gives all values of bn needed. Hence the second solution is
y2 (x) = y1 (x) ln (x) + b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x x2
x3
x4
x5
6
= 1− +
−
+
−
+O x
ln (x)
4 64 2304 147456 14745600
x 3x2 11x3
25x4
137x5
+ −
+
−
+
+ O x6
2
64
6912 884736 442368000
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x x2
x3
x4
x5
6
= c1 1 − +
−
+
−
+O x
4 64 2304 147456 14745600
x x2
x3
x4
x5
x 3x2
6
+ c2
1− +
−
+
−
+O x
ln (x) + −
4 64 2304 147456 14745600
2
64
3
4
5
11x
25x
137x
6
+
−
+
+O x
6912 884736 442368000
Hence the final solution is
y = yh
x x2
x3
x4
x5
6
= c1 1 − +
−
+
−
+O x
4 64 2304 147456 14745600
x x2
x3
x4
x5
x 3x2 11x3
6
+ c2
1− +
−
+
−
+O x
ln (x) + −
+
4 64 2304 147456 14745600
2
64
6912
4
5
25x
137x
6
−
+
+O x
884736 442368000
chapter 2 . book solved problems
3 Maple. Time used: 0.036 (sec). Leaf size: 44
Order:=6;
ode:=4*diff(diff(y(x),x),x)*x+y(x)+4*diff(y(x),x) = 0;
dsolve(ode,y(x),type='series',x=0);
343
1
1 2
1 3
1
1
4
5
6
y = (c2 ln (x) + c1 ) 1 − x + x −
x +
x −
x +O x
4
64
2304
147456
14745600
1
3
11 3
25
137
+
x − x2 +
x −
x4 +
x 5 + O x 6 c2
2
64
6912
884736
442368000
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
4x dx
y(x) + 4 dx
y(x) + y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
d d
y(x) = − y(x)
− dx x
dx dx
4x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
d d
dx
y(x)
+
+ y(x)
=0
dx dx
x
4x
◦
◦
Check to see if x0 = 0 is a regular singular point
Define functions
1
P2 (x) = x1 , P3 (x) = 4x
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=1
x=0
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
4x dx
y(x)
+ 4 dx
y(x) + y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d
Convert dx
y(x) to series expansion
chapter 2 . book solved problems
d
y(x) =
dx
◦
◦
∞
P
344
ak (k + r) xk+r−1
k=0
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + 1 + r) xk+r
dx
k=−1
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODEwith series expansions
∞
k+r
P
2
2 −1+r
4a0 r x
+
4ak+1 (k + 1 + r) + ak x
=0
k=0
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
4r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term in the series must be 0, giving the recursion relation
4ak+1 (k + 1)2 + ak = 0
Recursion relation that defines series solution to ODE
ak
ak+1 = − 4(k+1)
2
•
Recursion relation for r = 0
ak
ak+1 = − 4(k+1)
2
•
Solution
for r = 0
∞
P
a
k
y(x) =
ak xk , ak+1 = − 4(k+1)
2
k=0
3 Mathematica. Time used: 0.003 (sec). Leaf size: 117
ode=4*x*D[y[x],{x,2}]+4*D[y[x],x]+y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
x5
x4
x3
x2 x
137x5
25x4
11x3
−
+
−
+
− + 1 + c2
−
+
14745600 147456 2304 64 4
442368000 884736 6912
2
5
4
3
3x
x
x
x
x2 x
x
−
+ −
+
−
+
− + 1 log(x) +
64
14745600 147456 2304 64 4
2
3 Sympy. Time used: 0.204 (sec). Leaf size: 32
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(4*x*Derivative(y(x), (x, 2)) + y(x) + 4*Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C1 −
x5
x4
x3
x2 x
+
−
+
− + 1 + O x6
14745600 147456 2304 64 4
chapter 2 . book solved problems
2.4.6
345
Problem 6
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
Internal problem ID [8608]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 6
Date solved : Tuesday, November 11, 2025 at 12:04:06 PM
CAS classification : [[_Emden, _Fowler]]
xy 00 + y 0 + 36y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + y 0 + 36y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
36
q(x) =
x
p(x) =
Table 2.62: Table p(x), q(x) singularites.
p(x) = x1
singularity
type
x=0
“regular”
q(x) = 36
x
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + y 0 + 36y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
346
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x +
n=0
!
(n + r) an xn+r−1
+ 36
n=0
∞
X
!
an xn+r
= 0 (1)
n=0
Which simplifies to
∞
X
!
xn+r−1 an (n + r) (n + r − 1)
∞
X
+
n=0
!
(n + r) an xn+r−1
+
n=0
∞
X
!
36an xn+r
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
36an x
n+r
n =0
=
∞
X
36an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n+r) (n+r −1) +
n=0
∞
X
!
(n+r) an xn+r−1 +
∞
X
n=0
!
36an−1 xn+r−1
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + (n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
x−1+r r2 = 0
Since the above is true for all x then the indicial equation becomes
r2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r2 = 0
= 0 (2B)
chapter 2 . book solved problems
347
Solving for r gives the roots of the indicial equation as [0, 0].
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
y1 (x) =
∞
X
an xn+r
(1A)
n=0
Now the second solution y2 is found using
∞
X
y2 (x) = y1 (x) ln (x) +
!
bn xn+r
(1B)
n=1
Then the general solution will be
y = c1 y1 (x) + c2 y2 (x)
In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), a0 is never zero, and is arbitrary
and is typically taken as a0 = 1, and {c1 , c2 } are two arbitray constants of integration which
can be found from initial conditions. We start by finding the first solution y1 (x). Eq (2B)
derived above is now used to find all an coefficients. The case n = 0 is skipped since it was
used to find the roots of the indicial equation. a0 is arbitrary and taken as a0 = 1. For 1 ≤ n
the recursive equation is
an (n + r) (n + r − 1) + an (n + r) + 36an−1 = 0
(3)
Solving for an from recursive equation (4) gives
36an−1
an = − 2
n + 2nr + r2
Which for the root r = 0 becomes
(4)
36an−1
(5)
n2
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
an = −
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 = −
36
(r + 1)2
Which for the root r = 0 becomes
a1 = −36
And the table now becomes
n
a0
a1
an,r
1
36
− (r+1)
2
an
1
−36
For n = 2, using the above recursive equation gives
a2 =
1296
(r + 1)2 (r + 2)2
Which for the root r = 0 becomes
a2 = 324
chapter 2 . book solved problems
348
And the table now becomes
n
a0
a1
an,r
1
36
− (r+1)
2
an
1
−36
a2
1296
(r+1)2 (r+2)2
324
For n = 3, using the above recursive equation gives
a3 = −
46656
(r + 1) (r + 2)2 (r + 3)2
2
Which for the root r = 0 becomes
a3 = −1296
And the table now becomes
n
a0
a1
an,r
1
36
− (r+1)
2
an
1
−36
a2
1296
(r+1)2 (r+2)2
46656
− (r+1)2 (r+2)
2
(r+3)2
324
a3
−1296
For n = 4, using the above recursive equation gives
a4 =
1679616
(r + 1) (r + 2)2 (r + 3)2 (r + 4)2
2
Which for the root r = 0 becomes
a4 = 2916
And the table now becomes
n
a0
a1
an,r
1
36
− (r+1)
2
an
1
−36
a2
1296
(r+1)2 (r+2)2
46656
− (r+1)2 (r+2)
2
(r+3)2
1679616
(r+1)2 (r+2)2 (r+3)2 (r+4)2
324
a3
a4
−1296
2916
For n = 5, using the above recursive equation gives
a5 = −
60466176
(r + 1) (r + 2) (r + 3)2 (r + 4)2 (r + 5)2
2
2
Which for the root r = 0 becomes
a5 = −
104976
25
And the table now becomes
n
a0
a1
an,r
1
36
− (r+1)
2
an
1
−36
a2
1296
(r+1)2 (r+2)2
46656
− (r+1)2 (r+2)
2
(r+3)2
1679616
(r+1)2 (r+2)2 (r+3)2 (r+4)2
− (r+1)2 (r+2)260466176
(r+3)2 (r+4)2 (r+5)2
324
a3
a4
a5
−1296
2916
− 104976
25
chapter 2 . book solved problems
349
Using the above table, then the first solution y1 (x) becomes
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
= 2916x4 − 1296x3 + 324x2 − 36x + 1 −
104976x5
+ O x6
25
Now the second solution is found. The second solution is given by
!
∞
X
y2 (x) = y1 (x) ln (x) +
bn xn+r
n=1
Where bn is found using
d
an,r
dr
And the above is then evaluated at r = 0. The above table for an,r is used for this purpose.
Computing the derivatives gives the following table
bn =
n
b0
b1
bn,r
1
36
− (r+1)
2
an
1
−36
b2
1296
(r+1)2 (r+2)2
46656
− (r+1)2 (r+2)
2
(r+3)2
1679616
(r+1)2 (r+2)2 (r+3)2 (r+4)2
− (r+1)2 (r+2)260466176
(r+3)2 (r+4)2 (r+5)2
324
b3
b4
b5
−1296
2916
− 104976
25
d
bn,r = dr
an,r
N/A since bn starts from 1
72
(r+1)3
−5184r−7776
(r+1)3 (r+2)3
279936r2 +1119744r+1026432
3
(r+1)3 (r+2)3 (r+3)
13436928 r2 +5r+5 r+ 52
− (r+1)3 (r+2)3 (r+3)3 (r+4)3
604661760r4 +7255941120r3 +30837749760r2 +54419558400r+33135464448
(r+1)3 (r+2)3 (r+3)3 (r+4)3 (r+5)3
The above table gives all values of bn needed. Hence the second solution is
y2 (x) = y1 (x) ln (x) + b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
104976x5
4
3
2
6
= 2916x − 1296x + 324x − 36x + 1 −
+O x
ln (x)
25
2396952x5
− 12150x4 + 4752x3 − 972x2 + 72x +
+ O x6
125
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
104976x5
4
3
2
6
= c1 2916x − 1296x + 324x − 36x + 1 −
+O x
25
104976x5
4
3
2
6
+ c2
2916x − 1296x + 324x − 36x + 1 −
+O x
ln (x) − 12150x4
25
2396952x5
3
2
6
+ 4752x − 972x + 72x +
+O x
125
Hence the final solution is
y = yh
104976x5
4
3
2
6
= c1 2916x − 1296x + 324x − 36x + 1 −
+O x
25
104976x5
4
3
2
6
+ c2
2916x − 1296x + 324x − 36x + 1 −
+O x
ln (x) − 12150x4
25
2396952x5
3
2
6
+ 4752x − 972x + 72x +
+O x
125
bn (r
N/A
72
−97
4752
−12
23969
125
chapter 2 . book solved problems
3 Maple. Time used: 0.030 (sec). Leaf size: 44
Order:=6;
ode:=diff(diff(y(x),x),x)*x+diff(y(x),x)+36*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
350
104976 5
2
3
4
6
y = (c2 ln (x) + c1 ) 1 − 36x + 324x − 1296x + 2916x −
x +O x
25
2396952 5
+ 72x − 972x2 + 4752x3 − 12150x4 +
x + O x 6 c2
125
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
x dx
y(x) + dx
y(x) + 36y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
d d
y(x) = − 36y(x)
− dx x
dx dx
x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
d d
dx
y(x)
+
+ 36y(x)
=0
dx dx
x
x
◦
◦
Check to see if x0 = 0 is a regular singular point
Define functions
P2 (x) = x1 , P3 (x) = 36
x
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=1
x=0
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x)
+ dx
y(x) + 36y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d
Convert dx
y(x) to series expansion
chapter 2 . book solved problems
d
y(x) =
dx
◦
◦
∞
P
351
ak (k + r) xk+r−1
k=0
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + 1 + r) xk+r
dx
k=−1
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
∞
k+r
P
2
2 −1+r
a0 r x
+
ak+1 (k + 1 + r) + 36ak x
=0
k=0
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1)2 + 36ak = 0
Recursion relation that defines series solution to ODE
36ak
ak+1 = − (k+1)
2
•
Recursion relation for r = 0
36ak
ak+1 = − (k+1)
2
•
Solution
for r = 0
∞
P
36a
y(x) =
ak xk , ak+1 = − (k+1)k 2
k=0
3 Mathematica. Time used: 0.003 (sec). Leaf size: 93
ode=x*D[y[x],{x,2}]+D[y[x],x]+36*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
104976x5
2396952x5
4
3
2
y(x) → c1 −
+ 2916x − 1296x + 324x − 36x + 1 + c2
− 12150x4
25
125
104976x5
3
2
4
3
2
+ 4752x − 972x + −
+ 2916x − 1296x + 324x − 36x + 1 log(x) + 72x
25
3 Sympy. Time used: 0.198 (sec). Leaf size: 34
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*Derivative(y(x), (x, 2)) + 36*y(x) + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
104976x5
4
3
2
y(x) = C1 −
+ 2916x − 1296x + 324x − 36x + 1 + O x6
25
chapter 2 . book solved problems
2.4.7
352
Problem 7
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
Internal problem ID [8609]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 7
Date solved : Tuesday, November 11, 2025 at 12:04:07 PM
CAS classification : [[_Emden, _Fowler]]
y 00 + k 2 x2 y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.41)
(2.42)
(2)
(3)
chapter 2 . book solved problems
353
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = −k 2 x2 y
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −k 2 x(y 0 x + 2y)
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= k 2 x4 k 2 y − 4y 0 x − 2y
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= k 2 k 2 x4 − 6 y 0 + 8x3 k 2 y
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= −x2 −12y 0 x + y k 2 x4 − 30 k 4
(7)
chapter 2 . book solved problems
354
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = 0
F1 = 0
F2 = −2k 2 y(0)
F3 = −6y 0 (0) k 2
F4 = 0
Substituting all the above in (7) and simplifying gives the solution as
y=
k 2 x4
1−
12
1 2 5 0
y(0) + x − k x y (0) + O x6
20
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
n(n − 1) an xn−2 + k 2 x2
an x n = 0
n=2
(1)
n=0
Which simplifies to
∞
X
∞
X
!
n(n − 1) an xn−2
+
n=2
!
xn+2 k 2 an
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an xn−2 =
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
∞
X
x
n+2 2
k an =
n =0
∞
X
an−2 k 2 xn
n=2
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
(n + 2) an+2 (n + 1) xn
n=0
+
∞
X
!
an−2 k 2 xn
=0
(3)
n=2
For 2 ≤ n, the recurrence equation is
(n + 2) an+2 (n + 1) + an−2 k 2 = 0
(4)
chapter 2 . book solved problems
355
Solving for an+2 , gives
an+2 = −
an−2 k 2
(n + 2) (n + 1)
(5)
For n = 2 the recurrence equation gives
a0 k 2 + 12a4 = 0
Which after substituting the earlier terms found becomes
a4 = −
a0 k 2
12
For n = 3 the recurrence equation gives
a1 k 2 + 20a5 = 0
Which after substituting the earlier terms found becomes
a5 = −
a1 k 2
20
For n = 4 the recurrence equation gives
a2 k 2 + 30a6 = 0
Which after substituting the earlier terms found becomes
a6 = 0
For n = 5 the recurrence equation gives
a3 k 2 + 42a7 = 0
Which after substituting the earlier terms found becomes
a7 = 0
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
y = a0 + a1 x −
1
1
a0 k 2 x 4 − a1 k 2 x 5 + . . .
12
20
Collecting terms, the solution becomes
y=
k 2 x4
1−
12
1 2 5
a0 + x − k x a1 + O x 6
20
At x = 0 the solution above becomes
y=
k 2 x4
1−
12
1 2 5 0
y(0) + x − k x y (0) + O x6
20
(3)
chapter 2 . book solved problems
3 Maple. Time used: 0.006 (sec). Leaf size: 35
Order:=6;
ode:=diff(diff(y(x),x),x)+k^2*x^2*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
y=
k 2 x4
1−
12
356
1 2 5 0
y(0) + x − k x y (0) + O x6
20
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
•
Let’s solve
d d
y(x) + k 2 x2 y(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Assume series solution for y(x)
∞
P
y(x) =
ak x k
k=0
Rewrite ODE with series expansions
◦ Convert x2 · y(x) to series expansion
∞
P
x2 · y(x) =
ak xk+2
k=0
◦ Shift index using k− >k − 2
∞
P
x2 · y(x) =
ak−2 xk
k=2
d d
◦ Convert dx
y(x) to series expansion
dx
∞
P
d d
y(x) =
ak k(k − 1) xk−2
dx dx
k=2
◦ Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=0
Rewrite ODEwith series expansions
∞
P
2
k
6a3 x + 2a2 +
(ak+2 (k + 2) (k + 1) + k ak−2 ) x = 0
k=2
•
•
•
•
The coefficients of each power of x must be 0
[2a2 = 0, 6a3 = 0]
Solve for the dependent coefficient(s)
{a2 = 0, a3 = 0}
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 + k 2 ak−2 = 0
Shift index using k− >k + 2
chapter 2 . book solved problems
•
357
(k + 2)2 + 3k + 8 ak+4 + k 2 ak = 0
Recursion
relation that defines the series solution to the ODE
∞
P
k2 ak
y(x) =
ak xk , ak+4 = − k2 +7k+12
, a2 = 0, a3 = 0
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 34
ode=D[y[x],{x,2}]+k^2*x^2*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
k 2 x5
x−
20
k 2 x4
+ c1 1 −
12
3 Sympy. Time used: 0.196 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
k = symbols("k")
y = Function("y")
ode = Eq(k**2*x**2*y(x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
2 4
2 4
k x
k x
y(x) = C2 −
+ 1 + C1 x −
+ 1 + O x6
12
20
chapter 2 . book solved problems
2.4.8
358
Problem 8
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
Internal problem ID [8610]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 8
Date solved : Tuesday, November 11, 2025 at 12:04:07 PM
CAS classification : [[_Emden, _Fowler]]
y 00 + x4 k 2 y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.43)
(2.44)
(2)
(3)
chapter 2 . book solved problems
359
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = −x4 k 2 y
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −k 2 x3 (y 0 x + 4y)
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= x2 k 2 −8y 0 x + y k 2 x6 − 12
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= y 0 k 2 x7 + 16x6 k 2 y − 36y 0 x − 24y x k 2
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= k 2 −x12 k 4 y + 24y 0 k 2 x7 + 148x6 k 2 y − 96y 0 x − 24y
(7)
chapter 2 . book solved problems
360
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = 0
F1 = 0
F2 = 0
F3 = 0
F4 = −24k 2 y(0)
Substituting all the above in (7) and simplifying gives the solution as
y = y(0) + xy 0 (0) + O x6
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
y 00 =
∞
X
n=1
∞
X
nan xn−1
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
n(n − 1) an xn−2 + x4 k 2
an x n = 0
n=2
(1)
n=0
Which simplifies to
∞
X
!
n(n − 1) an x
n−2
∞
X
+
n=2
!
x
n+4 2
k an
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
∞
X
x
n+4 2
k an =
n =0
∞
X
an−4 k 2 xn
n=4
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
(n + 2) an+2 (n + 1) xn
n=0
+
∞
X
!
an−4 k 2 xn
=0
(3)
n=4
For 4 ≤ n, the recurrence equation is
(n + 2) an+2 (n + 1) + an−4 k 2 = 0
(4)
chapter 2 . book solved problems
361
Solving for an+2 , gives
an+2 = −
an−4 k 2
(n + 2) (n + 1)
(5)
For n = 4 the recurrence equation gives
a0 k 2 + 30a6 = 0
Which after substituting the earlier terms found becomes
a6 = −
a0 k 2
30
For n = 5 the recurrence equation gives
a1 k 2 + 42a7 = 0
Which after substituting the earlier terms found becomes
a7 = −
a1 k 2
42
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
y = a1 x + a0 + . . .
Collecting terms, the solution becomes
y = a1 x + a0 + O x 6
(3)
At x = 0 the solution above becomes
y = y(0) + xy 0 (0) + O x6
3 Maple. Time used: 0.005 (sec). Leaf size: 15
Order:=6;
ode:=diff(diff(y(x),x),x)+k^2*x^4*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
y = y(0) + y 0 (0) x + O x6
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
chapter 2 . book solved problems
362
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
•
Let’s solve
d d
y(x) + k 2 x4 y(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Assume series solution for y(x)
∞
P
y(x) =
ak x k
k=0
Rewrite ODE with series expansions
◦ Convert x4 · y(x) to series expansion
∞
P
x4 · y(x) =
ak xk+4
k=0
◦ Shift index using k− >k − 4
∞
P
x4 · y(x) =
ak−4 xk
k=4
d d
◦ Convert dx
y(x) to series expansion
dx
∞
P
d d
y(x) =
ak k(k − 1) xk−2
dx dx
k=2
◦ Shift index using k− >k + 2
∞
P
d d
y(x)
=
ak+2 (k + 2) (k + 1) xk
dx dx
k=0
Rewrite ODE with series expansions
∞
P
3
2
2
k
20a5 x + 12a4 x + 6a3 x + 2a2 +
(ak+2 (k + 2) (k + 1) + k ak−4 ) x = 0
k=4
•
•
•
•
•
The coefficients of each power of x must be 0
[2a2 = 0, 6a3 = 0, 12a4 = 0, 20a5 = 0]
Solve for the dependent coefficient(s)
{a2 = 0, a3 = 0, a4 = 0, a5 = 0}
Each term in the series must be 0, giving the recursion relation
(k 2 + 3k + 2) ak+2 + k 2 ak−4 = 0
Shift index using k− >k + 4
(k + 4)2 + 3k + 14 ak+6 + k 2 ak = 0
Recursion
relation that defines the series solution to the ODE ∞
P
k2 ak
y(x) =
ak xk , ak+6 = − k2 +11k+30
, a2 = 0, a3 = 0, a4 = 0, a5 = 0
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 10
ode=D[y[x],{x,2}]+k^2*x^4*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2 x + c1
chapter 2 . book solved problems
3 Sympy. Time used: 0.193 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
k = symbols("k")
y = Function("y")
ode = Eq(k**2*x**4*y(x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
2 6
2 6
k x
k x
y(x) = C2 −
+ 1 + C1 x −
+ 1 + O x6
30
42
363
chapter 2 . book solved problems
2.4.9
364
Problem 9
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373
Internal problem ID [8611]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.5. Bessel
Functions Y(x). General Solution page 200
Problem number : 9
Date solved : Tuesday, November 11, 2025 at 12:04:08 PM
CAS classification : [_Lienard]
xy 00 − 5y 0 + yx = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 − 5y 0 + yx = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
p(x) = −
5
x
q(x) = 1
Table 2.66: Table p(x), q(x) singularites.
p(x) = − x5
singularity
type
x=0
“regular”
q(x) = 1
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 − 5y 0 + yx = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
365
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x − 5
n=0
!
(n + r) an xn+r−1
∞
X
+
n=0
!
an xn+r
x = 0 (1)
n=0
Which simplifies to
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
−5(n + r) an x
n+r−1
+
n =0
∞
X
!
x1+n+r an
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
x
1+n+r
n =0
an =
∞
X
an−2 xn+r−1
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
−5(n + r) an x
n+r−1
+
n =0
∞
X
!
an−2 xn+r−1
n=2
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) − 5(n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) − 5ra0 x−1+r = 0
Or
x−1+r r(−1 + r) − 5r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
r x−1+r (−6 + r) = 0
Since the above is true for all x then the indicial equation becomes
r(−6 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 6
r2 = 0
Since a0 6= 0 then the indicial equation becomes
r x−1+r (−6 + r) = 0
= 0 (2B)
chapter 2 . book solved problems
366
Solving for r gives the roots of the indicial equation as [6, 0].
Since r1 − r2 = 6 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) + xr2
!
bn x n
n=0
Or
∞
X
y1 (x) = x6
!
an x n
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
an xn+6
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x n
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) − 5an (n + r) + an−2 = 0
(3)
Solving for an from recursive equation (4) gives
an−2
an = − 2
n + 2nr + r2 − 6n − 6r
(4)
Which for the root r = 6 becomes
an = −
an−2
n (n + 6)
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 6
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
1
a2 = − 2
r − 2r − 8
Which for the root r = 6 becomes
a2 = −
And the table now becomes
1
16
chapter 2 . book solved problems
n
a0
a1
a2
367
an,r
1
0
1
− r2 −2r−8
an
1
0
1
− 16
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
1
− r2 −2r−8
0
an
1
0
1
− 16
0
For n = 4, using the above recursive equation gives
1
a4 =
r4 − 20r2 + 64
Which for the root r = 6 becomes
a4 =
1
640
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
1
− r2 −2r−8
0
an
1
0
1
− 16
0
1
r4 −20r2 +64
1
640
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
1
− r2 −2r−8
0
an
1
0
1
− 16
0
1
r4 −20r2 +64
1
640
0
0
For n = 6, using the above recursive equation gives
a6 = −
1
(r4 − 20r2 + 64) r (r + 6)
Which for the root r = 6 becomes
a6 = −
And the table now becomes
1
46080
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
a6
368
an,r
1
0
1
− r2 −2r−8
0
an
1
0
1
− 16
0
1
r4 −20r2 +64
1
640
0
1
− (r4 −20r2 +64)r(r+6)
0
1
− 46080
Using the above table, then the solution y1 (x) is
y1 (x) = x6 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 + a7 x7 . . .
x2
x4
x6
6
7
=x 1−
+
−
+O x
16 640 46080
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 6. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a6 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a6
=−
1
(r4 − 20r2 + 64) r (r + 6)
Therefore
lim −
r→r2
1
1
=
lim
−
(r4 − 20r2 + 64) r (r + 6) r→0 (r4 − 20r2 + 64) r (r + 6)
= undefined
Since the limit does not exist then the log term is needed. Therefore the second solution has
the form
!
∞
X
y2 (x) = Cy1 (x) ln (x) +
bn xn+r2
n=0
Therefore
d
Cy1 (x)
y2 (x) = Cy10 (x) ln (x) +
+
dx
x
Cy1 (x)
= Cy10 (x) ln (x) +
+
x
∞
X
bn xn+r2 (n + r2 )
n=0
∞
X
!
x
!
x−1+n+r2 bn (n + r2 )
n=0
∞
d2
2Cy10 (x) Cy1 (x) X
00
y2 (x) = Cy1 (x) ln (x) +
−
+
dx2
x
x2
n=0
2Cy10 (x) Cy1 (x)
= Cy100 (x) ln (x) +
−
+
x
x2
∞
X
n=0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!
x−2+n+r2 bn (n + r2 ) (−1 + n + r2 )
!
chapter 2 . book solved problems
369
Substituting these back into the given ode xy 00 − 5y 0 + yx = 0 gives
∞
2Cy10 (x) Cy1 (x) X
00
Cy1 (x) ln (x) +
−
+
x
x2
n=0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!
∞
X
5Cy1 (x)
bn xn+r2 (n + r2 )
0
− 5Cy1 (x) ln (x) −
−5
x
x
n=0
!!
∞
X
+ Cy1 (x) ln (x) +
bn xn+r2
x=0
!!
x
n=0
Which can be written as
2y10 (x) y1 (x)
− 2
x
x
!!
∞
2
n+r
n+r
X bn x 2 (n + r2 )
bn x 2 (n + r2 )
−
x
2
2
x
x
n=0
!
!
∞
∞
n+r2
X
X
b
x
(n
+
r
)
n
2
bn xn+r2 x − 5
=0
x
n=0
n=0
(y1 (x) x + y100 (x) x − 5y10 (x)) ln (x) + x
+
+
5y1 (x)
−
x
C
(7)
But since y1 (x) is a solution to the ode, then
y1 (x) x + y100 (x) x − 5y10 (x) = 0
Eq (7) simplifes to
0
2y1 (x) y1 (x)
5y1 (x)
x
− 2
−
C
x
x
x
!!
∞
2
n+r2
n+r2
X
bn x
(n + r2 )
bn x
(n + r2 )
+
−
x
2
x
x2
n=0
!
!
∞
∞
n+r2
X
X
b
x
(n
+
r
)
n
2
+
bn xn+r2 x − 5
=0
x
n=0
n=0
∞
P
Substituting y1 =
(8)
an xn+r1 into the above gives
n=0
∞
∞
P −1+n+r1
P
n+r1
2
x
an (n + r1 ) x − 6
an x
C
n=0
n=0
(9)
∞
P −1+n+r2
−2+n+r2
2
x
bn (n + r2 ) (−1 + n + r2 )
x −5
x
bn (n + r2 ) x
x
∞
P
+
bn xn+r2
+
n=0
∞
P
n=0
n=0
x
=0
Since r1 = 6 and r2 = 0 then the above becomes
∞
∞
P 5+n
P
n+6
2
x an (n + 6) x − 6
an x
C
n=0
∞
P
+
n=0
n=0
bn x n
+
∞x
P
(10)
xn−2 bn n(−1 + n)
n=0
∞
P −1+n
2
x −5
x
bn n x
n=0
x
=0
chapter 2 . book solved problems
370
Which simplifies to
∞
X
!
2x
5+n
an (n + 6) C
∞
X
+
n=0
−6C x
5+n
an +
n =0
∞
X
+
∞
X
!
x
1+n
bn
(2A)
n=0
!
n x−1+n bn (−1 + n)
+
n=0
∞
X
−5x−1+n bn n = 0
n =0
The next step is to make all powers of x be −1 + n in each summation term. Going over each
summation term above with power of x in it which is not already x−1+n and adjusting the
power and the corresponding index gives
∞
X
∞
X
2x5+n an (n + 6) C =
n =0
∞
X
2Ca−6+n n x−1+n
n=6
−6C x
5+n
∞
X
an =
n =0
−6Ca−6+n x−1+n
n=6
∞
X
x
1+n
∞
X
bn =
n =0
bn−2 x−1+n
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to −1 + n.
∞
X
!
2Ca−6+n n x−1+n
n=6
+
+
∞
X
−6Ca−6+n x
n =6
∞
X
+
∞
X
!
bn−2 x−1+n
n=2
!
n x−1+n bn (−1 + n)
n=0
−1+n
+
∞
X
(2B)
−5x−1+n bn n = 0
n =0
For n = 0 in Eq. (2B), we choose arbitray value for b0 as b0 = 1. For n = 1, Eq (2B) gives
−5b1 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
−5b1 = 0
Solving the above for b1 gives
b1 = 0
For n = 2, Eq (2B) gives
b0 − 8b2 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
1 − 8b2 = 0
Solving the above for b2 gives
b2 =
1
8
For n = 3, Eq (2B) gives
b1 − 9b3 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
−9b3 = 0
chapter 2 . book solved problems
371
Solving the above for b3 gives
b3 = 0
For n = 4, Eq (2B) gives
b2 − 8b4 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
1
− 8b4 = 0
8
Solving the above for b4 gives
1
b4 =
64
For n = 5, Eq (2B) gives
b3 − 5b5 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
−5b5 = 0
Solving the above for b5 gives
b5 = 0
For n = N , where N = 6 which is the difference between the two roots, we are free to choose
b6 = 0. Hence for n = 6, Eq (2B) gives
6C +
1
=0
64
Which is solved for C. Solving for C gives
C=−
1
384
Now that we found all bn and C, we can calculate the second solution from
!
∞
X
y2 (x) = Cy1 (x) ln (x) +
bn xn+r2
n=0
1
Using the above value found for C = − 384
and all bn , then the second solution becomes
1
x2
x4
x6
x2 x4
6
7
y2 (x) = −
x 1−
+
−
+O x
ln (x) + 1 +
+
+ O x7
384
16 640 46080
8
64
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x2
x4
x6
6
7
= c1 x 1 −
+
−
+O x
16 640 46080
1
x2
x4
x6
x2 x4
6
7
7
+ c2 −
x 1−
+
−
+O x
ln (x) + 1 +
+
+O x
384
16 640 46080
8
64
Hence the final solution is
y = yh
x2
x4
x6
7
= c1 x 1 −
+
−
+O x
16 640 46080


2
x4
x6
x6 1 − x16 + 640
− 46080
+ O(x7 ) ln (x)
2
4
x
x
+ c2  −
+1+
+
+ O x7 
384
8
64
6
chapter 2 . book solved problems
3 Maple. Time used: 0.030 (sec). Leaf size: 32
Order:=6;
ode:=diff(diff(y(x),x),x)*x-5*diff(y(x),x)+x*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
372
1 2
1 4
6
y = c1 x 1 − x +
x +O x
+ c2 −86400 − 10800x2 − 1350x4 + O x6
16
640
6
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
•
Let’s solve
d d
d
x dx
y(x)
− 5 dx
y(x) + xy(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative d d
y(x) = −y(x) +
dx dx
•
•
•
•
•
•
•
•
d
y(x)
dx
x
Group termswith y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(x) −
dx dx
•
5
5
d
y(x)
dx
x
+ y(x) = 0
Simplify ODE
d d
d
x2 dx
y(x)
−
5
y(x)
x + x2 y(x) = 0
dx
dx
Make a change of variables
y(x) = x3 u(x)
d
Compute dx
y(x)
d
2
3 d
y(x)
=
3x
u(x)
+
x
u(x)
dx
dx
d d
Compute dx
y(x)
dx
d d
2 d
3 d d
y(x)
=
6xu(x)
+
6x
u(x)
+
x
u(x)
dx dx
dx
dx dx
Apply change of variables to the ODE
2
d d
d
x2 u(x) + dx
u(x)
x
+
u(x)
x − 9u(x) = 0
dx
dx
ODE is now of the Bessel form
Solution to Bessel ODE
u(x) = C 1 BesselJ (3, x) + C 2 BesselY (3, x)
Make the change from y(x) back to y(x)
y(x) = (C 1 BesselJ (3, x) + C 2 BesselY (3, x)) x3
chapter 2 . book solved problems
3 Mathematica. Time used: 0.006 (sec). Leaf size: 44
ode=x*D[y[x],{x,2}]-5*D[y[x],x]+x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
373
4
10
x
x2
x
x8
6
y(x) → c1
+
+ 1 + c2
−
+x
64
8
640 16
7 Sympy
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*y(x) + x*Derivative(y(x), (x, 2)) - 5*Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
ValueError : Expected Expr or iterable but got None
chapter 2 . book solved problems
2.5
374
Chapter 5. Series Solutions of ODEs. REVIEW
QUESTIONS. page 201
Local contents
2.5.1 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.2 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.3 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.4 Problem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.5 Problem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.6 Problem 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.7 Problem 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.8 Problem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.9 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.10 Problem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
375
381
388
395
402
411
420
429
437
447
chapter 2 . book solved problems
2.5.1
375
Problem 11
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
Internal problem ID [8612]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 11
Date solved : Tuesday, November 11, 2025 at 12:04:09 PM
CAS classification : [[_2nd_order, _missing_x]]
y 00 + 4y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
X xn+2 dn f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
y(x) = y(x0 ) + (x − x0 ) y 0 (x0 ) +
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.45)
(2.46)
(2)
(3)
chapter 2 . book solved problems
376
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
0
∂x
∂y
∂y
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
To find y(x) series solution around x = 0. Hence
F0 = −4y
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
= −4y 0
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
= 16y
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
= 16y 0
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
= −64y
(7)
chapter 2 . book solved problems
377
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = −4y(0)
F1 = −4y 0 (0)
F2 = 16y(0)
F3 = 16y 0 (0)
F4 = −64y(0)
Substituting all the above in (7) and simplifying gives the solution as
2 4
2 3
2 5 0
2
y = 1 − 2x + x y(0) + x − x + x y (0) + O x6
3
3
15
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
y 00 =
∞
X
n=1
∞
X
nan xn−1
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
!
!
∞
∞
X
X
n(n − 1) an xn−2 + 4
an x n = 0
n=2
(1)
n=0
Which simplifies to
∞
X
!
n(n − 1) an xn−2
+
n=2
∞
X
!
4an xn
=0
(2)
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
n(n − 1) an xn−2 =
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
(n + 2) an+2 (n + 1) xn
n=0
+
∞
X
!
4an xn
=0
(3)
n=0
For 0 ≤ n, the recurrence equation is
(n + 2) an+2 (n + 1) + 4an = 0
(4)
Solving for an+2 , gives
an+2 = −
4an
(n + 2) (n + 1)
(5)
chapter 2 . book solved problems
378
For n = 0 the recurrence equation gives
2a2 + 4a0 = 0
Which after substituting the earlier terms found becomes
a2 = −2a0
For n = 1 the recurrence equation gives
6a3 + 4a1 = 0
Which after substituting the earlier terms found becomes
a3 = −
2a1
3
For n = 2 the recurrence equation gives
12a4 + 4a2 = 0
Which after substituting the earlier terms found becomes
a4 =
2a0
3
For n = 3 the recurrence equation gives
20a5 + 4a3 = 0
Which after substituting the earlier terms found becomes
a5 =
2a1
15
For n = 4 the recurrence equation gives
30a6 + 4a4 = 0
Which after substituting the earlier terms found becomes
a6 = −
4a0
45
For n = 5 the recurrence equation gives
42a7 + 4a5 = 0
Which after substituting the earlier terms found becomes
a7 = −
4a1
315
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
chapter 2 . book solved problems
379
Substituting the values for an found above, the solution becomes
2
2
2
y = a0 + a1 x − 2a0 x2 − a1 x3 + a0 x4 + a1 x5 + . . .
3
3
15
Collecting terms, the solution becomes
y=
2 4
2 3
2 5
1 − 2x + x a0 + x − x + x a1 + O x6
3
3
15
2
(3)
At x = 0 the solution above becomes
2 4
2 3
2 5 0
2
y = 1 − 2x + x y(0) + x − x + x y (0) + O x6
3
3
15
Figure 2.8: Slope field y 00 + 4y = 0
3 Maple. Time used: 0.009 (sec). Leaf size: 39
Order:=6;
ode:=4*y(x)+diff(diff(y(x),x),x) = 0;
dsolve(ode,y(x),type='series',x=0);
2 4
2 3
2 5 0
2
y = 1 − 2x + x y(0) + x − x + x y (0) + O x6
3
3
15
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful
Maple step by step
•
Let’s solve
d d
y(x) + 4y(x) = 0
dx dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
chapter 2 . book solved problems
•
•
Characteristic polynomial of ODE
r2 + 4 = 0
Use quadratic
formula to solve for r
√
0±
•
•
•
•
•
380
−16
r=
2
Roots of the characteristic polynomial
r = (−2 I, 2 I)
1st solution of the ODE
y1 (x) = cos (2x)
2nd solution of the ODE
y2 (x) = sin (2x)
General solution of the ODE
y(x) = C 1 y1 (x) + C 2 y2 (x)
Substitute in solutions
y(x) = C 1 cos (2x) + C 2 sin (2x)
3 Mathematica. Time used: 0.001 (sec). Leaf size: 40
ode=D[y[x],{x,2}]+4*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
3 Sympy. Time used: 0.166 (sec). Leaf size: 32
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(4*y(x) + Derivative(y(x), (x, 2)),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_ordinary",x0=0,n=6)
y(x) = C2
4
2x5 2x3
2x
2
−
+ x + c1
− 2x + 1
15
3
3
2x4
2x2
2
− 2x + 1 + C1 x 1 −
+ O x6
3
3
chapter 2 . book solved problems
2.5.2
381
Problem 12
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
Internal problem ID [8613]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 12
Date solved : Tuesday, November 11, 2025 at 12:04:09 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
xy 00 + (1 − 2x) y 0 + y(x − 1) = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + (1 − 2x) y 0 + y(x − 1) = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
2x − 1
x
x−1
q(x) =
x
p(x) = −
Table 2.69: Table p(x), q(x) singularites.
q(x) = x−1
x
singularity
type
x=0
“regular”
p(x) = − 2x−1
x
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + (1 − 2x) y 0 + y(x − 1) = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
382
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
(n + r) (n + r − 1) an xn+r−2 x
(1)
n=0
∞
X
+ (1 − 2x)
!
(n + r) an xn+r−1
+
n=0
∞
X
!
an xn+r
(x − 1) = 0
n=0
Which simplifies to
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
+
n=0
∞
X
−2xn+r an (n + r)
(2A)
n =0
+
∞
X
∞
X
!
(n + r) an x
n+r−1
+
n=0
∞
X
!
x
1+n+r
an
+
n=0
−an xn+r = 0
n =0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
−2x
n+r
∞
X
an (n + r) =
−2an−1 (n + r − 1) xn+r−1
n =0
n=1
∞
X
x
1+n+r
∞
X
an =
n =0
∞
X
an−2 xn+r−1
n=2
−an x
n+r
=
n =0
∞
X
−an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
n=0
+
∞
X
!
(n + r) an xn+r−1
n=0
+
+
∞
X
n =1
∞
X
−2an−1 (n + r − 1) xn+r−1
an−2 xn+r−1
n=2
!
+
∞
X
−an−1 xn+r−1 = 0
n =1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + (n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + r x−1+r a0 = 0
(2B)
chapter 2 . book solved problems
383
Since a0 6= 0 then the above simplifies to
x−1+r r2 = 0
Since the above is true for all x then the indicial equation becomes
r2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r2 = 0
Solving for r gives the roots of the indicial equation as [0, 0].
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
y1 (x) =
∞
X
an xn+r
(1A)
n=0
Now the second solution y2 is found using
y2 (x) = y1 (x) ln (x) +
∞
X
!
bn x
n+r
(1B)
n=1
Then the general solution will be
y = c1 y1 (x) + c2 y2 (x)
In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), a0 is never zero, and is arbitrary
and is typically taken as a0 = 1, and {c1 , c2 } are two arbitray constants of integration which
can be found from initial conditions. We start by finding the first solution y1 (x). Eq (2B)
derived above is now used to find all an coefficients. The case n = 0 is skipped since it
was used to find the roots of the indicial equation. a0 is arbitrary and taken as a0 = 1.
Substituting n = 1 in Eq. (2B) gives
a1 =
2r + 1
(1 + r)2
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) − 2an−1 (n + r − 1) + an (n + r) + an−2 − an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an =
2nan−1 + 2ran−1 − an−2 − an−1
n2 + 2nr + r2
(4)
Which for the root r = 0 becomes
an =
(2n − 1) an−1 − an−2
n2
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
2r+1
(1+r)2
an
1
1
chapter 2 . book solved problems
384
For n = 2, using the above recursive equation gives
a2 =
3r2 + 6r + 2
(1 + r)2 (r + 2)2
Which for the root r = 0 becomes
a2 =
1
2
And the table now becomes
n
a0
a1
an,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
a2
an
1
1
1
2
For n = 3, using the above recursive equation gives
a3 =
4r3 + 18r2 + 22r + 6
(1 + r)2 (r + 2)2 (r + 3)2
Which for the root r = 0 becomes
a3 =
1
6
And the table now becomes
n
a0
a1
a2
a3
an,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
4r3 +18r2 +22r+6
(1+r)2 (r+2)2 (r+3)2
an
1
1
1
2
1
6
For n = 4, using the above recursive equation gives
5r4 + 40r3 + 105r2 + 100r + 24
a4 =
(1 + r)2 (r + 2)2 (r + 3)2 (r + 4)2
Which for the root r = 0 becomes
a4 =
1
24
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
4r3 +18r2 +22r+6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
an
1
1
1
2
1
6
1
24
For n = 5, using the above recursive equation gives
a5 =
6r5 + 75r4 + 340r3 + 675r2 + 548r + 120
(1 + r)2 (r + 2)2 (r + 3)2 (r + 4)2 (r + 5)2
Which for the root r = 0 becomes
a5 =
And the table now becomes
1
120
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
385
an,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
4r3 +18r2 +22r+6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
6r5 +75r4 +340r3 +675r2 +548r+120
(1+r)2 (r+2)2 (r+3)2 (r+4)2 (r+5)2
an
1
1
1
2
1
6
1
24
1
120
Using the above table, then the first solution y1 (x) becomes
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
=1+x+
x2 x3 x4
x5
+
+
+
+ O x6
2
6
24 120
Now the second solution is found. The second solution is given by
!
∞
X
y2 (x) = y1 (x) ln (x) +
bn xn+r
n=1
Where bn is found using
d
an,r
dr
And the above is then evaluated at r = 0. The above table for an,r is used for this purpose.
Computing the derivatives gives the following table
bn =
n
b0
b1
b2
b3
b4
b5
bn,r
1
2r+1
(1+r)2
3r2 +6r+2
(1+r)2 (r+2)2
4r3 +18r2 +22r+6
(1+r)2 (r+2)2 (r+3)2
5r4 +40r3 +105r2 +100r+24
(1+r)2 (r+2)2 (r+3)2 (r+4)2
6r5 +75r4 +340r3 +675r2 +548r+120
(1+r)2 (r+2)2 (r+3)2 (r+4)2 (r+5)2
an
1
1
d
bn,r = dr
an,r
N/A since bn starts from 1
2r
− (1+r)
3
bn (r = 0
N/A
0
1
2
−6r3 −18r2 −14r
(1+r)3 (r+2)3
12 r4 +8r3 + 47
r2 +30r+ 85
r
2
6
− (1+r)3 (r+2)
3
(r+3)3
20 r6 +15r5 + 183
r4 +290r3 + 5031
r2 +453r+166 r
2
10
−
(1+r)3 (r+2)3 (r+3)3 (r+4)3
30 r8 +24r7 + 739
r6 +1410r5 +4915r4 +10668r3 +14063r2 +10290r+ 48076
r
3
15
−
3
3
3
3
3
(1+r) (r+2) (r+3) (r+4) (r+5)
0
1
6
1
24
1
120
The above table gives all values of bn needed. Hence the second solution is
y2 (x) = y1 (x) ln (x) + b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x2 x3 x 4
x5
6
= 1+x+
+
+
+
+O x
ln (x) + O x6
2
6
24 120
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x2 x3 x4
x5
6
= c1 1 + x +
+
+
+
+O x
2
6
24 120
x 2 x3 x 4
x5
6
6
+ c2
1+x+
+
+
+
+O x
ln (x) + O x
2
6
24 120
Hence the final solution is
y = yh
x2 x3 x4
x5
6
= c1 1 + x +
+
+
+
+O x
2
6
24 120
x2 x3 x4
x5
6
6
+ c2
1+x+
+
+
+
+O x
ln (x) + O x
2
6
24 120
0
0
0
chapter 2 . book solved problems
3 Maple. Time used: 0.029 (sec). Leaf size: 39
Order:=6;
ode:=diff(diff(y(x),x),x)*x+(1-2*x)*diff(y(x),x)+(x-1)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
386
1 2 1 3
1 4
1 5
y = 1+x+ x + x + x +
x (ln (x) c2 + c1 ) + O x6
2
6
24
120
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Group is reducible, not completely reducible
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x dx
y(x)
+
(1
−
2x)
y(x)
+ (x − 1) y(x) = 0
dx
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d d
y(x) = − (x−1)y(x)
+
dx dx
x
•
(−1+2x)
d
y(x)
dx
x
Group terms with y(x) on
the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
◦
◦
(x · P2 (x))
◦
•
•
d
y(x)
dx
x
=1
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
(−1+2x)
+ (x−1)y(x)
=0
x
Check to see if x0 = 0 is a regular singular point
Define functions
x−1
P2 (x) = − −1+2x
,
P
(x)
=
3
x
x
x · P2 (x) is analytic at x = 0
d d
y(x) −
dx dx
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x)
+
(1
−
2x)
y(x)
+ (x − 1) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..1
∞
P
xm · y(x) =
ak xk+r+m
k=0
chapter 2 . book solved problems
◦
◦
387
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert xm · dx
y(x) to series expansion for m = 0..1
∞
P
d
xm · dx
y(x) =
ak (k + r) xk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
xm · dx
y(x) =
ak+1−m (k + 1 − m + r) xk+r
k=−1+m
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
∞
P
2
2
a0 r2 x−1+r + a1 (1 + r) − a0 (1 + 2r) xr +
ak+1 (k + 1 + r) − ak (2k + 2r + 1) + ak−1 xk+r
k=1
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term must be 0
a1 (1 + r)2 − a0 (1 + 2r) = 0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1)2 + (−2k − 1) ak + ak−1 = 0
Shift index using k− >k + 1
ak+2 (k + 2)2 + (−2k − 3) ak+1 + ak = 0
Recursion relation that defines series solution to ODE
−ak +3ak+1
ak+2 = 2kak+1(k+2)
2
•
Recursion relation for r = 0
−ak +3ak+1
ak+2 = 2kak+1(k+2)
2
•
Solution
for r = 0
∞
P
2kak+1 −ak +3ak+1
k
y(x) =
ak x , ak+2 =
, a 1 − a0 = 0
(k+2)2
k=0
3 Mathematica. Time used: 0.004 (sec). Leaf size: 74
ode=x*D[y[x],{x,2}]+(1-2*x)*D[y[x],x]+(x-1)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
5
x5
x4 x3 x2
x
x4 x3 x2
+
+
+
+ x + 1 + c2
+
+
+
+ x + 1 log(x)
120 24
6
2
120 24
6
2
3 Sympy. Time used: 0.246 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*Derivative(y(x), (x, 2)) + (1 - 2*x)*Derivative(y(x), x) + (x - 1)*y(x
),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
y(x) = C1
x5
x4 x 3 x2
+
+
+
+ x + 1 + O x6
120 24
6
2
chapter 2 . book solved problems
2.5.3
388
Problem 13
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
Internal problem ID [8614]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 13
Date solved : Tuesday, November 11, 2025 at 12:04:10 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
(x − 1)2 y 00 − (x − 1) y 0 − 35y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X
x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.47)
(2.48)
(2)
(3)
chapter 2 . book solved problems
389
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
390
To find y(x) series solution around x = 0. Hence
y 0 x − y 0 + 35y
F0 =
(x − 1)2
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
(35x − 35) y 0 − 35y
=
(x − 1)3
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
(−70x + 70) y 0 + 1330y
=
(x − 1)4
dF2
F3 =
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
1470(x − 1) y 0 − 7770y
=
(x − 1)5
dF3
F4 =
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
(−12180x + 12180) y 0 + 90300y
=
(x − 1)6
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
F0 = 35y(0) − y 0 (0)
F1 = 35y(0) + 35y 0 (0)
F2 = 1330y(0) + 70y 0 (0)
F3 = 7770y(0) + 1470y 0 (0)
F4 = 90300y(0) + 12180y 0 (0)
Substituting all the above in (7) and simplifying gives the solution as
35 2 35 3 665 4 259 5
1 2 35 3 35 4 49 5 0
y = 1+ x + x +
x +
x y(0)+ x− x + x + x + x y (0)+O x6
2
6
12
4
2
6
12
4
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
x2 − 2x + 1 y 00 + (1 − x) y 0 − 35y = 0
Let the solution be represented as power series of the form
y=
∞
X
an x n
n=0
Then
0
y =
∞
X
nan xn−1
n=1
y 00 =
∞
X
n=2
n(n − 1) an xn−2
chapter 2 . book solved problems
391
Substituting the above back into the ode gives
!
!
!
∞
∞
∞
X
X
X
x2 − 2x + 1
n(n − 1) an xn−2 + (1 − x)
nan xn−1 − 35
an xn = 0 (1)
n=2
n=1
n=0
Which simplifies to
∞
X
!
xn an n(n − 1)
n=2
+
∞
X
+
−2n xn−1 an (n − 1) +
∞
X
!
n(n − 1) an xn−2
(2)
n=2
! n =2
∞
∞
∞
X
X
X
nan xn−1 +
(−nan xn ) +
(−35an xn ) = 0
n=1
n =1
n =0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
−2n x
n−1
∞
X
an (n − 1) =
(−2(n + 1) an+1 n xn )
n =2
n=1
∞
X
n(n − 1) an x
n−2
=
n =2
∞
X
(n + 2) an+2 (n + 1) xn
n=0
∞
X
nan x
n−1
n =1
=
∞
X
(n + 1) an+1 xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
xn an n(n − 1)
n=2
+
+
∞
X
∞
X
(−2(n + 1) an+1 n xn ) +
n =1
∞
X
n=0
(n + 2) an+2 (n + 1) xn
n=0
!
(n + 1) an+1 xn
!
+
∞
X
(−nan xn ) +
n =1
∞
X
(3)
(−35an xn ) = 0
n =0
n = 0 gives
2a2 + a1 − 35a0 = 0
a2 =
35a0 a1
−
2
2
n = 1 gives
−2a2 + 6a3 − 36a1 = 0
Which after substituting earlier equations, simplifies to
a3 =
35a0 35a1
+
6
6
For 2 ≤ n, the recurrence equation is
nan (n − 1) − 2(n + 1) an+1 n + (n + 2) an+2 (n + 1) + (n + 1) an+1 − nan − 35an = 0 (4)
chapter 2 . book solved problems
392
Solving for an+2 , gives
an+2 = −
=−
n2 an − 2n2 an+1 − 2nan − nan+1 − 35an + an+1
(n + 2) (n + 1)
(n2 − 2n − 35) an (−2n2 − n + 1) an+1
−
(n + 2) (n + 1)
(n + 2) (n + 1)
For n = 2 the recurrence equation gives
−35a2 − 9a3 + 12a4 = 0
Which after substituting the earlier terms found becomes
a4 =
665a0 35a1
+
12
12
For n = 3 the recurrence equation gives
−32a3 − 20a4 + 20a5 = 0
Which after substituting the earlier terms found becomes
a5 =
259a0 49a1
+
4
4
For n = 4 the recurrence equation gives
−27a4 − 35a5 + 30a6 = 0
Which after substituting the earlier terms found becomes
a6 =
1505a0 203a1
+
12
12
For n = 5 the recurrence equation gives
−20a5 − 54a6 + 42a7 = 0
Which after substituting the earlier terms found becomes
a7 =
2305a0 331a1
+
12
12
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
35a0 a1
35a0 35a1
2
y = a0 + a1 x +
−
x +
+
x3
2
2
6
6
665a0 35a1
259a0 49a1
4
+
+
x +
+
x5 + . . .
12
12
4
4
(5)
chapter 2 . book solved problems
393
Collecting terms, the solution becomes
35 2 35 3 665 4 259 5
1 2 35 3 35 4 49 5
y = 1+ x + x +
x +
x a0 + x − x + x + x + x a1 + O x 6
2
6
12
4
2
6
12
4
(3)
At x = 0 the solution above becomes
35 2 35 3 665 4 259 5
1 2 35 3 35 4 49 5 0
y = 1+ x + x +
x +
x y(0)+ x− x + x + x + x y (0)+O x6
2
6
12
4
2
6
12
4
3 Maple. Time used: 0.011 (sec). Leaf size: 59
Order:=6;
ode:=(x-1)^2*diff(diff(y(x),x),x)-(x-1)*diff(y(x),x)-35*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
35 2 35 3 665 4 259 5
1 2 35 3 35 4 49 5 0
y = 1+ x + x +
x +
x y(0)+ x− x + x + x + x y (0)+O x6
2
6
12
4
2
6
12
4
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
<- LODE of Euler type successful
Maple step by step
Let’s solve
d d
d
(x − 1)2 dx
y(x) − (x − 1) dx
y(x) − 35y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
35y(x)
d d
dx
y(x)
=
+
2
dx dx
x−1
(x−1)
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
35y(x)
d d
dx
y(x)
−
− (x−1)
2 = 0
dx dx
x−1
◦
Check to see if x0 = 1 is a regular singular point
Define
functions
h
i
1
35
P2 (x) = − x−1
, P3 (x) = − (x−1)
2
◦
(x − 1) · P2 (x) is analytic at x = 1
((x − 1) · P2 (x))
= −1
x=1
◦
(x − 1)2 · P3 (x) is analytic at x = 1
(x − 1)2 · P3 (x)
= −35
◦
x = 1is a regular singular point
Check to see if x0 = 1 is a regular singular point
x0 = 1
Multiply by denominators
d d
d
(x − 1)2 dx
y(x) + (1 − x) dx
y(x) − 35y(x) = 0
dx
Change variables using x = u + 1 so that the regular singular point is at u = 0
d d
d
u2 du
y(u) − u du
y(u) − 35y(u) = 0
du
x=1
•
•
chapter 2 . book solved problems
•
Assume series solution for y(u)
∞
P
y(u) =
ak uk+r
Rewrite DE with series expansions
d
Convert u · du
y(u) to series expansion
∞
P
d
u · du
y(u) =
ak (k + r) uk+r
k=0
d d
Convert u2 · du
y(u) to series expansion
du
∞
P
d d
u2 · du
y(u)
=
ak (k + r) (k + r − 1) uk+r
du
394
k=0
◦
◦
k=0
Rewrite DE with series expansions
∞
P
ak (k + r + 5) (k + r − 7) uk+r = 0
k=0
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r=0
Each term in the series must be 0, giving the recursion relation
ak (k + 5) (k − 7) = 0
Recursion relation that defines series solution to ODE
ak = 0
Recursion relation for r = 0
ak = 0
Solution
for r = 0
∞
P
k
y(u) =
ak u , a k = 0
k=0
•
Revert
the change of variablesu = x − 1
∞
P
y(x) =
ak (x − 1)k , ak = 0
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 70
ode=(x-1)^2*D[y[x],{x,2}]-(x-1)*D[y[x],x]-35*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
259x5 665x4 35x3 35x2
49x5 35x4 35x3 x2
+
+
+
+ 1 + c2
+
+
−
+x
4
12
6
2
4
12
6
2
7 Sympy
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((x - 1)**2*Derivative(y(x), (x, 2)) - (x - 1)*Derivative(y(x), x) - 35*y
(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
IndexError : list index out of range
chapter 2 . book solved problems
2.5.4
395
Problem 14
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401
Internal problem ID [8615]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 14
Date solved : Tuesday, November 11, 2025 at 12:04:10 PM
CAS classification : [[_Emden, _Fowler]]
16(1 + x)2 y 00 + 3y = 0
Using series expansion around x = 0.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let
y 00 = f (x, y, y 0 )
Assuming expansion is at x0 = 0 (we can always shift the actual expansion point to 0 by
change of variables) and assuming f (x, y, y 0 ) is analytic at x0 which must be the case for an
ordinary point. Let initial conditions be y(x0 ) = y0 and y 0 (x0 ) = y00 . Using Taylor series gives
(x − x0 )2 00
(x − x0 )3 000
y(x) = y(x0 ) + (x − x0 ) y (x0 ) +
y (x0 ) +
y (x0 ) + · · ·
2
3!
x2
x3
= y0 + xy00 + f |x0 ,y0 ,y0 + f 0 |x0 ,y0 ,y0 + · · ·
0
0
2
3!
∞
n+2
n
X
x
d f
= y0 + xy00 +
(n + 2) ! dxn x0 ,y0 ,y0
n=0
0
0
But
df
∂f dx ∂f dy
∂f dy 0
=
+
+ 0
dx
∂x dx ∂y dx ∂y dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
2
df
d df
=
2
dx
dx dx
∂ df
∂ df
∂ df
0
=
+
y + 0
f
∂x dx
∂y dx
∂y dx
d3 f
d d2 f
=
dx3
dx dx2
∂ d2 f
∂ d2 f
∂ d2 f
0
=
+
y + 0
f
∂x dx2
∂y dx2
∂y dx2
..
.
(1)
(2.50)
(2.51)
(2)
(3)
chapter 2 . book solved problems
396
And so on. Hence if we name F0 = f (x, y, y 0 ) then the above can be written as
F0 = f (x, y, y 0 )
df
F1 =
dx
dF0
=
dx
∂f
∂f 0 ∂f 00
=
+
y + 0y
∂x ∂y
∂y
∂f
∂f 0 ∂f
=
+
y + 0f
∂x ∂y
∂y
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
d d
F2 =
f
dx dx
d
=
(F1 )
dx
∂
∂F1
∂F1
0
=
F1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂F1
∂F1
0
=
F1 +
y +
F0
∂x
∂y
∂y 0
..
.
d
Fn =
(Fn−1 )
dx
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
y 00
∂x
∂y
∂y 0
∂
∂Fn−1
∂Fn−1
0
=
Fn−1 +
y +
F0
∂x
∂y
∂y 0
(4)
(5)
(6)
Therefore (6) can be used from now on along with
y(x) = y0 + xy00 +
∞
X
xn+2
Fn |x0 ,y0 ,y0
0
(n
+
2)
!
n=0
(7)
chapter 2 . book solved problems
397
To find y(x) series solution around x = 0. Hence
3y
16 (1 + x)2
dF0
F1 =
dx
∂F0 ∂F0 0 ∂F0
=
+
y +
F0
∂x
∂y
∂y 0
(−3x − 3) y 0 + 6y
=
16 (1 + x)3
dF1
F2 =
dx
∂F1 ∂F1 0 ∂F1
=
+
y +
F1
∂x
∂y
∂y 0
F0 = −
=
− 279y
+ 3(1+x)y
256
4
0
(1 + x)4
dF2
dx
∂F2 ∂F2 0 ∂F2
=
+
y +
F2
∂x
∂y
∂y 0
F3 =
=
855(1+x)y 0
135y
−
32
256
5
(1 + x)
dF3
dx
∂F3 ∂F3 0 ∂F3
=
+
y +
F3
∂x
∂y
∂y 0
F4 =
=
− 83835y
+ 1125(1+x)y
4096
64
0
(1 + x)6
And so on. Evaluating all the above at initial conditions x = 0 and y(0) = y(0) and
y 0 (0) = y 0 (0) gives
3y(0)
16
3y(0) 3y 0 (0)
F1 =
−
8
16
279y(0) 3y 0 (0)
F2 = −
+
256
4
135y(0) 855y 0 (0)
F3 =
−
32
256
83835y(0) 1125y 0 (0)
F4 = −
+
4096
64
F0 = −
Substituting all the above in (7) and simplifying gives the solution as
3 2 1 3
93 4
9 5
1 3 1 4
57 5 0
y = 1− x + x −
x +
x y(0) + x − x + x −
x y (0) + O x6
32
16
2048
256
32
32
2048
Since the expansion point x = 0 is an ordinary point, then this can also be solved using the
standard power series method. The ode is normalized to be
16x2 + 32x + 16 y 00 + 3y = 0
Let the solution be represented as power series of the form
y=
∞
X
n=0
an x n
chapter 2 . book solved problems
398
Then
∞
X
0
y =
n=1
∞
X
y 00 =
nan xn−1
n(n − 1) an xn−2
n=2
Substituting the above back into the ode gives
16x2 + 32x + 16
∞
X
!
n(n − 1) an xn−2
+3
n=2
∞
X
!
an x n
=0
(1)
n=0
Which simplifies to
∞
X
!
16xn an n(n − 1)
∞
X
+
n=2
+
∞
X
!
32n xn−1 an (n − 1)
!n=2
16n(n − 1) an xn−2
+
n=2
∞
X
(2)
!
3an xn
=0
n=0
The next step is to make all powers of x be n in each summation term. Going over each
summation term above with power of x in it which is not already xn and adjusting the power
and the corresponding index gives
∞
X
32n xn−1 an (n − 1) =
n =2
∞
X
∞
X
32(n + 1) an+1 n xn
n=1
16n(n − 1) an x
n−2
=
n =2
∞
X
16(n + 2) an+2 (n + 1) xn
n=0
Substituting all the above in Eq (2) gives the following equation where now all powers of x
are the same and equal to n.
∞
X
!
16xn an n(n − 1)
+
n=2
+
∞
X
!
32(n + 1) an+1 n xn
(3)
n=1
∞
X
!
16(n + 2) an+2 (n + 1) xn
+
n=0
∞
X
!
3an xn
=0
n=0
n = 0 gives
32a2 + 3a0 = 0
a2 = −
3a0
32
n = 1 gives
64a2 + 96a3 + 3a1 = 0
Which after substituting earlier equations, simplifies to
a3 =
a0
a1
−
16 32
For 2 ≤ n, the recurrence equation is
16nan (n − 1) + 32(n + 1) an+1 n + 16(n + 2) an+2 (n + 1) + 3an = 0
(4)
chapter 2 . book solved problems
399
Solving for an+2 , gives
an+2 = −
=−
16n2 an + 32n2 an+1 − 16nan + 32nan+1 + 3an
16 (n + 2) (n + 1)
(5)
(16n2 − 16n + 3) an (32n2 + 32n) an+1
−
16 (n + 2) (n + 1)
16 (n + 2) (n + 1)
For n = 2 the recurrence equation gives
35a2 + 192a3 + 192a4 = 0
Which after substituting the earlier terms found becomes
a4 = −
93a0
a1
+
2048 32
For n = 3 the recurrence equation gives
99a3 + 384a4 + 320a5 = 0
Which after substituting the earlier terms found becomes
9a0
57a1
−
256 2048
a5 =
For n = 4 the recurrence equation gives
195a4 + 640a5 + 480a6 = 0
Which after substituting the earlier terms found becomes
a6 = −
1863a0 25a1
+
65536
1024
For n = 5 the recurrence equation gives
323a5 + 960a6 + 672a7 = 0
Which after substituting the earlier terms found becomes
a7 =
777a0 1409a1
−
32768
65536
And so on. Therefore the solution is
y=
∞
X
an xn
n=0
= a3 x 3 + a2 x 2 + a1 x + a0 + . . .
Substituting the values for an found above, the solution becomes
3a0 x2 a0
a1 3
93a0
a1
9a0
57a1
4
y = a0 + a1 x −
+
−
x + −
+
x +
−
x5 + . . .
32
16 32
2048 32
256 2048
Collecting terms, the solution becomes
3 2 1 3
93 4
9 5
1 3 1 4
57 5
y = 1− x + x −
x +
x a0 + x − x + x −
x a1 + O x6 (3)
32
16
2048
256
32
32
2048
chapter 2 . book solved problems
400
At x = 0 the solution above becomes
3 2 1 3
93 4
9 5
1 3 1 4
57 5 0
y = 1− x + x −
x +
x y(0) + x − x + x −
x y (0) + O x6
32
16
2048
256
32
32
2048
3 Maple. Time used: 0.013 (sec). Leaf size: 54
Order:=6;
ode:=16*(x+1)^2*diff(diff(y(x),x),x)+3*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
3 2 1 3
93 4
9 5
1 3 1 4
57 5 0
y = 1− x + x −
x +
x y(0) + x − x + x −
x y (0) + O x6
32
16
2048
256
32
32
2048
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
<- LODE of Euler type successful
Maple step by step
Let’s solve
d d
16(x + 1)2 dx
y(x)
+ 3y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
3y(x)
d d
y(x) = − 16(x+1)
2
dx dx
•
•
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
3y(x)
d d
y(x) + 16(x+1)
2 = 0
dx dx
Check to see if x0 = −1 is a regular singular point
Define
functions
h
i
◦
3
P2 (x) = 0, P3 (x) = 16(x+1)
2
◦
(x + 1) · P2 (x) is analytic at x = −1
((x + 1) · P2 (x))
◦
=0
x=−1
(x + 1)2 · P3 (x) is analytic at x = −1
3
(x + 1)2 · P3 (x)
= 16
x=−1
◦
•
•
•
x = −1is a regular singular point
Check to see if x0 = −1 is a regular singular point
x0 = −1
Multiply by denominators
d d
16(x + 1)2 dx
y(x) + 3y(x) = 0
dx
Change variables using x = u − 1 so that the regular singular point is at u = 0
d d
16u2 du
y(u) + 3y(u) = 0
du
Assume series solution for y(u)
∞
P
y(u) =
ak uk+r
k=0
◦
Rewrite DE with series expansions
d d
Convert u2 · du
y(u) to series expansion
du
∞
P
d d
u2 · du
y(u) =
ak (k + r) (k + r − 1) uk+r
du
k=0
chapter 2 . book solved problems
401
Rewrite DE with series expansions
∞
P
ak (4k + 4r − 1) (4k + 4r − 3) uk+r = 0
k=0
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r=0
Each term in the series must be 0, giving the recursion relation
ak (4k − 1) (4k − 3) = 0
Recursion relation that defines series solution to ODE
ak = 0
Recursion relation for r = 0
ak = 0
Solution
for r = 0
∞
P
k
y(u) =
ak u , a k = 0
k=0
•
Revert
the change of variablesu = x + 1
∞
P
y(x) =
ak (x + 1)k , ak = 0
k=0
3 Mathematica. Time used: 0.001 (sec). Leaf size: 63
ode=16*(x+1)^2*D[y[x],{x,2}]+3*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
5
57x5 x4 x3
9x
93x4 x3 3x2
y(x) → c2 −
+
−
+ x + c1
−
+
−
+1
2048 32 32
256 2048 16
32
7 Sympy
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(16*(x + 1)**2*Derivative(y(x), (x, 2)) + 3*y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
IndexError : list index out of range
chapter 2 . book solved problems
2.5.5
402
Problem 15
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410
Internal problem ID [8616]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 15
Date solved : Tuesday, November 11, 2025 at 12:04:11 PM
CAS classification : [_Bessel]
x2 y 00 + y 0 x + x2 − 5 y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
x2 y 00 + y 0 x + x2 − 5 y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
x2 − 5
q(x) =
x2
p(x) =
Table 2.73: Table p(x), q(x) singularites.
2
q(x) = x x−5
2
singularity
type
x=0
“regular”
p(x) = x1
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
x2 y 00 + y 0 x + x2 − 5 y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
403
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
x2
∞
X
!
(n + r) (n + r − 1) an xn+r−2 +
n=0
∞
X
!
(n + r) an xn+r−1
∞
X
2
x+ x −5
an xn+r
n=0
!
=0
n=0
(1)
Which simplifies to
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
xn+r an (n + r)
(2A)
n=0
∞
X
!
xn+r+2 an
+
n=0
∞
X
−5an xn+r = 0
n =0
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives
∞
X
x
n+r+2
an =
n =0
∞
X
an−2 xn+r
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r.
∞
X
!
xn+r an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
xn+r an (n + r)
n=0
∞
X
!
an−2 x
n=2
n+r
+
∞
X
−5an xn+r = 0
n =0
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r an (n + r) (n + r − 1) + xn+r an (n + r) − 5an xn+r = 0
When n = 0 the above becomes
xr a0 r(−1 + r) + xr a0 r − 5a0 xr = 0
Or
(xr r(−1 + r) + xr r − 5xr ) a0 = 0
Since a0 6= 0 then the above simplifies to
r 2 − 5 xr = 0
Since the above is true for all x then the indicial equation becomes
r2 − 5 = 0
(2B)
chapter 2 . book solved problems
404
Solving for r gives the roots of the indicial equation as
√
r1 = 5
√
r2 = − 5
Since a0 6= 0 then the indicial equation becomes
r 2 − 5 xr = 0
√ √
Solving for r gives the roots of the indicial equation as
5, − 5 .
√
Since r1 − r2 = 2 5 is not an integer, then we can construct two linearly independent
solutions
!
∞
X
y1 (x) = xr1
an x n
y2 (x) = xr2
n=0
∞
X
!
bn x n
n=0
Or
y1 (x) =
y2 (x) =
∞
X
n=0
∞
X
√
an xn+ 5
√
bn xn− 5
n=0
We start by finding y1 (x). Eq (2B) derived above is now used to find all an coefficients. The
case n = 0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary
and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + an (n + r) + an−2 − 5an = 0
(3)
Solving for an from recursive equation (4) gives
an−2
an = − 2
n + 2nr + r2 − 5
Which for the root r =
√
5 becomes
an = −
an−2
√
n 2 5+n
At this point, it is a good idea to keep track of an in a table both before substituting r =
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
1
a2 = − 2
r + 4r − 1
Which for the root r =
(4)
√
5 becomes
a2 = −
1
√
4+4 5
(5)
√
5
chapter 2 . book solved problems
405
And the table now becomes
n
a0
a1
a2
an,r
1
0
1
− r2 +4r−1
an
1
0
− 4+41√5
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
1
− r2 +4r−1
0
an
1
0
− 4+41√5
0
For n = 4, using the above recursive equation gives
a4 =
Which for the root r =
1
(r2 + 4r − 1) (r2 + 8r + 11)
√
5 becomes
a4 =
32
√
1
√ 5+1 2+ 5
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
1
− r2 +4r−1
0
an
1
0
− 4+41√5
0
1
(r2 +4r−1)(r2 +8r+11)
1
√
√ 32
5+1 2+ 5
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
1
− r2 +4r−1
0
an
1
0
− 4+41√5
0
1
(r2 +4r−1)(r2 +8r+11)
1
√
√ 32
5+1 2+ 5
a5
0
0
Using the above table, then the solution y1 (x) is
√
y1 (x) = x 5 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x2
x4
√ +
√
√ + O x6
1−
4 + 4 5 32 5 + 1 2 + 5
!
√
=x 5
chapter 2 . book solved problems
406
Now the second solution y2 (x) is found. Eq (2B) derived above is now used to find all bn
coefficients. The case n = 0 is skipped since it was used to find the roots of the indicial
equation. b0 is arbitrary and taken as b0 = 1. Substituting n = 1 in Eq. (2B) gives
b1 = 0
For 2 ≤ n the recursive equation is
bn (n + r) (n + r − 1) + bn (n + r) + bn−2 − 5bn = 0
(3)
Solving for bn from recursive equation (4) gives
bn−2
bn = − 2
n + 2nr + r2 − 5
(4)
√
Which for the root r = − 5 becomes
bn = −
bn−2
√
n −2 5 + n
(5)
√
At this point, it is a good idea to keep track of bn in a table both before substituting r = − 5
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
0
bn
1
0
For n = 2, using the above recursive equation gives
1
b2 = − 2
r + 4r − 1
√
Which for the root r = − 5 becomes
b2 =
1
√
−4 + 4 5
And the table now becomes
n
b0
b1
b2
bn,r
1
0
1
− r2 +4r−1
bn
1
0
1√
−4+4 5
For n = 3, using the above recursive equation gives
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
0
1
− r2 +4r−1
0
bn
1
0
1√
−4+4 5
0
For n = 4, using the above recursive equation gives
b4 =
1
(r2 + 4r − 1) (r2 + 8r + 11)
chapter 2 . book solved problems
407
√
Which for the root r = − 5 becomes
b4 =
1
√
√ 32 5 − 1 −2 + 5
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
1
− r2 +4r−1
0
bn
1
0
1
(r2 +4r−1)(r2 +8r+11)
1
√
√ 32
5−1 −2+ 5
1√
−4+4 5
0
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
1
− r2 +4r−1
0
bn
1
0
1
(r2 +4r−1)(r2 +8r+11)
1
√
√ 32
5−1 −2+ 5
b5
0
0
1√
−4+4 5
0
Using the above table, then the solution y2 (x) is
√
y2 (x) = x 5 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
!
x2
x4
√ +
√
√ + O x6
1+
−4 + 4 5 32 5 − 1 −2 + 5
√
= x− 5
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
√
= c1 x 5
+ c2 x
x2
x4
√ +
√
√ + O x6
1−
4 + 4 5 32 5 + 1 2 + 5
√
− 5
!
x2
x4
√ +
√
√ + O x6
1+
−4 + 4 5 32 5 − 1 −2 + 5
Hence the final solution is
y = yh
√
= c1 x 5
+ c2 x
!
x2
x4
√ +
√
√ + O x6
1−
4 + 4 5 32 5 + 1 2 + 5
√
− 5
x2
x4
√ +
√
√ + O x6
1+
−4 + 4 5 32 5 − 1 −2 + 5
!
!
chapter 2 . book solved problems
408
3 Maple. Time used: 0.030 (sec). Leaf size: 89
Order:=6;
ode:=x^2*diff(diff(y(x),x),x)+diff(y(x),x)*x+(x^2-5)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
√
− 5
1
1
1
√ x2 +
√ √
x 4 + O x6
y = c1 x
1+
32 −2 + 5
−4 + 4 5
5−1
!
√
1
1
1
√ x2 +
√ √
x4 + O x6
+ c2 x 5 1 −
32
4+4 5
2+ 5
5+1
!
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
Let’s solve
d d
d
x2 dx
y(x) + x dx
y(x) + (x2 − 5) y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
•
d
y(x)
x2 −5 y(x)
d d
dx
y(x)
=
−
−
2
dx dx
x
x
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
x2 −5 y(x)
d d
dx
y(x)
+
+
=0
dx dx
x
x2
◦
Check to see if x0 = 0 is a regular singular point
Define
functions
h
i
2
P2 (x) = x1 , P3 (x) = x x−5
2
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=1
x=0
= −5
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x2 dx
y(x)
+
x
y(x)
+ (x2 − 5) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
chapter 2 . book solved problems
◦
409
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..2
∞
P
xm · y(x) =
ak xk+r+m
k=0
◦
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert x · dx
y(x) to series expansion
∞
P
d
x · dx
y(x) =
ak (k + r) xk+r
k=0
d d
Convert x2 · dx
y(x)
to series expansion
dx
∞
P
d d
x2 · dx
y(x) =
ak (k + r) (k + r − 1) xk+r
dx
k=0
Rewrite ODE with series expansions r
2
2
a0 (r − 5) x + a1 (r + 2r − 4) x
1+r
+
∞
P
2
2
(ak (k + 2kr + r − 5) + ak−2 ) x
k+r
=0
k=2
•
•
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r2 − 5 = 0
Values of r that satisfy the indicial equation
√
√
r∈
5, − 5
Each term must be 0
a1 (r2 + 2r − 4) = 0
Solve for the dependent coefficient(s)
a1 = 0
Each term in the series must be 0, giving the recursion relation
ak (k 2 + 2kr + r2 − 5) + ak−2 = 0
Shift index using k− >k + 2
ak+2 (k + 2)2 + 2(k + 2) r + r2 − 5 + ak = 0
Recursion relation that defines series solution to ODE
ak+2 = − k2 +2kr+ra2k+4k+4r−1
√
Recursion relation for r = 5
ak
√
ak+2 = − k2 +2k√5+4+4k+4
5
√
Solution
for
r
=
5
∞
√
P
a
k+ 5
k
√ , a1 = 0
y(x) =
ak x
, ak+2 = − k2 +2k√5+4+4k+4
5
k=0
√
Recursion relation for r = − 5
ak
√
ak+2 = − k2 −2k√5+4+4k−4
5
√
Solution
for
r
=
−
5
∞
√
P
a
k− 5
k
y(x) =
ak x
, ak+2 = − k2 −2k√5+4+4k−4√5 , a1 = 0
k=0
•
Combine
and rename
solutions
∞ parameters
∞
√
√
P
P
ak
k+ 5
k− 5
√ , a1 = 0, bk+2 = −
√ bk
y(x) =
ak x
+
bk x
, ak+2 = − k2 +2k√5+4+4k+4
5
k2 −2k 5+4+4
k=0
k=0
chapter 2 . book solved problems
410
3 Mathematica. Time used: 0.003 (sec). Leaf size: 210
ode=x^2*D[y[x],{x,2}]+x*D[y[x],x]+(x^2-5)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c2
x4
√
√ √ √
√ √ −3 − 5 + 1 − 5 2 − 5
−1 − 5 + 3 − 5 4 − 5
!
√
x2
√
√ √ + 1 x− 5
−
−3 − 5 + 1 − 5 2 − 5
+ c1
−3 +
√
x4
√ √ √
√ √ 5+ 1+ 5 2+ 5
−1 + 5 + 3 + 5 4 + 5
2
−
−3 +
√
!
√
x
√ √ +1 x 5
5+ 1+ 5 2+ 5
7 Sympy
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x**2*Derivative(y(x), (x, 2)) + x*Derivative(y(x), x) + (x**2 - 5)*y(x)
,0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
NotImplementedError : Not sure of sign of 6 - x0
chapter 2 . book solved problems
2.5.6
411
Problem 16
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419
Internal problem ID [8617]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 16
Date solved : Tuesday, November 11, 2025 at 12:04:11 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
x2 y 00 + 2y 0 x3 + x2 − 2 y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
x2 y 00 + 2y 0 x3 + x2 − 2 y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
p(x) = 2x
q(x) =
x2 − 2
x2
Table 2.75: Table p(x), q(x) singularites.
p(x) = 2x
singularity
type
x=∞
“regular”
x = −∞ “regular”
2
q(x) = x x−2
2
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [∞, −∞, 0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
x2 y 00 + 2y 0 x3 + x2 − 2 y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
412
Then
∞
X
0
y =
(n + r) an xn+r−1
n=0
∞
X
00
y =
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
x2
!
(n + r) (n + r − 1) an xn+r−2
n=0
∞
X
+2
!
(n + r) an x
n+r−1
(1)
∞
X
x + x −2
an xn+r
3
!
2
n=0
=0
n=0
Which simplifies to
∞
X
!
x
n+r
an (n + r) (n + r − 1)
+
n=0
+
∞
X
!
2x
2+n+r
an (n + r)
(2A)
n=0
∞
X
!
x2+n+r an
+
n=0
∞
X
−2an xn+r = 0
n =0
The next step is to make all powers of x be n + r in each summation term. Going over each
summation term above with power of x in it which is not already xn+r and adjusting the
power and the corresponding index gives
∞
X
2x
2+n+r
an (n + r) =
n =0
∞
X
2an−2 (n + r − 2) xn+r
n=2
∞
X
x
2+n+r
an =
n =0
∞
X
an−2 xn+r
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r.
∞
X
!
xn+r an (n + r) (n + r − 1)
n=0
+
+
∞
X
!
2an−2 (n + r − 2) xn+r
n=2
∞
X
!
an−2 xn+r
n=2
+
∞
X
−2an xn+r = 0
n =0
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r an (n + r) (n + r − 1) − 2an xn+r = 0
When n = 0 the above becomes
xr a0 r(−1 + r) − 2a0 xr = 0
Or
(xr r(−1 + r) − 2xr ) a0 = 0
Since a0 6= 0 then the above simplifies to
r 2 − r − 2 xr = 0
(2B)
chapter 2 . book solved problems
413
Since the above is true for all x then the indicial equation becomes
r2 − r − 2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 2
r2 = −1
Since a0 6= 0 then the indicial equation becomes
r 2 − r − 2 xr = 0
Solving for r gives the roots of the indicial equation as [2, −1].
Since r1 − r2 = 3 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = Cy1 (x) ln (x) + xr2
∞
X
!
bn x n
n=0
Or
∞
X
y1 (x) = x2
!
an x n
n=0
∞
P
y2 (x) = Cy1 (x) ln (x) +
bn x n
n=0
x
Or
y1 (x) =
∞
X
an xn+2
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn xn−1
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + 2an−2 (n + r − 2) + an−2 − 2an = 0
(3)
Solving for an from recursive equation (4) gives
an−2 (2n + 2r − 3)
an = − 2
n + 2nr + r2 − n − r − 2
(4)
Which for the root r = 2 becomes
an =
an−2 (−2n − 1)
n (n + 3)
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 2
and after as more terms are found using the above recursive equation.
n
a0
a1
an,r
1
0
an
1
0
chapter 2 . book solved problems
414
For n = 2, using the above recursive equation gives
a2 =
−2r − 1
r (r + 3)
Which for the root r = 2 becomes
a2 = −
1
2
And the table now becomes
n
a0
a1
a2
an,r
1
0
−2r−1
r(r+3)
an
1
0
− 12
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
a3
an,r
1
0
−2r−1
r(r+3)
0
an
1
0
− 12
0
For n = 4, using the above recursive equation gives
a4 =
4r2 + 12r + 5
r (r + 3) (r + 5) (r + 2)
Which for the root r = 2 becomes
a4 =
9
56
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
0
an
1
0
− 12
0
4r2 +12r+5
r(r+3)(r+5)(r+2)
9
56
−2r−1
r(r+3)
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
0
an
1
0
− 12
0
4r2 +12r+5
r(r+3)(r+5)(r+2)
9
56
0
0
−2r−1
r(r+3)
chapter 2 . book solved problems
415
Using the above table, then the solution y1 (x) is
y1 (x) = x2 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x2 9x4
2
6
=x 1−
+
+O x
2
56
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 3. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a3 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a3
=0
Therefore
lim 0 = lim 0
r→r2
r→−1
=0
The limit is 0. Since the limit exists then the log term is not needed and we can set C = 0.
Therefore the second solution has the form
y2 (x) =
∞
X
bn xn+r
n=0
=
∞
X
bn xn−1
n=0
Eq (3) derived above is used to find all bn coefficients. The case n = 0 is skipped since
it was used to find the roots of the indicial equation. b0 is arbitrary and taken as b0 = 1.
Substituting n = 1 in Eq(3) gives
b1 = 0
For 2 ≤ n the recursive equation is
bn (n + r) (n + r − 1) + 2bn−2 (n + r − 2) + bn−2 − 2bn = 0
(4)
Which for for the root r = −1 becomes
bn (n − 1) (n − 2) + 2bn−2 (n − 3) + bn−2 − 2bn = 0
(4A)
Solving for bn from the recursive equation (4) gives
bn−2 (2n + 2r − 3)
bn = − 2
n + 2nr + r2 − n − r − 2
(5)
Which for the root r = −1 becomes
bn = −
bn−2 (2n − 5)
n2 − 3n
(6)
At this point, it is a good idea to keep track of bn in a table both before substituting r = −1
and after as more terms are found using the above recursive equation.
n
b0
b1
bn,r
1
0
bn
1
0
chapter 2 . book solved problems
416
For n = 2, using the above recursive equation gives
b2 = −
1 + 2r
r (r + 3)
Which for the root r = −1 becomes
b2 = −
1
2
And the table now becomes
n
b0
b1
b2
bn,r
1
0
bn
1
0
− 12
−2r−1
r(r+3)
For n = 3, using the above recursive equation gives
b3 = 0
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
0
bn
1
0
− 12
0
−2r−1
r(r+3)
0
For n = 4, using the above recursive equation gives
4r2 + 12r + 5
b4 =
r (r + 3) (r2 + 7r + 10)
Which for the root r = −1 becomes
b4 =
3
8
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
0
bn
1
0
− 12
0
4r2 +12r+5
r(r+3)(r+5)(r+2)
3
8
−2r−1
r(r+3)
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
0
0
bn
1
0
− 12
0
4r2 +12r+5
r(r+3)(r+5)(r+2)
3
8
0
0
−2r−1
r(r+3)
chapter 2 . book solved problems
417
Using the above table, then the solution y2 (x) is
y2 (x) = x2 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
2
4
1 − x2 + 3x8 + O(x6 )
=
x
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x2
3x4
6
c2 1 − 2 + 8 + O(x )
x2 9x4
= c1 x 2 1 −
+
+ O x6 +
2
56
x
Hence the final solution is
y = yh
= c1 x 2 1 −
2
4
x
9x
+
+O x
2
56
6
+
2
4
c2 1 − x2 + 3x8 + O(x6 )
x
3 Maple. Time used: 0.024 (sec). Leaf size: 35
Order:=6;
ode:=x^2*diff(diff(y(x),x),x)+2*x^3*diff(y(x),x)+(x^2-2)*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
c2 12 − 6x2 + 92 x4 + O (x6 )
1 2
9 4
6
y = c1 x 1 − x + x + O x
+
2
56
x
2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
•
•
Let’s solve
d d
3 d
x2 dx
y(x)
+
2x
y(x)
+ (x2 − 2) y(x) = 0
dx
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
x2 −2 y(x)
d d
d
y(x)
=
−
− 2x dx
y(x)
2
dx dx
x
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
x2 −2 y(x)
d d
d
y(x)
+
2x
y(x)
+
=0
dx dx
dx
x2
chapter 2 . book solved problems
◦
418
Check to see if x0 = 0 is a regular singular point
Define
functions
h
i
2
P2 (x) = 2x, P3 (x) = x x−2
2
◦
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
=0
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
= −2
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
3 d
x2 dx
y(x)
+
2x
y(x)
+ (x2 − 2) y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert xm · y(x) to series expansion for m = 0..2
∞
P
xm · y(x) =
ak xk+r+m
k=0
◦
◦
Shift index using k− >k − m
∞
P
xm · y(x) =
ak−m xk+r
k=m
d
Convert x3 · dx
y(x) to series expansion
∞
P
d
x3 · dx
y(x) =
ak (k + r) xk+r+2
k=0
◦
◦
Shift index using k− >k − 2
∞
P
d
x3 · dx
y(x) =
ak−2 (k − 2 + r) xk+r
k=2
d d
Convert x2 · dx
y(x) to series expansion
dx
∞
P
d d
x2 · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r
dx
k=0
Rewrite ODE with series expansions
r
a0 (1 + r) (−2 + r) x + a1 (2 + r) (−1 + r) x
1+r
+
∞
P
(ak (k + r + 1) (k − 2 + r) + ak−2 (2k − 3 + 2
k=2
•
•
•
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
(1 + r) (−2 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {−1, 2}
Each term must be 0
a1 (2 + r) (−1 + r) = 0
Solve for the dependent coefficient(s)
a1 = 0
Each term in the series must be 0, giving the recursion relation
ak (k + r + 1) (k − 2 + r) + 2 k − 32 + r ak−2 = 0
Shift index using k− >k + 2
ak+2 (k + 3 + r) (k + r) + 2 k + 12 + r ak = 0
Recursion relation that defines series solution to ODE
(2k+2r+1)ak
ak+2 = − (k+3+r)(k+r)
Recursion relation for r = −1
(2k−1)ak
ak+2 = − (k+2)(k−1)
Solution
for r = −1
∞
P
(2k−1)ak
k−1
y(x) =
ak x , ak+2 = − (k+2)(k−1) , a1 = 0
k=0
•
Recursion relation for r = 2
chapter 2 . book solved problems
•
419
(2k+5)ak
ak+2 = − (k+5)(k+2)
Solution
for r = 2
∞
P
(2k+5)ak
k+2
y(x) =
ak x , ak+2 = − (k+5)(k+2) , a1 = 0
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
(2k−1)ak
(2k+5)bk
k−1
k+2
y(x) =
ak x
+
bk x
, ak+2 = − (k+2)(k−1) , a1 = 0, bk+2 = − (k+5)(k+2) , b1 = 0
k=0
k=0
3 Mathematica. Time used: 0.008 (sec). Leaf size: 44
ode=x^2*D[y[x],{x,2}]+2*x^3*D[y[x],x]+(x^2-2)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
y(x) → c1
3x3 x 1
− +
8
2 x
+ c2
9x6 x4
−
+ x2
56
2
3 Sympy. Time used: 0.329 (sec). Leaf size: 41
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(2*x**3*Derivative(y(x), x) + x**2*Derivative(y(x), (x, 2)) + (x**2 - 2)*
y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
2
C − 7x6 + 3x4 − x2 + 1
1
48
8
2
x
y(x) = C2 x2
−1 +
+ O x6
2
x
chapter 2 . book solved problems
2.5.7
420
Problem 17
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428
Internal problem ID [8618]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 17
Date solved : Tuesday, November 11, 2025 at 12:04:12 PM
CAS classification : [_Laguerre]
xy 00 − (1 + x) y 0 + y = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + (−x − 1) y 0 + y = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
p(x) = −
q(x) =
1+x
x
1
x
Table 2.77: Table p(x), q(x) singularites.
p(x) = − 1+x
x
singularity
type
x=0
“regular”
q(x) = x1
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + (−x − 1) y 0 + y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
421
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
(n + r) (n + r − 1) an xn+r−2
∞
X
x + (−x − 1)
n=0
!
(n + r) an xn+r−1 +
n=0
∞
X
!
an xn+r
=0
n=0
(1)
Which simplifies to
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
+
n=0
+
∞
X
∞
X
−xn+r an (n + r)
n =0
∞
X
−(n + r) an xn+r−1 +
an xn+r
n =0
(2A)
!
=0
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
−x
n+r
∞
X
an (n + r) =
−an−1 (n + r − 1) xn+r−1
n =0
n=1
∞
X
an x
n+r
=
n =0
∞
X
an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
n=0
+
∞
X
−(n + r) an xn+r−1 +
n =0
+
∞
X
−an−1 (n + r − 1) xn+r−1
n =1
∞
X
!
an−1 xn+r−1
=0
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) − (n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) − ra0 x−1+r = 0
Or
x−1+r r(−1 + r) − r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
r x−1+r (−2 + r) = 0
(2B)
chapter 2 . book solved problems
422
Since the above is true for all x then the indicial equation becomes
r(−2 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 2
r2 = 0
Since a0 6= 0 then the indicial equation becomes
r x−1+r (−2 + r) = 0
Solving for r gives the roots of the indicial equation as [2, 0].
Since r1 − r2 = 2 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = Cy1 (x) ln (x) + xr2
∞
X
!
bn x n
n=0
Or
∞
X
y1 (x) = x2
!
an x n
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
an xn+2
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x
n
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. For 1 ≤ n the recursive equation is
an (n + r) (n + r − 1) − an−1 (n + r − 1) − an (n + r) + an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an =
an−1
n+r
Which for the root r = 2 becomes
(4)
an−1
(5)
n+2
At this point, it is a good idea to keep track of an in a table both before substituting r = 2
and after as more terms are found using the above recursive equation.
an =
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 =
1
1+r
chapter 2 . book solved problems
423
Which for the root r = 2 becomes
1
3
a1 =
And the table now becomes
n
a0
a1
an,r
1
an
1
1
1+r
1
3
For n = 2, using the above recursive equation gives
a2 =
1
(1 + r) (2 + r)
Which for the root r = 2 becomes
a2 =
1
12
And the table now becomes
n
a0
a1
a2
an,r
1
an
1
1
1+r
1
3
1
12
1
(1+r)(2+r)
For n = 3, using the above recursive equation gives
a3 =
1
(2 + r) (3 + r) (1 + r)
Which for the root r = 2 becomes
a3 =
1
60
And the table now becomes
n
a0
a1
a2
a3
an,r
1
an
1
1
1+r
1
3
1
12
1
60
1
(1+r)(2+r)
1
(2+r)(3+r)(1+r)
For n = 4, using the above recursive equation gives
a4 =
1
(3 + r) (1 + r) (2 + r) (4 + r)
Which for the root r = 2 becomes
a4 =
1
360
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
an
1
1
1+r
1
3
1
12
1
60
1
360
1
(1+r)(2+r)
1
(2+r)(3+r)(1+r)
1
(3+r)(1+r)(2+r)(4+r)
For n = 5, using the above recursive equation gives
a5 =
1
(1 + r) (2 + r) (4 + r) (3 + r) (5 + r)
chapter 2 . book solved problems
424
Which for the root r = 2 becomes
a5 =
1
2520
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
an
1
1
1+r
1
3
1
12
1
60
1
360
1
2520
1
(1+r)(2+r)
1
(2+r)(3+r)(1+r)
1
(3+r)(1+r)(2+r)(4+r)
1
(1+r)(2+r)(4+r)(3+r)(5+r)
Using the above table, then the solution y1 (x) is
y1 (x) = x2 a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x x2 x3
x4
x5
2
6
=x 1+ +
+
+
+
+O x
3 12 60 360 2520
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 2. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a2 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a2
=
1
(1 + r) (2 + r)
Therefore
1
1
= lim
r→r2 (1 + r) (2 + r)
r→0 (1 + r) (2 + r)
1
=
2
lim
The limit is 12 . Since the limit exists then the log term is not needed and we can set C = 0.
Therefore the second solution has the form
y2 (x) =
∞
X
bn xn+r
n=0
=
∞
X
bn x n
n=0
Eq (3) derived above is used to find all bn coefficients. The case n = 0 is skipped since it was
used to find the roots of the indicial equation. b0 is arbitrary and taken as b0 = 1. For 1 ≤ n
the recursive equation is
bn (n + r) (n + r − 1) − bn−1 (n + r − 1) − (n + r) bn + bn−1 = 0
(4)
Which for for the root r = 0 becomes
bn n(n − 1) − bn−1 (n − 1) − nbn + bn−1 = 0
(4A)
Solving for bn from the recursive equation (4) gives
bn =
bn−1
n+r
(5)
chapter 2 . book solved problems
425
Which for the root r = 0 becomes
bn−1
(6)
n
At this point, it is a good idea to keep track of bn in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
bn =
n
b0
bn,r
1
bn
1
For n = 1, using the above recursive equation gives
b1 =
1
1+r
Which for the root r = 0 becomes
b1 = 1
And the table now becomes
n
b0
b1
bn,r
1
bn
1
1
1
1+r
For n = 2, using the above recursive equation gives
b2 =
1
(1 + r) (2 + r)
Which for the root r = 0 becomes
b2 =
1
2
And the table now becomes
n
b0
b1
b2
bn,r
1
1
1+r
bn
1
1
1
(1+r)(2+r)
1
2
For n = 3, using the above recursive equation gives
b3 =
1
(2 + r) (3 + r) (1 + r)
Which for the root r = 0 becomes
b3 =
1
6
And the table now becomes
n
b0
b1
b2
b3
bn,r
1
1
1+r
bn
1
1
1
(1+r)(2+r)
1
(2+r)(3+r)(1+r)
1
2
1
6
For n = 4, using the above recursive equation gives
b4 =
1
(3 + r) (1 + r) (2 + r) (4 + r)
chapter 2 . book solved problems
Which for the root r = 0 becomes
b4 =
426
1
24
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
1
1+r
bn
1
1
1
(1+r)(2+r)
1
(2+r)(3+r)(1+r)
1
(3+r)(1+r)(2+r)(4+r)
1
2
1
6
1
24
For n = 5, using the above recursive equation gives
b5 =
1
(1 + r) (2 + r) (4 + r) (3 + r) (5 + r)
Which for the root r = 0 becomes
b5 =
1
120
And the table now becomes
n
b0
b1
b2
b3
b4
b5
bn,r
1
1
1+r
bn
1
1
1
(1+r)(2+r)
1
(2+r)(3+r)(1+r)
1
(3+r)(1+r)(2+r)(4+r)
1
(1+r)(2+r)(4+r)(3+r)(5+r)
1
2
1
6
1
24
1
120
Using the above table, then the solution y2 (x) is
y2 (x) = b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x2 x3 x4
x5
=1+x+
+
+
+
+ O x6
2
6
24 120
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x x2 x3 x4
x5
x2 x3 x 4 x5
2
6
6
= c1 x 1+ + + +
+
+O x
+c2 1+x+ + + +
+O x
3 12 60 360 2520
2
6 24 120
Hence the final solution is
y = yh
x x 2 x3
x4
x5
x2 x3 x4
x5
6
6
= c1 x 1 + + + +
+
+O x
+ c2 1 + x + + + +
+O x
3 12 60 360 2520
2
6 24 120
2
3 Maple. Time used: 0.030 (sec). Leaf size: 44
Order:=6;
ode:=y(x)-(x+1)*diff(y(x),x)+diff(diff(y(x),x),x)*x = 0;
dsolve(ode,y(x),type='series',x=0);
1
1 2
1 3
1 4
1 5
6
y = c1 x 1 + x + x + x +
x +
x +O x
3
12
60
360
2520
1 3
1 4
1 5
2
6
+ c2 −2 − 2x − x − x − x − x + O x
3
12
60
2
chapter 2 . book solved problems
427
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Reducible group (found an exponential solution)
Reducible group (found another exponential solution)
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x dx
y(x) − (x + 1) dx
y(x) + y(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative •
•
d d
y(x) = − y(x)
+
dx dx
x
•
(x+1)
d
y(x)
dx
x
Group terms with
y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
◦
◦
(x · P2 (x))
◦
•
•
d
y(x)
dx
x
= −1
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
(x+1)
+ y(x)
=0
x
Check to see if x0 = 0 is a regular singular point
Define functions
1
P2 (x) = − x+1
,
P
(x)
=
3
x
x
x · P2 (x) is analytic at x = 0
d d
y(x) −
dx dx
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x)
+
(−x
−
1)
y(x)
+ y(x) = 0
dx
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d
Convert xm · dx
y(x) to series expansion for m = 0..1
∞
P
d
xm · dx
y(x) =
ak (k + r) xk+r−1+m
k=0
◦
◦
Shift index using k− >k + 1 − m
∞
P
d
xm · dx
y(x) =
ak+1−m (k + 1 − m + r) xk+r
k=−1+m
d d
Convert x · dx
y(x)
to series expansion
dx
∞
P
d d
x · dx
y(x) =
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
chapter 2 . book solved problems
a0 r(−2 + r) x
−1+r
+
∞
P
(ak+1 (k + 1 + r) (k + r − 1) − ak (k + r − 1)) x
428
k+r
=0
k=0
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r(−2 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {0, 2}
Each term in the series must be 0, giving the recursion relation
(k + r − 1) (ak+1 (k + 1 + r) − ak ) = 0
Recursion relation that defines series solution to ODE
ak
ak+1 = k+1+r
Recursion relation for r = 0
ak
ak+1 = k+1
Solution
for r = 0
∞
P
a
k
y(x) =
ak xk , ak+1 = k+1
k=0
•
•
Recursion relation for r = 2
ak
ak+1 = k+3
Solution
for r = 2
∞
P
ak
k+2
y(x) =
ak x , ak+1 = k+3
k=0
•
Combine
parameters
solutions
andrename
∞
∞
P
P
ak
bk
k
k+2
y(x) =
ak x +
bk x
, ak+1 = k+1 , bk+1 = k+3
k=0
k=0
3 Mathematica. Time used: 0.014 (sec). Leaf size: 66
ode=x*D[y[x],{x,2}]-(x+1)*D[y[x],x]+y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
4
6
x
x3 x2
x
x5 x4 x3
2
y(x) → c1
+
+
+ x + 1 + c2
+
+
+
+x
24
6
2
360 60 12
3
3 Sympy. Time used: 0.249 (sec). Leaf size: 31
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(x*Derivative(y(x), (x, 2)) - (x + 1)*Derivative(y(x), x) + y(x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
3
x
x2 x
y(x) = C2 (x + 1) + C1 x
+
+ + 1 + O x6
60 12 3
2
chapter 2 . book solved problems
2.5.8
429
Problem 18
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436
Internal problem ID [8619]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 18
Date solved : Tuesday, November 11, 2025 at 12:04:13 PM
CAS classification : [[_Emden, _Fowler]]
xy 00 + 3y 0 + 4yx3 = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + 3y 0 + 4yx3 = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
3
x
q(x) = 4x2
p(x) =
Table 2.79: Table p(x), q(x) singularites.
q(x) = 4x2
singularity
type
x=∞
“regular”
x = −∞ “regular”
p(x) = x3
singularity
type
x=0
“regular”
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0, ∞, −∞]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + 3y 0 + 4yx3 = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
430
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
∞
X
!
∞
X
(n + r) (n + r − 1) an xn+r−2 x + 3
n=0
!
(n + r) an xn+r−1 + 4
n=0
∞
X
!
an xn+r
x3 = 0 (1)
n=0
Which simplifies to
∞
X
!
xn+r−1 an (n + r) (n + r − 1) +
n=0
∞
X
!
3(n + r) an xn+r−1 +
n=0
∞
X
!
4x3+n+r an
= 0 (2A)
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
4x
3+n+r
n =0
an =
∞
X
4an−4 xn+r−1
n=4
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
xn+r−1 an (n+r) (n+r −1) +
n=0
∞
X
!
3(n+r) an xn+r−1 +
n=0
∞
X
!
4an−4 xn+r−1
n=4
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + 3(n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + 3ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + 3r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
r x−1+r (2 + r) = 0
Since the above is true for all x then the indicial equation becomes
r(2 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = −2
Since a0 6= 0 then the indicial equation becomes
r x−1+r (2 + r) = 0
= 0 (2B)
chapter 2 . book solved problems
431
Solving for r gives the roots of the indicial equation as [0, −2].
Since r1 − r2 = 2 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
y2 (x) = Cy1 (x) ln (x) + xr2
∞
X
!
bn x n
n=0
Or
y1 (x) =
∞
X
an x n
n=0
∞
P
bn x n
y2 (x) = Cy1 (x) ln (x) + n=0 2
x
Or
y1 (x) =
∞
X
an xn
n=0
y2 (x) = Cy1 (x) ln (x) +
∞
X
!
bn x
n−2
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. Substituting n = 1 in Eq. (2B) gives
a1 = 0
Substituting n = 2 in Eq. (2B) gives
a2 = 0
Substituting n = 3 in Eq. (2B) gives
a3 = 0
For 4 ≤ n the recursive equation is
an (n + r) (n + r − 1) + 3an (n + r) + 4an−4 = 0
(3)
Solving for an from recursive equation (4) gives
4an−4
an = − 2
n + 2nr + r2 + 2n + 2r
(4)
Which for the root r = 0 becomes
an = −
4an−4
n (n + 2)
(5)
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
n
a0
a1
a2
a3
an,r
1
0
0
0
an
1
0
0
0
For n = 4, using the above recursive equation gives
4
a4 = − 2
r + 10r + 24
chapter 2 . book solved problems
432
Which for the root r = 0 becomes
a4 = −
1
6
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
0
0
0
4
− r2 +10r+24
an
1
0
0
0
− 16
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
a3
a4
a5
an,r
1
0
0
0
4
− r2 +10r+24
0
an
1
0
0
0
− 16
0
Using the above table, then the solution y1 (x) is
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x4
=1−
+ O x6
6
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 2. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a2 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a2
=0
Therefore
lim 0 = lim 0
r→r2
r→−2
=0
The limit is 0. Since the limit exists then the log term is not needed and we can set C = 0.
Therefore the second solution has the form
y2 (x) =
=
∞
X
n=0
∞
X
n=0
bn xn+r
bn xn−2
chapter 2 . book solved problems
433
Eq (3) derived above is used to find all bn coefficients. The case n = 0 is skipped since
it was used to find the roots of the indicial equation. b0 is arbitrary and taken as b0 = 1.
Substituting n = 1 in Eq(3) gives
b1 = 0
Substituting n = 2 in Eq(3) gives
b2 = 0
Substituting n = 3 in Eq(3) gives
b3 = 0
For 4 ≤ n the recursive equation is
bn (n + r) (n + r − 1) + 3(n + r) bn + 4bn−4 = 0
(4)
Which for for the root r = −2 becomes
bn (n − 2) (n − 3) + 3(n − 2) bn + 4bn−4 = 0
(4A)
Solving for bn from the recursive equation (4) gives
4bn−4
bn = − 2
n + 2nr + r2 + 2n + 2r
(5)
Which for the root r = −2 becomes
4bn−4
bn = − 2
n − 2n
(6)
At this point, it is a good idea to keep track of bn in a table both before substituting r = −2
and after as more terms are found using the above recursive equation.
n
b0
b1
b2
b3
bn,r
1
0
0
0
bn
1
0
0
0
For n = 4, using the above recursive equation gives
4
b4 = − 2
r + 10r + 24
Which for the root r = −2 becomes
b4 = −
1
2
And the table now becomes
n
b0
b1
b2
b3
b4
bn,r
1
0
0
0
4
− r2 +10r+24
For n = 5, using the above recursive equation gives
b5 = 0
And the table now becomes
bn
1
0
0
0
− 12
chapter 2 . book solved problems
n
b0
b1
b2
b3
b4
b5
bn,r
1
0
0
0
4
− r2 +10r+24
0
434
bn
1
0
0
0
− 12
0
Using the above table, then the solution y2 (x) is
y2 (x) = 1 b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
4
1 − x2 + O(x6 )
=
x2
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x4
6
c2 1 − 2 + O(x )
x4
= c1 1 −
+ O x6 +
6
x2
Hence the final solution is
y = yh
x4
6
c2 1 − 2 + O(x )
x4
= c1 1 −
+ O x6 +
6
x2
3 Maple. Time used: 0.026 (sec). Leaf size: 28
Order:=6;
ode:=diff(diff(y(x),x),x)*x+3*diff(y(x),x)+4*x^3*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
y = c1
1 4
c2 (−2 + x4 + O (x6 ))
6
1− x +O x
+
6
x2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
A Liouvillian solution exists
Group is reducible or imprimitive
<- Kovacics algorithm successful
Maple step by step
Let’s solve
d d
d
x dx
y(x) + 3 dx
y(x) + 4x3 y(x) = 0
dx
chapter 2 . book solved problems
•
435
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
•
d d
y(x) = −4x2 y(x) −
dx dx
•
3
d
y(x)
dx
x
Group termswith y(x)
on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
3
d d
y(x) +
dx dx
◦
◦
+ 4x2 y(x) = 0
Check to see if x0 = 0 is a regular singular point
Define functions
P2 (x) = x3 , P3 (x) = 4x2
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
x
=3
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
d
y(x)
dx
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x)
+ 3 dx
y(x) + 4x3 y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert x3 · y(x) to series expansion
∞
P
x3 · y(x) =
ak xk+r+3
k=0
◦
Shift index using k− >k − 3
∞
P
x3 · y(x) =
ak−3 xk+r
◦
d
Convert dx
y(x) to series expansion
∞
P
d
y(x) =
ak (k + r) xk+r−1
dx
◦
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + 1 + r) xk+r
dx
k=−1
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x) =
ak (k + r) (k + r − 1) xk+r−1
dx
k=3
k=0
◦
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
a0 r(2 + r) x
−1+r
r
+ a1 (1 + r) (3 + r) x + a2 (2 + r) (4 + r) x
1+r
+ a3 (3 + r) (5 + r) x
2+r
+
∞
P
k=3
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r(2 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {−2, 0}
The coefficients of each power of x must be 0
[a1 (1 + r) (3 + r) = 0, a2 (2 + r) (4 + r) = 0, a3 (3 + r) (5 + r) = 0]
Solve for the dependent coefficient(s)
{a1 = 0, a2 = 0, a3 = 0}
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1 + r) (k + r + 3) + 4ak−3 = 0
Shift index using k− >k + 3
(ak+
chapter 2 . book solved problems
•
•
•
436
ak+4 (k + 4 + r) (k + 6 + r) + 4ak = 0
Recursion relation that defines series solution to ODE
4ak
ak+4 = − (k+4+r)(k+6+r)
Recursion relation for r = −2
4ak
ak+4 = − (k+2)(k+4)
Solution
for r = −2
∞
P
4ak
k−2
y(x) =
ak x , ak+4 = − (k+2)(k+4) , a1 = 0, a2 = 0, a3 = 0
k=0
•
•
Recursion relation for r = 0
4ak
ak+4 = − (k+4)(k+6)
Solution
for r = 0
∞
P
4ak
k
y(x) =
ak x , ak+4 = − (k+4)(k+6) , a1 = 0, a2 = 0, a3 = 0
k=0
•
Combine
and
solutions
rename
∞ parameters
∞
P
P
4ak
4bk
k−2
k
y(x) =
ak x
+
bk x , ak+4 = − (k+2)(k+4)
, a1 = 0, a2 = 0, a3 = 0, bk+4 = − (k+4)(k+6)
k=0
k=0
3 Mathematica. Time used: 0.006 (sec). Leaf size: 30
ode=x*D[y[x],{x,2}]+3*D[y[x],x]+4*x^3*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
x4
1
x2
y(x) → c2 1 −
+ c1 2 −
6
x
2
3 Sympy. Time used: 0.240 (sec). Leaf size: 26
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(4*x**3*y(x) + x*Derivative(y(x), (x, 2)) + 3*Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
C 1 − x4
4
1
2
x
y(x) = C2 1 −
+
+ O x6
2
6
x
chapter 2 . book solved problems
2.5.9
437
Problem 19
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446
Internal problem ID [8620]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 19
Date solved : Tuesday, November 11, 2025 at 12:04:13 PM
CAS classification : [[_Emden, _Fowler]]
y 00 +
y
=0
4x
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
y
y 00 +
=0
4x
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
p(x) = 0
1
q(x) =
4x
Table 2.81: Table p(x), q(x) singularites.
1
q(x) = 4x
singularity
type
x=0
“regular”
p(x) = 0
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
4xy 00 + y = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
n=0
an xn+r
chapter 2 . book solved problems
438
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
(n + r) (n + r − 1) an xn+r−2
n=0
Substituting the above back into the ode gives
4
∞
X
!
(n + r) (n + r − 1) an xn+r−2 x +
∞
X
n=0
!
an xn+r
=0
(1)
=0
(2A)
n=0
Which simplifies to
∞
X
!
4xn+r−1 an (n + r) (n + r − 1)
∞
X
+
n=0
!
an xn+r
n=0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
an x
n+r
n =0
=
∞
X
an−1 xn+r−1
n=1
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
4xn+r−1 an (n + r) (n + r − 1)
+
n=0
∞
X
!
an−1 xn+r−1
n=1
The indicial equation is obtained from n = 0. From Eq (2B) this gives
4xn+r−1 an (n + r) (n + r − 1) = 0
When n = 0 the above becomes
4x−1+r a0 r(−1 + r) = 0
Or
4x−1+r a0 r(−1 + r) = 0
Since a0 6= 0 then the above simplifies to
4x−1+r r(−1 + r) = 0
Since the above is true for all x then the indicial equation becomes
4r(−1 + r) = 0
Solving for r gives the roots of the indicial equation as
r1 = 1
r2 = 0
Since a0 6= 0 then the indicial equation becomes
4x−1+r r(−1 + r) = 0
=0
(2B)
chapter 2 . book solved problems
439
Solving for r gives the roots of the indicial equation as [1, 0].
Since r1 − r2 = 1 is an integer, then we can construct two linearly independent solutions
!
∞
X
y1 (x) = xr1
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) + xr2
!
bn x n
n=0
Or
y1 (x) = x
∞
X
!
an x n
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) +
!
bn x n
n=0
Or
y1 (x) =
∞
X
an xn+1
n=0
∞
X
y2 (x) = Cy1 (x) ln (x) +
!
bn x n
n=0
Where C above can be zero. We start by finding y1 . Eq (2B) derived above is now used to
find all an coefficients. The case n = 0 is skipped since it was used to find the roots of the
indicial equation. a0 is arbitrary and taken as a0 = 1. For 1 ≤ n the recursive equation is
4an (n + r) (n + r − 1) + an−1 = 0
(3)
Solving for an from recursive equation (4) gives
an = −
an−1
4 (n + r) (n + r − 1)
(4)
an−1
4 (n + 1) n
(5)
Which for the root r = 1 becomes
an = −
At this point, it is a good idea to keep track of an in a table both before substituting r = 1
and after as more terms are found using the above recursive equation.
n
a0
an,r
1
an
1
For n = 1, using the above recursive equation gives
a1 = −
1
4 (1 + r) r
Which for the root r = 1 becomes
a1 = −
1
8
And the table now becomes
n
a0
a1
an,r
1
1
− 4(1+r)r
an
1
− 18
chapter 2 . book solved problems
440
For n = 2, using the above recursive equation gives
a2 =
1
16 (1 + r)2 r (2 + r)
Which for the root r = 1 becomes
a2 =
1
192
And the table now becomes
n
a0
a1
a2
an,r
1
1
− 4(1+r)r
an
1
− 18
1
16(1+r)2 r(2+r)
1
192
For n = 3, using the above recursive equation gives
a3 = −
1
64 (1 + r) r (2 + r)2 (3 + r)
2
Which for the root r = 1 becomes
a3 = −
1
9216
And the table now becomes
n
a0
a1
a2
a3
an,r
1
1
− 4(1+r)r
an
1
− 18
1
16(1+r)2 r(2+r)
1
− 64(1+r)2 r(2+r)
2
(3+r)
1
192
1
− 9216
For n = 4, using the above recursive equation gives
a4 =
1
256 (1 + r) r (2 + r)2 (3 + r)2 (4 + r)
2
Which for the root r = 1 becomes
a4 =
1
737280
And the table now becomes
n
a0
a1
a2
a3
a4
an,r
1
1
− 4(1+r)r
an
1
− 18
1
16(1+r)2 r(2+r)
1
− 64(1+r)2 r(2+r)
2
(3+r)
1
256(1+r)2 r(2+r)2 (3+r)2 (4+r)
1
192
1
− 9216
1
737280
For n = 5, using the above recursive equation gives
a5 = −
1
1024 (1 + r) r (2 + r) (3 + r)2 (4 + r)2 (5 + r)
2
2
Which for the root r = 1 becomes
a5 = −
And the table now becomes
1
88473600
chapter 2 . book solved problems
n
a0
a1
a2
a3
a4
a5
441
an,r
1
1
− 4(1+r)r
an
1
− 18
1
16(1+r)2 r(2+r)
1
− 64(1+r)2 r(2+r)
2
(3+r)
1
2
2
256(1+r) r(2+r) (3+r)2 (4+r)
− 1024(1+r)2 r(2+r)21(3+r)2 (4+r)2 (5+r)
1
192
1
− 9216
1
737280
1
− 88473600
Using the above table, then the solution y1 (x) is
y1 (x) = x a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
x
x2
x3
x4
x5
6
=x 1− +
−
+
−
+O x
8 192 9216 737280 88473600
Now the second solution y2 (x) is found. Let
r1 − r2 = N
Where N is positive integer which is the difference between the two roots. r1 is taken as the
larger root. Hence for this problem we have N = 1. Now we need to determine if C is zero
or not. This is done by finding limr→r2 a1 (r). If this limit exists, then C = 0, else we need to
keep the log term and C 6= 0. The above table shows that
aN = a1
=−
1
4 (1 + r) r
Therefore
lim −
r→r2
1
1
= lim −
4 (1 + r) r r→0 4 (1 + r) r
= undefined
Since the limit does not exist then the log term is needed. Therefore the second solution has
the form
!
∞
X
y2 (x) = Cy1 (x) ln (x) +
bn xn+r2
n=0
Therefore
d
Cy1 (x)
y2 (x) = Cy10 (x) ln (x) +
+
dx
x
Cy1 (x)
= Cy10 (x) ln (x) +
+
x
∞
X
bn xn+r2 (n + r2 )
x
n=0
∞
X
!
x
−1+n+r2
bn (n + r2 )
n=0
∞
d2
2Cy10 (x) Cy1 (x) X
00
y2 (x) = Cy1 (x) ln (x) +
−
+
dx2
x
x2
n=0
2Cy10 (x) Cy1 (x)
= Cy100 (x) ln (x) +
−
+
2
x
!
x
∞
X
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!
x
−2+n+r2
bn (n + r2 ) (−1 + n + r2 )
n=0
Substituting these back into the given ode 4xy 00 + y = 0 gives
∞
2Cy10 (x) Cy1 (x) X
00
4 Cy1 (x) ln (x) +
−
+
x
x2
n=0
!
∞
X
+ Cy1 (x) ln (x) +
bn xn+r2 = 0
n=0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!!
x
!
chapter 2 . book solved problems
442
Which can be written as
0
2y1 (x) y1 (x)
00
(4y1 (x) x + y1 (x)) ln (x) + 4x
− 2
C
x
x
!!
∞
X
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
+4
−
x+
x2
x2
n=0
(7)
∞
X
!
bn xn+r2
=0
bn xn+r2 (n + r2 )2 bn xn+r2 (n + r2 )
−
x2
x2
!!
n=0
But since y1 (x) is a solution to the ode, then
4y100 (x) x + y1 (x) = 0
Eq (7) simplifes to
∞
X
2y10 (x) y1 (x)
4x
− 2
C +4
x
x
n=0
!
∞
X
+
bn xn+r2 = 0
x
(8)
n=0
∞
P
Substituting y1 =
an xn+r1 into the above gives
n=0
∞
∞
P −1+n+r1
P
n+r1
8
x
an (n + r1 ) x − 4
an x
C
n=0
n=0
(9)
x
∞
∞
P −2+n+r2
P
4
x
bn (n + r2 ) (−1 + n + r2 ) x2 +
bn xn+r2 x
+
n=0
n=0
=0
x
Since r1 = 1 and r2 = 0 then the above becomes
∞
∞
P n
P
n+1
8
x an (n + 1) x − 4
an x
C
n=0
n=0
n=0
+
x
=0
∞
∞
P −2+n
P
2
n
4
x
bn n(n − 1) x +
bn x x
n=0
x
(10)
Which simplifies to
∞
X
n=0
!
8xn an (n + 1) C
+
∞
X
(−4C xn an ) +
n =0
∞
X
!
4n xn−1 bn (n − 1) +
n=0
∞
X
!
bn x n
= 0 (2A)
n=0
The next step is to make all powers of x be n − 1 in each summation term. Going over each
summation term above with power of x in it which is not already xn−1 and adjusting the
power and the corresponding index gives
∞
X
n
8x an (n + 1) C =
n =0
∞
X
8Can−1 n xn−1
n=1
∞
X
n
(−4C x an ) =
n =0
∞
X
−4Can−1 xn−1
n=1
∞
X
n =0
bn x n =
∞
X
n=1
bn−1 xn−1
chapter 2 . book solved problems
443
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n − 1.
∞
X
n=1
!
8Can−1 n xn−1 +
∞
X
−4Can−1 x
n−1
n =1
+
∞
X
!
4n xn−1 bn (n − 1) +
n=0
∞
X
!
bn−1 xn−1
=0
n=1
(2B)
For n = 0 in Eq. (2B), we choose arbitray value for b0 as b0 = 1. For n = N , where N = 1
which is the difference between the two roots, we are free to choose b1 = 0. Hence for n = 1,
Eq (2B) gives
4C + 1 = 0
Which is solved for C. Solving for C gives
C=−
1
4
For n = 2, Eq (2B) gives
12Ca1 + b1 + 8b2 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
3
8b2 + = 0
8
Solving the above for b2 gives
3
b2 = −
64
For n = 3, Eq (2B) gives
20Ca2 + b2 + 24b3 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
7
24b3 −
=0
96
Solving the above for b3 gives
7
b3 =
2304
For n = 4, Eq (2B) gives
28Ca3 + b3 + 48b4 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
35
48b4 +
=0
9216
Solving the above for b4 gives
35
b4 = −
442368
For n = 5, Eq (2B) gives
36Ca4 + b4 + 80b5 = 0
Which when replacing the above values found already for bn and the values found earlier for
an and for C, gives
101
80b5 −
=0
1105920
Solving the above for b5 gives
101
b5 =
88473600
Now that we found all bn and C, we can calculate the second solution from
!
∞
X
y2 (x) = Cy1 (x) ln (x) +
bn xn+r2
n=0
chapter 2 . book solved problems
444
Using the above value found for C = − 14 and all bn , then the second solution becomes
1
x
x2
x3
x4
x5
6
y2 (x) = − x 1 − +
−
+
−
+O x
ln (x)
4
8 192 9216 737280 88473600
3x2
7x3
35x4
101x5
+1−
+
−
+
+ O x6
64
2304 442368 88473600
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x
x2
x3
x4
x5
6
= c1 x 1 − +
−
+
−
+O x
8 192 9216 737280 88473600
1
x
x2
x3
x4
x5
6
+ c2 − x 1 − +
−
+
−
+O x
ln (x) + 1
4
8 192 9216 737280 88473600
3x2
7x3
35x4
101x5
6
−
+
−
+
+O x
64
2304 442368 88473600
Hence the final solution is
y = yh
x
x2
x3
x4
x5
6
= c1 x 1 − +
−
+
−
+O x
8 192 9216 737280 88473600
 x2
x3
x4
x5
x 1 − x8 + 192
− 9216
+ 737280
− 88473600
+ O(x6 ) ln (x)
3x2
7x3
+ c2  −
+1−
+
4
64
2304
4
−
5
35x
101x
+
442368 88473600

+ O x6 
3 Maple. Time used: 0.018 (sec). Leaf size: 58
Order:=6;
ode:=diff(diff(y(x),x),x)+1/4*y(x)/x = 0;
dsolve(ode,y(x),type='series',x=0);
1
1 2
1 3
1
1
4
5
6
y = c1 x 1 − x +
x −
x +
x −
x +O x
8
192
9216
737280
88473600
1
1 2
1 3
1
1
4
5
6
+ c2 ln (x) − x + x −
x +
x −
x +O x
4
32
768
36864
2949120
3 2
7 3
35
101
4
5
6
+ 1− x +
x −
x +
x +O x
64
2304
442368
88473600
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
chapter 2 . book solved problems
<- Bessel successful
<- special function solution successful
Maple step by step
•
◦
◦
Let’s solve
d d
y(x) + y(x)
=0
dx dx
4x
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Check to see if x0 = 0 is a regular singular point
Define functions
1
P2 (x) = 0, P3 (x) = 4x
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
=0
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
4 dx
y(x)
x + y(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
d d
Convert x · dx
y(x)
to series expansion
dx
∞
P
d d
x · dx
y(x) =
ak (k + r) (k + r − 1) xk+r−1
dx
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + 1 + r) (k + r) xk+r
dx
k=−1
Rewrite ODE with series
∞expansions
P
−1+r
k+r
4a0 r(−1 + r) x
+
(4ak+1 (k + 1 + r) (k + r) + ak ) x
=0
k=0
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
4r(−1 + r) = 0
Values of r that satisfy the indicial equation
r ∈ {0, 1}
Each term in the series must be 0, giving the recursion relation
4ak+1 (k + 1 + r) (k + r) + ak = 0
Recursion relation that defines series solution to ODE
ak
ak+1 = − 4(k+1+r)(k+r)
Recursion relation for r = 0
ak
ak+1 = − 4(k+1)k
Solution
for r = 0
∞
P
a
k
y(x) =
ak xk , ak+1 = − 4(k+1)k
k=0
•
•
Recursion relation for r = 1
ak
ak+1 = − 4(k+2)(k+1)
Solution
for r = 1
∞
P
ak
k+1
y(x) =
ak x , ak+1 = − 4(k+2)(k+1)
•
Combine solutions and rename parameters
k=0
445
chapter 2 . book solved problems
y(x) =
∞
P
ak x
k
k=0
+
∞
P
k=0
bk x
k+1
446
ak
bk
, ak+1 = − 4(k+1)k
, bk+1 = − 4(k+2)(k+1)
3 Mathematica. Time used: 0.012 (sec). Leaf size: 85
ode=D[y[x],{x,2}]+1/(4*x)*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
x(x3 − 48x2 + 1152x − 9216) log(x)
36864
−47x4 + 1920x3 − 34560x2 + 110592x + 442368
+
442368
5
4
3
2
x
x
x
x
+ c2
−
+
−
+x
737280 9216 192
8
y(x) → c1
3 Sympy. Time used: 0.186 (sec). Leaf size: 29
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(Derivative(y(x), (x, 2)) + y(x)/(4*x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
x4
x3
x2
x
y(x) = C1 x
−
+
− + 1 + O x6
737280 9216 192 8
chapter 2 . book solved problems
2.5.10
447
Problem 20
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453
Internal problem ID [8621]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 5. Series Solutions of ODEs. REVIEW QUESTIONS. page 201
Problem number : 20
Date solved : Tuesday, November 11, 2025 at 12:04:14 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
xy 00 + y 0 − yx = 0
Using series expansion around x = 0.
The type of the expansion point is first determined. This is done on the homogeneous part
of the ODE.
xy 00 + y 0 − yx = 0
The following is summary of singularities for the above ode. Writing the ode as
y 00 + p(x)y 0 + q(x)y = 0
Where
1
x
q(x) = −1
p(x) =
Table 2.83: Table p(x), q(x) singularites.
p(x) = x1
singularity
type
x=0
“regular”
q(x) = −1
singularity type
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : [0]
Irregular singular points : [∞]
Since x = 0 is regular singular point, then Frobenius power series is used. The ode is
normalized to be
xy 00 + y 0 − yx = 0
Let the solution be represented as Frobenius power series of the form
y=
∞
X
an xn+r
n=0
Then
0
y =
∞
X
(n + r) an xn+r−1
n=0
00
y =
∞
X
n=0
(n + r) (n + r − 1) an xn+r−2
chapter 2 . book solved problems
448
Substituting the above back into the ode gives
∞
X
∞
X
!
(n + r) (n + r − 1) an x
n+r−2
x+
n=0
∞
X
!
(n + r) an x
n+r−1
−
n=0
!
an x
n+r
x = 0 (1)
n=0
Which simplifies to
∞
X
!
x
n+r−1
an (n + r) (n + r − 1)
+
n=0
∞
X
!
(n + r) an x
n+r−1
+
n=0
∞
X
−x1+n+r an = 0 (2A)
n =0
The next step is to make all powers of x be n + r − 1 in each summation term. Going over
each summation term above with power of x in it which is not already xn+r−1 and adjusting
the power and the corresponding index gives
∞
X
∞
X
−x1+n+r an =
−an−2 xn+r−1
n =0
n=2
Substituting all the above in Eq (2A) gives the following equation where now all powers of x
are the same and equal to n + r − 1.
∞
X
!
x
n+r−1
an (n + r) (n + r − 1) +
n=0
∞
X
!
(n + r) an x
n+r−1
+
n=0
∞
X
−an−2 xn+r−1 = 0 (2B)
n =2
The indicial equation is obtained from n = 0. From Eq (2B) this gives
xn+r−1 an (n + r) (n + r − 1) + (n + r) an xn+r−1 = 0
When n = 0 the above becomes
x−1+r a0 r(−1 + r) + ra0 x−1+r = 0
Or
x−1+r r(−1 + r) + r x−1+r a0 = 0
Since a0 6= 0 then the above simplifies to
x−1+r r2 = 0
Since the above is true for all x then the indicial equation becomes
r2 = 0
Solving for r gives the roots of the indicial equation as
r1 = 0
r2 = 0
Since a0 6= 0 then the indicial equation becomes
x−1+r r2 = 0
Solving for r gives the roots of the indicial equation as [0, 0].
Since the root of the indicial equation is repeated, then we can construct two linearly
independent solutions. The first solution has the form
y1 (x) =
∞
X
n=0
an xn+r
(1A)
chapter 2 . book solved problems
449
Now the second solution y2 is found using
∞
X
y2 (x) = y1 (x) ln (x) +
!
bn xn+r
(1B)
n=1
Then the general solution will be
y = c1 y1 (x) + c2 y2 (x)
In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), a0 is never zero, and is arbitrary
and is typically taken as a0 = 1, and {c1 , c2 } are two arbitray constants of integration which
can be found from initial conditions. We start by finding the first solution y1 (x). Eq (2B)
derived above is now used to find all an coefficients. The case n = 0 is skipped since it
was used to find the roots of the indicial equation. a0 is arbitrary and taken as a0 = 1.
Substituting n = 1 in Eq. (2B) gives
a1 = 0
For 2 ≤ n the recursive equation is
an (n + r) (n + r − 1) + an (n + r) − an−2 = 0
(3)
Solving for an from recursive equation (4) gives
an =
an−2
2
n + 2nr + r2
Which for the root r = 0 becomes
(4)
an−2
(5)
n2
At this point, it is a good idea to keep track of an in a table both before substituting r = 0
and after as more terms are found using the above recursive equation.
an =
n
a0
a1
an,r
1
0
an
1
0
For n = 2, using the above recursive equation gives
a2 =
1
(r + 2)2
Which for the root r = 0 becomes
a2 =
1
4
And the table now becomes
n
a0
a1
a2
an,r
1
0
an
1
0
1
(r+2)2
1
4
For n = 3, using the above recursive equation gives
a3 = 0
And the table now becomes
n
a0
a1
a2
an,r
1
0
an
1
0
1
(r+2)2
1
4
a3
0
0
chapter 2 . book solved problems
450
For n = 4, using the above recursive equation gives
a4 =
1
(r + 2) (r + 4)2
2
Which for the root r = 0 becomes
a4 =
1
64
And the table now becomes
n
a0
a1
a2
an,r
1
0
an
1
0
1
(r+2)2
1
4
a3
a4
0
0
1
(r+2)2 (r+4)2
1
64
For n = 5, using the above recursive equation gives
a5 = 0
And the table now becomes
n
a0
a1
a2
an,r
1
0
an
1
0
1
(r+2)2
1
4
a3
a4
0
0
1
(r+2)2 (r+4)2
1
64
a5
0
0
Using the above table, then the first solution y1 (x) becomes
y1 (x) = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 . . .
=1+
x2 x4
+
+ O x6
4
64
Now the second solution is found. The second solution is given by
!
∞
X
y2 (x) = y1 (x) ln (x) +
bn xn+r
n=1
Where bn is found using
d
an,r
dr
And the above is then evaluated at r = 0. The above table for an,r is used for this purpose.
Computing the derivatives gives the following table
bn =
1
4
d
bn,r = dr
an,r
N/A since bn starts from 1
0
2
− (r+2)
3
bn (r = 0)
N/A
0
− 14
0
0
0
1
(r+2)2 (r+4)2
1
64
−4r−12
(r+2)3 (r+4)3
0
3
− 128
0
0
0
0
n
b0
b1
b2
bn,r
1
0
an
1
0
1
(r+2)2
b3
b4
b5
chapter 2 . book solved problems
451
The above table gives all values of bn needed. Hence the second solution is
y2 (x) = y1 (x) ln (x) + b0 + b1 x + b2 x2 + b3 x3 + b4 x4 + b5 x5 + b6 x6 . . .
x2 x4
x2 3x4
6
= 1+
+
+O x
ln (x) −
−
+ O x6
4
64
4
128
Therefore the homogeneous solution is
yh (x) = c1 y1 (x) + c2 y2 (x)
x2 x4
x2 x4
x2 3x4
6
6
6
= c1 1 + +
+O x
+ c2
1+ +
+O x
ln (x) − −
+O x
4
64
4
64
4
128
Hence the final solution is
y = yh
x2 x4
x2 x4
x2 3x4
6
6
6
= c1 1 +
+
+O x
+ c2
1+
+
+O x
ln (x) −
−
+O x
4
64
4
64
4
128
3 Maple. Time used: 0.025 (sec). Leaf size: 32
Order:=6;
ode:=diff(diff(y(x),x),x)*x+diff(y(x),x)-x*y(x) = 0;
dsolve(ode,y(x),type='series',x=0);
1 2
1 4
1 2
3 4
6
6
y = (c2 ln (x) + c1 ) 1 + x + x + O x
+ − x −
x +O x
c2
4
64
4
128
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
checking if the LODE is of Euler type
trying a symmetry of the form [xi=0, eta=F(x)]
checking if the LODE is missing y
-> Trying a Liouvillian solution using Kovacics algorithm
<- No Liouvillian solutions exists
-> Trying a solution in terms of special functions:
-> Bessel
<- Bessel successful
<- special function solution successful
Maple step by step
•
•
Let’s solve
d d
d
x dx
y(x) + dx
y(x) − xy(x) = 0
dx
Highest derivative means the order of the ODE is 2
d d
y(x)
dx dx
Isolate 2nd derivative
d
y(x)
d d
y(x) = y(x) − dx x
dx dx
•
Group terms with y(x) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d
y(x)
d d
dx
y(x)
+
− y(x) = 0
dx dx
x
chapter 2 . book solved problems
◦
◦
Check to see if x0 = 0 is a regular singular point
Define functions
P2 (x) = x1 , P3 (x) = −1
x · P2 (x) is analytic at x = 0
(x · P2 (x))
◦
=1
x=0
x2 · P3 (x) is analytic at x = 0
(x2 · P3 (x))
◦
•
•
=0
x=0
x = 0is a regular singular point
Check to see if x0 = 0 is a regular singular point
x0 = 0
Multiply by denominators
d d
d
x dx
y(x) + dx
y(x) − xy(x) = 0
dx
Assume series solution for y(x)
∞
P
y(x) =
ak xk+r
k=0
◦
Rewrite ODE with series expansions
Convert x · y(x) to series expansion
∞
P
x · y(x) =
ak xk+r+1
k=0
◦
Shift index using k− >k − 1
∞
P
x · y(x) =
ak−1 xk+r
◦
d
Convert dx
y(x) to series expansion
∞
P
d
y(x) =
ak (k + r) xk+r−1
dx
◦
Shift index using k− >k + 1
∞
P
d
y(x)
=
ak+1 (k + r + 1) xk+r
dx
k=−1
d d
Convert x · dx
y(x) to series expansion
dx
∞
P
d d
x · dx
y(x)
=
ak (k + r) (k + r − 1) xk+r−1
dx
k=1
k=0
◦
k=0
◦
Shift index using k− >k + 1
∞
P
d d
x · dx
y(x)
=
ak+1 (k + r + 1) (k + r) xk+r
dx
k=−1
Rewrite ODE with series expansions
∞
k+r
P
2 r
2
2 −1+r
a0 r x
+ a1 (1 + r) x +
ak+1 (k + r + 1) − ak−1 x
=0
k=1
•
•
•
•
•
•
a0 cannot be 0 by assumption, giving the indicial equation
r2 = 0
Values of r that satisfy the indicial equation
r=0
Each term must be 0
a1 (1 + r)2 = 0
Each term in the series must be 0, giving the recursion relation
ak+1 (k + 1)2 − ak−1 = 0
Shift index using k− >k + 1
ak+2 (k + 2)2 − ak = 0
Recursion relation that defines series solution to ODE
ak
ak+2 = (k+2)
2
•
Recursion relation for r = 0
ak
ak+2 = (k+2)
2
•
Solution
for r = 0
∞
P
ak
k
y(x) =
ak x , ak+2 = (k+2)2 , a1 = 0
k=0
452
chapter 2 . book solved problems
3 Mathematica. Time used: 0.002 (sec). Leaf size: 60
ode=x*D[y[x],{x,2}]+D[y[x],x]-x*y[x]==0;
ic={};
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
453
4
4
x
x2
3x4 x2
x
x2
y(x) → c1
+
+ 1 + c2 −
−
+
+
+ 1 log(x)
64
4
128
4
64
4
3 Sympy. Time used: 0.222 (sec). Leaf size: 19
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-x*y(x) + x*Derivative(y(x), (x, 2)) + Derivative(y(x), x),0)
ics = {}
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
4
x
x2
y(x) = C1
+
+ 1 + O x6
64
4
chapter 2 . book solved problems
2.6
454
Chapter 6. Laplace Transforms. Problem set 6.2,
page 216
Local contents
2.6.1 Problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.2 Problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.3 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.4 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.5 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.6 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.7 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.8 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.9 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.10 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.11 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.12 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.13 Problem 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.14 Problem 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.15 Problem 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
455
458
461
464
468
471
475
479
482
486
490
494
497
500
504
chapter 2 . book solved problems
2.6.1
455
Problem 1
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
Internal problem ID [8622]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 1
Date solved : Tuesday, November 11, 2025 at 12:04:15 PM
CAS classification : [[_linear, ‘class A‘]]
26y
97 sin (2t)
=
5
5
y0 +
y(0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
The given ode now becomes an algebraic equation in the Laplace domain
sY (s) − y(0) +
26Y (s)
194
=
5
5 (s2 + 4)
Replacing initial condition gives
sY (s) +
26Y (s)
194
=
5
5 (s2 + 4)
Solving for Y (s) gives
Y (s) =
194
(s2 + 4) (5s + 26)
Applying partial fractions decomposition results in
− 58 − 13i
− 58 + 13i
5
8
8
Y (s) =
+
+
26
s − 2i
s + 2i
4 s+ 5
The inverse Laplace of each term above is now found, which gives
!
26t
5
5 e− 5
−1
=
L
4
4 s + 26
5
5 13i 5 13i 2it
−1 − 8 − 8
L
= − −
e
s − 2i
8
8
5 13i 5 13i −2it
−1 − 8 + 8
L
= − +
e
s + 2i
8
8
(1)
chapter 2 . book solved problems
456
Adding the above results and simplifying gives
26t
5 e− 5
5 cos (2t) 13 sin (2t)
y=
−
+
4
4
4
97 sin(2t)
(b) Slope field y 0 + 26y
5 =
5
(a) Solution plot
3 Maple. Time used: 0.107 (sec). Leaf size: 23
ode:=diff(y(t),t)+26/5*y(t) = 97/5*sin(2*t);
ic:=[y(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
26t
5 e− 5
5 cos (2t) 13 sin (2t)
y=
−
+
4
4
4
Maple trace
Methods for first order ODEs:
--- Trying classification methods --trying a quadrature
trying 1st order linear
<- 1st order linear successful
Maple step by step
Let’s
solve
h
i
26y(t)
97 sin(2t)
d
y(t)
+
=
,
y(0)
=
0
dt
5
5
•
•
•
•
Highest derivative means the order of the ODE is 1
d
y(t)
dt
Solve for the highest derivative
d
y(t) = − 26y(t)
+ 97 sin(2t)
dt
5
5
Group terms with y(t) on the lhs of the ODE and the rest on the rhs of the ODE
d
y(t) + 26y(t)
= 97 sin(2t)
dt
5
5
The ODE is linear; multiply by an integrating factor µ(t)
µ(t)
d
y(t) + 26y(t)
dt
5
= 97µ(t)5sin(2t)
•
d
Assume
the lhs of the
ODE is the total derivativedt (y(t) µ(t))
µ(t) dtd y(t) + 26y(t)
= dtd y(t) µ(t) + y(t) dtd µ(t)
5
•
Isolate dtd µ(t)
d
µ(t) = 26µ(t)
dt
5
chapter 2 . book solved problems
•
•
Solve to find the integrating factor
26t
µ(t) = e 5
Integrate both sides with respect to t
R d
R
(y(t) µ(t)) dt = 97µ(t)5sin(2t) dt + C 1
dt
Evaluate the integral on the lhs
R
y(t) µ(t) = 97µ(t)5sin(2t) dt + C 1
Solve for y(t)
•
5
y(t) =
µ(t)
26t
Substitute µ(t) = e 5
•
•
R 97µ(t) sin(2t)
dt+C 1
R 97 e 26t
5 sin(2t)
y(t) =
•
dt+C 1
26t
e 5
Evaluate the integrals on the rhs
y(t) =
•
5
−
26t
26t
5 e 5 cos(2t)
13 e 5 sin(2t)
+
+C 1
4
4
26t
e 5
•
Simplify
26t
y(t) = e− 5 C 1 − 5 cos(2t)
+ 13 sin(2t)
4
4
Use initial condition y(0) = 0
0 = C 1 − 54
Solve for _C1
C 1 = 54
Substitute _C1 = 54 into general solution and simplify
•
y(t) = 5 e 4 5 − 5 cos(2t)
+ 13 sin(2t)
4
4
Solution to the IVP
•
•
457
− 26t
− 26t
y(t) = 5 e 4 5 − 5 cos(2t)
+ 13 sin(2t)
4
4
3 Mathematica. Time used: 0.06 (sec). Leaf size: 37
ode=D[y[t],t]+52/10*y[t]==194/10*Sin[2*t];
ic={y[0]==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
−26t/5
y(t) → e
Z t
97 26K[1]
e 5 sin(2K[1])dK[1]
0 5
3 Sympy. Time used: 0.102 (sec). Leaf size: 29
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(26*y(t)/5 - 97*sin(2*t)/5 + Derivative(y(t), t),0)
ics = {y(0): 0}
dsolve(ode,func=y(t),ics=ics)
26t
y(t) =
13 sin (2t) 5 cos (2t) 5e− 5
−
+
4
4
4
chapter 2 . book solved problems
2.6.2
458
Problem 2
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460
Internal problem ID [8623]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 2
Date solved : Tuesday, November 11, 2025 at 12:04:15 PM
CAS classification : [_quadrature]
y 0 + 2y = 0
y(0) =
3
2
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
The given ode now becomes an algebraic equation in the Laplace domain
sY (s) − y(0) + 2Y (s) = 0
Replacing initial condition gives
sY (s) −
3
+ 2Y (s) = 0
2
Solving for Y (s) gives
Y (s) =
3
2 (s + 2)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
3
−1
=L
2 (s + 2)
−2t
3e
=
2
(1)
chapter 2 . book solved problems
459
(b) Slope field y 0 + 2y = 0
(a) Solution plot
3 Maple. Time used: 0.097 (sec). Leaf size: 10
ode:=diff(y(t),t)+2*y(t) = 0;
ic:=[y(0) = 3/2];
dsolve([ode,op(ic)],y(t),method='laplace');
y=
3 e−2t
2
Maple trace
Methods for first order ODEs:
--- Trying classification methods --trying a quadrature
trying 1st order linear
<- 1st order linear successful
Maple step by step
•
•
•
Let’s solve
d
3
y(t)
+
2y(t)
=
0,
y(0)
=
dt
2
Highest derivative means the order of the ODE is 1
d
y(t)
dt
Solve for the highest derivative
d
y(t) = −2y(t)
dt
Separate variables
d
y(t)
dt
= −2
Integrate both sides with respect to t
R dtd y(t)
R
dt = (−2) dt + C 1
y(t)
Evaluate integral
ln (y(t)) = −2t + C 1
Solve for y(t)
y(t) = e−2t+C 1
Redefine the integration constant(s)
y(t) = C 1 e−2t
Use initial condition y(0) = 32
3
= C1
2
Solve for _C1
C 1 = 32
y(t)
•
•
•
•
•
•
chapter 2 . book solved problems
•
•
460
Substitute _C1 = 32 into general solution and simplify
−2t
y(t) = 3 e2
Solution to the IVP
−2t
y(t) = 3 e2
3 Mathematica. Time used: 0.027 (sec). Leaf size: 44
ode=D[y[t],t]+52/10*y[t]==194/10*Sin[2*t];
ic={y[0]==15/10};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Z t
1 −26t/5
97 26K[1]
y(t) → e
2
e 5 sin(2K[1])dK[1] + 3
2
0 5
3 Sympy. Time used: 0.062 (sec). Leaf size: 10
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(2*y(t) + Derivative(y(t), t),0)
ics = {y(0): 3/2}
dsolve(ode,func=y(t),ics=ics)
y(t) =
3e−2t
2
chapter 2 . book solved problems
2.6.3
461
Problem 3
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463
Internal problem ID [8624]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 3
Date solved : Tuesday, November 11, 2025 at 12:04:16 PM
CAS classification : [[_2nd_order, _missing_x]]
y 00 − y 0 − 6y = 0
y(0) = 11
y 0 (0) = 28
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) − sY (s) + y(0) − 6Y (s) = 0
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 17 − 11s − sY (s) − 6Y (s) = 0
Solving the above equation for Y (s) results in
Y (s) =
11s + 17
s2 − s − 6
Applying partial fractions decomposition results in
Y (s) =
1
10
+
s+2 s−3
The inverse Laplace of each term above is now found, which gives
L
L
−1
−1
1
s+2
10
s−3
= e−2t
= 10 e3t
(1)
chapter 2 . book solved problems
462
Adding the above results and simplifying gives
y = 10 e3t + e−2t
Solving for c1 and c2 from the given initial conditions results in
y = 10 e3t + e−2t
Simplifying the solution gives
y = 10 e3t + e−2t
Figure 2.11: Solutions plot
3 Maple. Time used: 0.104 (sec). Leaf size: 15
ode:=diff(diff(y(t),t),t)-diff(y(t),t)-6*y(t) = 0;
ic:=[y(0) = 11, D(y)(0) = 28];
dsolve([ode,op(ic)],y(t),method='laplace');
y = 10 e3t + e−2t
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful
Maple step by step
Let’s
solve
d d
y(t) − dtd y(t) − 6y(t) = 0, y(0) = 11,
dt dt
•
•
•
d
y(t)
dt
= 28
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of ODE
r2 − r − 6 = 0
Factor the characteristic polynomial
chapter 2 . book solved problems
•
•
•
•
•
◦
◦
◦
463
(r + 2) (r − 3) = 0
Roots of the characteristic polynomial
r = (−2, 3)
1st solution of the ODE
y1 (t) = e−2t
2nd solution of the ODE
y2 (t) = e3t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t)
Substitute in solutions
y(t) = C 1 e−2t + C 2 e3t
Check validity of solution y(t) = _C1e−2t + _C2e3t
Use initial condition y(0) = 11
11 = _C1 + _C2
Compute derivative of the solution
d
y(t) = −2_C1 e−2t + 3_C2 e3t
dt
Use the initial condition dtd y(t)
= 28
{t=0}
◦
◦
•
28 = −2_C1 + 3_C2
Solve for _C1 and _C2
{_C1 = 1, _C2 = 10}
Substitute constant values into general solution and simplify
y(t) = e−2t + 10 e3t
Solution to the IVP
y(t) = e−2t + 10 e3t
3 Mathematica. Time used: 0.009 (sec). Leaf size: 18
ode=D[y[t],{t,2}]-D[y[t],t]-6*y[t]==0;
ic={y[0]==11,Derivative[1][y][0] ==28};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → e−2t + 10e3t
3 Sympy. Time used: 0.100 (sec). Leaf size: 15
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-6*y(t) - Derivative(y(t), t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 11, Subs(Derivative(y(t), t), t, 0): 28}
dsolve(ode,func=y(t),ics=ics)
y(t) = 10e3t + e−2t
chapter 2 . book solved problems
2.6.4
464
Problem 4
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467
Internal problem ID [8625]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 4
Date solved : Tuesday, November 11, 2025 at 12:04:17 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 + 9y = 10 e−t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 9Y (s) =
10
1+s
Substituting given initial conditions into Eq (1) gives
s2 Y (s) + 9Y (s) =
10
1+s
Solving the above equation for Y (s) results in
Y (s) =
10
(1 + s) (s2 + 9)
Applying partial fractions decomposition results in
− 12 − 6i
− 12 + 6i
1
Y (s) =
+
+
1+s
s − 3i
s + 3i
The inverse Laplace of each term above is now found, which gives
1
−1
L
= e−t
1+s
1 i 1 i
−1 − 2 − 6
L
= − −
e3it
s − 3i
2 6
1 i 1 i
−1 − 2 + 6
L
= − +
e−3it
s + 3i
2 6
(1)
chapter 2 . book solved problems
465
Adding the above results and simplifying gives
y = e−t − cos (3t) +
sin (3t)
3
Solving for c1 and c2 from the given initial conditions results in
y = e−t − cos (3t) +
sin (3t)
3
y = e−t − cos (3t) +
sin (3t)
3
Simplifying the solution gives
Figure 2.12: Solutions plot
3 Maple. Time used: 0.118 (sec). Leaf size: 21
ode:=diff(diff(y(t),t),t)+9*y(t) = 10*exp(-t);
ic:=[y(0) = 0, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
y = e−t − cos (3t) +
sin (3t)
3
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
chapter 2 . book solved problems
466
Let’s
solve
d d
y(t) + 9y(t) = 10 e−t , y(0) = 0,
dt dt
•
d
y(t)
dt
=0
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 9 = 0
Use quadratic
formula to solve for r
√
•
•
0±
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
−36
r=
2
Roots of the characteristic polynomial
r = (−3 I, 3 I)
1st solution of the homogeneous ODE
y1 (t) = cos (3t)
2nd solution of the homogeneous ODE
y2 (t) = sin (3t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 cos (3t) + C 2 sin (3t) + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R
(t)
y1 (t)f (t)
−t
yp (t) = −y1 (t) W (yy21(t)f
dt
+
y
(t)
dt,
f
(t)
=
10
e
2
(t),y2 (t))
W (y1 (t),y2 (t))
Wronskian of solutions of the homogeneous equation
cos (3t)
sin (3t)
W (y1 (t) , y2 (t)) =
−3 sin (3t) 3 cos (3t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 3
Substitute functions
into equation forR yp (t)
R
10 cos(3t) sin(3t)e−t dt
10 sin(3t) cos(3t)e−t dt
yp (t) = −
+
3
3
Compute integrals
yp (t) = e−t
Substitute particular solution into general solution to ODE
y(t) = C 1 cos (3t) + C 2 sin (3t) + e−t
Check validity of solution y(t) = _C1 cos (3t) + _C2 sin (3t) + e−t
Use initial condition y(0) = 0
0 = _C1 + 1
Compute derivative of the solution
d
y(t) = −3_C1 sin (3t) + 3_C2 cos (3t) − e−t
dt
Use the initial condition dtd y(t)
=0
{t=0}
◦
◦
•
0 = −1 + 3_C2
Solve for _C1 and _C2
_C1 = −1, _C2 = 13
Substitute constant values into general solution and simplify
y(t) = − cos (3t) + sin(3t)
+ e−t
3
Solution to the IVP
y(t) = − cos (3t) + sin(3t)
+ e−t
3
chapter 2 . book solved problems
3 Mathematica. Time used: 0.012 (sec). Leaf size: 25
ode=D[y[t],{t,2}]+9*y[t]==10*Exp[-t];
ic={y[0]==0,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → e−t +
1
sin(3t) − cos(3t)
3
3 Sympy. Time used: 0.052 (sec). Leaf size: 19
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(9*y(t) + Derivative(y(t), (t, 2)) - 10*exp(-t),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
y(t) =
467
sin (3t)
− cos (3t) + e−t
3
chapter 2 . book solved problems
2.6.5
468
Problem 5
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 469
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470
Internal problem ID [8626]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 5
Date solved : Tuesday, November 11, 2025 at 12:04:17 PM
CAS classification : [[_2nd_order, _missing_x]]
y 00 −
y
=0
4
y(0) = 12
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) −
Y (s)
=0
4
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 12s −
Y (s)
=0
4
Solving the above equation for Y (s) results in
Y (s) =
48s
4s2 − 1
Applying partial fractions decomposition results in
Y (s) =
6
6
1 +
s+ 2
s − 12
The inverse Laplace of each term above is now found, which gives
−1
6
s + 12
6
s − 12
L
L
−1
t
= 6 e− 2
t
= 6 e2
(1)
chapter 2 . book solved problems
469
Adding the above results and simplifying gives
t
y = 12 cosh
2
Solving for c1 and c2 from the given initial conditions results in
t
y = 12 cosh
2
Simplifying the solution gives
t
y = 12 cosh
2
Figure 2.13: Solutions plot
3 Maple. Time used: 0.086 (sec). Leaf size: 10
ode:=diff(diff(y(t),t),t)-1/4*y(t) = 0;
ic:=[y(0) = 12, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
t
y = 12 cosh
2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful
Maple step by step
Let’s
solve
d d
y(t) − y(t)
= 0, y(0) = 12,
dt dt
4
•
d
y(t)
dt
=0
{t=0}
Highest derivative means the order of the ODE is 2
chapter 2 . book solved problems
470
d d
y(t)
dt dt
•
•
•
•
•
•
•
◦
◦
◦
Characteristic polynomial of ODE
r2 − 14 = 0
Factor the characteristic polynomial
(2r−1)(2r+1)
=0
4
Roots of the characteristic polynomial
r = − 12 , 12
1st solution of the ODE
t
y1 (t) = e− 2
2nd solution of the ODE
t
y2 (t) = e 2
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t)
Substitute in solutions
t
t
y(t) = C 1 e− 2 + C 2 e 2
t
t
Check validity of solution y(t) = _C1e− 2 + _C2e 2
Use initial condition y(0) = 12
12 = _C1 + _C2
Compute derivative of the solution
t
t
_C1 e− 2 + _C2 e 2
d
y(t)
=
−
dt
2
2
Use the initial condition dtd y(t)
=0
{t=0}
0 = − _C1 + _C2
2
◦
◦
•
2
Solve for _C1 and _C2
{_C1 = 6, _C2 = 6}
Substitute constant values into general solution and simplify
y(t) = 12 cosh 2t
Solution to the IVP
y(t) = 12 cosh 2t
3 Mathematica. Time used: 0.009 (sec). Leaf size: 19
ode=D[y[t],{t,2}]-1/4*y[t]==0;
ic={y[0]==12,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → 6e−t/2 et + 1
3 Sympy. Time used: 0.048 (sec). Leaf size: 15
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-y(t)/4 + Derivative(y(t), (t, 2)),0)
ics = {y(0): 12, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
t
t
y(t) = 6e 2 + 6e− 2
chapter 2 . book solved problems
2.6.6
471
Problem 6
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474
Internal problem ID [8627]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 6
Date solved : Tuesday, November 11, 2025 at 12:04:17 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 − 6y 0 + 5y = 29 cos (2t)
y(0) =
16
5
y 0 (0) =
31
5
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) − 6sY (s) + 6y(0) + 5Y (s) =
29s
s2 + 4
Substituting given initial conditions into Eq (1) gives
s2 Y (s) + 13 −
16s
29s
− 6sY (s) + 5Y (s) = 2
5
s +4
Solving the above equation for Y (s) results in
Y (s) =
16s3 − 65s2 + 209s − 260
5 (s2 + 4) (s2 − 6s + 5)
Applying partial fractions decomposition results in
1
1
+ 6i5
− 6i5
1
2
10
10
Y (s) =
+
+
+
s−1 s−5
s − 2i
s + 2i
(1)
chapter 2 . book solved problems
472
The inverse Laplace of each term above is now found, which gives
L
L
L
−1
L
−1
−1
−1
1
s−1
2
s−5
+ 6i5
10
s − 2i
1
1
− 6i5
s + 2i
10
= et
= 2 e5t
1
6i
+
10
5
1
6i
−
10
5
=
=
e2it
e−2it
Adding the above results and simplifying gives
y=
cos (2t) 12 sin (2t)
−
+ et + 2 e5t
5
5
Solving for c1 and c2 from the given initial conditions results in
y=
cos (2t) 12 sin (2t)
−
+ et + 2 e5t
5
5
Simplifying the solution gives
y=
cos (2t) 12 sin (2t)
−
+ et + 2 e5t
5
5
Figure 2.14: Solutions plot
3 Maple. Time used: 0.099 (sec). Leaf size: 25
ode:=diff(diff(y(t),t),t)-6*diff(y(t),t)+5*y(t) = 29*cos(2*t);
ic:=[y(0) = 16/5, D(y)(0) = 31/5];
dsolve([ode,op(ic)],y(t),method='laplace');
y = et + 2 e5t +
Maple trace
cos (2t) 12 sin (2t)
−
5
5
chapter 2 . book solved problems
473
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) − 6 dtd y(t) + 5y(t) = 29 cos (2t) , y(0) = 16
,
dt dt
5
•
•
•
•
•
•
•
•
◦
◦
◦
◦
•
◦
◦
◦
{t=0}
= 31
5
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 − 6r + 5 = 0
Factor the characteristic polynomial
(r − 1) (r − 5) = 0
Roots of the characteristic polynomial
r = (1, 5)
1st solution of the homogeneous ODE
y1 (t) = et
2nd solution of the homogeneous ODE
y2 (t) = e5t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 et + C 2 e5t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function i
h variation ofRparameters to find yp here
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 29 cos (2t)
Wronskian of solutions of the homogeneous equation
t
e e5t
W (y1 (t) , y2 (t)) =
et 5 e5t
Compute Wronskian
W (y1 (t) , y2 (t)) = 4 e6t
Substitute functions
into equation
for yp (t)
R
R
29 et
◦
d
y(t)
dt
cos(2t)e−5t dte4t − e−t cos(2t)dt
yp (t) =
4
Compute integrals
yp (t) = cos(2t)
− 12 sin(2t)
5
5
Substitute particular solution into general solution to ODE
y(t) = C 1 et + C 2 e5t + cos(2t)
− 12 sin(2t)
5
5
Check validity of solution y(t) = _C1et + _C2e5t + cos(2t)
− 12 sin(2t)
5
5
Use initial condition y(0) = 16
5
16
1
=
_C1
+
_C2
+
5
5
Compute derivative of the solution
d
y(t) = _C1 et + 5_C2 e5t − 2 sin(2t)
− 24 cos(2t)
dt
5
5
d
31
Use the initial condition dt y(t)
= 5
{t=0}
31
= _C1 + 5_C2 − 24
5
5
chapter 2 . book solved problems
◦
◦
•
474
Solve for _C1 and _C2
{_C1 = 1, _C2 = 2}
Substitute constant values into general solution and simplify
y(t) = et + 2 e5t + cos(2t)
− 12 sin(2t)
5
5
Solution to the IVP
y(t) = et + 2 e5t + cos(2t)
− 12 sin(2t)
5
5
3 Mathematica. Time used: 0.012 (sec). Leaf size: 32
ode=D[y[t],{t,2}]-6*D[y[t],t]+5*y[t]==29*Cos[2*t];
ic={y[0]==32/10,Derivative[1][y][0] ==62/10};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → et + 2e5t −
12
1
sin(2t) + cos(2t)
5
5
3 Sympy. Time used: 0.151 (sec). Leaf size: 27
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(5*y(t) - 29*cos(2*t) - 6*Derivative(y(t), t) + Derivative(y(t), (t, 2))
,0)
ics = {y(0): 16/5, Subs(Derivative(y(t), t), t, 0): 31/5}
dsolve(ode,func=y(t),ics=ics)
y(t) = 2e5t + et −
12 sin (2t) cos (2t)
+
5
5
chapter 2 . book solved problems
2.6.7
475
Problem 7
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478
Internal problem ID [8628]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 7
Date solved : Tuesday, November 11, 2025 at 12:04:18 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 + 7y 0 + 12y = 21 e3t
y(0) =
7
2
y 0 (0) = −10
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 7sY (s) − 7y(0) + 12Y (s) =
Substituting given initial conditions into Eq (1) gives
s2 Y (s) −
29 7s
21
−
+ 7sY (s) + 12Y (s) =
2
2
s−3
Solving the above equation for Y (s) results in
7s2 + 8s − 45
Y (s) =
2 (s − 3) (s2 + 7s + 12)
Applying partial fractions decomposition results in
Y (s) =
1
1
5
+
+
2s + 6 2s − 6 2 (s + 4)
The inverse Laplace of each term above is now found, which gives
1
e−3t
−1
L
=
2s + 6
2
1
e3t
−1
L
=
2s − 6
2
5
5 e−4t
−1
L
=
2 (s + 4)
2
21
s−3
(1)
chapter 2 . book solved problems
476
Adding the above results and simplifying gives
y = cosh (3t) +
5 e−4t
2
Solving for c1 and c2 from the given initial conditions results in
y = cosh (3t) +
5 e−4t
2
y = cosh (3t) +
5 e−4t
2
Simplifying the solution gives
Figure 2.15: Solutions plot
3 Maple. Time used: 0.101 (sec). Leaf size: 15
ode:=diff(diff(y(t),t),t)+7*diff(y(t),t)+12*y(t) = 21*exp(3*t);
ic:=[y(0) = 7/2, D(y)(0) = -10];
dsolve([ode,op(ic)],y(t),method='laplace');
5 e−4t
y = cosh (3t) +
2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
chapter 2 . book solved problems
477
Let’s
solve
d d
y(t) + 7 dtd y(t) + 12y(t) = 21 e3t , y(0) = 72 ,
dt dt
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
d
y(t)
dt
= −10
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 7r + 12 = 0
Factor the characteristic polynomial
(r + 4) (r + 3) = 0
Roots of the characteristic polynomial
r = (−4, −3)
1st solution of the homogeneous ODE
y1 (t) = e−4t
2nd solution of the homogeneous ODE
y2 (t) = e−3t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−4t + C 2 e−3t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 21 e3t
Wronskian of solutions of the homogeneous equation
e−4t
e−3t
W (y1 (t) , y2 (t)) =
−4 e−4t −3 e−3t
Compute Wronskian
W (y1 (t) , y2 (t)) = e−7t
Substitute functions into equation for yp (t)
R
R
yp (t) = −21 e−4t e7t dt + 21 e−3t e6t dt
Compute integrals
3t
yp (t) = e2
Substitute particular solution into general solution to ODE
3t
y(t) = C 1 e−4t + C 2 e−3t + e2
3t
Check validity of solution y(t) = _C1e−4t + _C2e−3t + e2
Use initial condition y(0) = 72
7
= _C1 + _C2 + 12
2
Compute derivative of the solution
3t
d
y(t) = −4_C1 e−4t − 3_C2 e−3t + 3 e2
dt
Use the initial condition dtd y(t)
= −10
{t=0}
◦
◦
•
−10 = −4_C1 − 3_C2 + 32
Solve for _C1 and _C2
_C1 = 52 , _C2 = 12
Substitute constant values into general solution and simplify
−4t
y(t) = 5 e2 + cosh (3t)
Solution to the IVP
−4t
y(t) = 5 e2 + cosh (3t)
chapter 2 . book solved problems
478
3 Mathematica. Time used: 0.012 (sec). Leaf size: 28
ode=D[y[t],{t,2}]+7*D[y[t],t]+12*y[t]==21*Exp[3*t];
ic={y[0]==32/10,Derivative[1][y][0] ==62/10};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) →
1 −4t
e
155et + 5e7t − 128
10
3 Sympy. Time used: 0.138 (sec). Leaf size: 26
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(12*y(t) - 21*exp(3*t) + 7*Derivative(y(t), t) + Derivative(y(t), (t, 2))
,0)
ics = {y(0): 7/2, Subs(Derivative(y(t), t), t, 0): -10}
dsolve(ode,func=y(t),ics=ics)
y(t) =
e3t e−3t 5e−4t
+
+
2
2
2
chapter 2 . book solved problems
2.6.8
479
Problem 8
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481
Internal problem ID [8629]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 8
Date solved : Tuesday, November 11, 2025 at 12:04:18 PM
CAS classification : [[_2nd_order, _missing_x]]
y 00 − 4y 0 + 4y = 0
y(0) =
81
10
y 0 (0) =
39
10
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) − 4sY (s) + 4y(0) + 4Y (s) = 0
Substituting given initial conditions into Eq (1) gives
s2 Y (s) +
57 81s
−
− 4sY (s) + 4Y (s) = 0
2
10
Solving the above equation for Y (s) results in
Y (s) =
− 57
+ 81s
2
10
s2 − 4s + 4
Applying partial fractions decomposition results in
Y (s) =
81
123
−
10 (s − 2) 10 (s − 2)2
The inverse Laplace of each term above is now found, which gives
L
−1
L
−
−1
81
10 (s − 2)
123
10 (s − 2)2
=
81 e2t
10
=−
123t e2t
10
(1)
chapter 2 . book solved problems
480
Adding the above results and simplifying gives
y=−
3(41t − 27) e2t
10
Solving for c1 and c2 from the given initial conditions results in
y=−
3(41t − 27) e2t
10
y=−
3(41t − 27) e2t
10
Simplifying the solution gives
Figure 2.16: Solutions plot
3 Maple. Time used: 0.099 (sec). Leaf size: 15
ode:=diff(diff(y(t),t),t)-4*diff(y(t),t)+4*y(t) = 0;
ic:=[y(0) = 81/10, D(y)(0) = 39/10];
dsolve([ode,op(ic)],y(t),method='laplace');
y=−
3(41t − 27) e2t
10
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful
Maple step by step
Let’s
solve
d d
y(t) − 4 dtd y(t) + 4y(t) = 0, y(0) = 81
,
dt dt
10
•
d
y(t)
dt
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
= 39
10
chapter 2 . book solved problems
•
•
•
•
•
•
•
481
Characteristic polynomial of ODE
r2 − 4r + 4 = 0
Factor the characteristic polynomial
(r − 2)2 = 0
Root of the characteristic polynomial
r=2
1st solution of the ODE
y1 (t) = e2t
Repeated root, multiply y1 (t) by t to ensure linear independence
y2 (t) = t e2t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t)
Substitute in solutions
y(t) = C 1 e2t + C 2 t e2t
Check validity of solution y(t) = _C1e2t + _C2te2t
◦ Use initial condition y(0) = 81
10
81
=
_C1
10
◦ Compute derivative of the solution
d
y(t) = 2_C1 e2t + _C2 e2t + 2_C2t e2t
dt
◦ Use the initial condition dtd y(t)
= 39
10
{t=0}
39
= 2_C1 + _C2
10
•
◦ Solve for _C1 and _C2
_C1 = 81
, _C2 = − 123
10
10
◦ Substitute constant values into general solution and simplify
2t
y(t) = − 3 e (−27+41t)
10
Solution to the IVP
2t
y(t) = − 3 e (−27+41t)
10
3 Mathematica. Time used: 0.009 (sec). Leaf size: 19
ode=D[y[t],{t,2}]-4*D[y[t],t]+4*y[t]==0;
ic={y[0]==81/10,Derivative[1][y][0] ==39/10};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → −
3 Sympy. Time used: 0.100 (sec). Leaf size: 15
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(4*y(t) - 4*Derivative(y(t), t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 81/10, Subs(Derivative(y(t), t), t, 0): 39/10}
dsolve(ode,func=y(t),ics=ics)
y(t) =
3 2t
e (41t − 27)
10
81 123t
−
10
10
e2t
chapter 2 . book solved problems
2.6.9
482
Problem 9
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485
Internal problem ID [8630]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 9
Date solved : Tuesday, November 11, 2025 at 12:04:18 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 − 4y 0 + 3y = 6t − 8
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) − 4sY (s) + 4y(0) + 3Y (s) =
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 4sY (s) + 3Y (s) =
6
8
−
2
s
s
Solving the above equation for Y (s) results in
2(−3 + 4s)
Y (s) = − 2 2
s (s − 4s + 3)
Applying partial fractions decomposition results in
Y (s) =
2
1
1
−
+
2
s
s−3 s−1
The inverse Laplace of each term above is now found, which gives
2
−1
L
= 2t
s2
1
−1
L
−
= −e3t
s−3
1
−1
L
= et
s−1
6
8
−
2
s
s
(1)
chapter 2 . book solved problems
483
Adding the above results and simplifying gives
y = 2t − 2 e2t sinh (t)
Solving for c1 and c2 from the given initial conditions results in
y = 2t − 2 e2t sinh (t)
Simplifying the solution gives
y = 2t − e3t + et
Figure 2.17: Solutions plot
3 Maple. Time used: 0.104 (sec). Leaf size: 16
ode:=diff(diff(y(t),t),t)-4*diff(y(t),t)+3*y(t) = 6*t-8;
ic:=[y(0) = 0, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
y = 2t − e3t + et
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) − 4 dtd y(t) + 3y(t) = 6t − 8, y(0) = 0,
dt dt
d
y(t)
dt
=0
{t=0}
chapter 2 . book solved problems
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
484
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 − 4r + 3 = 0
Factor the characteristic polynomial
(r − 1) (r − 3) = 0
Roots of the characteristic polynomial
r = (1, 3)
1st solution of the homogeneous ODE
y1 (t) = et
2nd solution of the homogeneous ODE
y2 (t) = e3t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 et + C 2 e3t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 6t − 8
Wronskian of solutions of the homogeneous equation
t
e e3t
W (y1 (t) , y2 (t)) =
et 3 e3t
Compute Wronskian
W (y1 (t) , y2 (t)) = 2 e4t
Substitute functions into equation for yp (t)
R
R
yp (t) = et − (3t − 4) e−t dt + (3t − 4) e−3t dte2t
Compute integrals
yp (t) = 2t
Substitute particular solution into general solution to ODE
y(t) = C 1 et + C 2 e3t + 2t
Check validity of solution y(t) = _C1et + _C2e3t + 2t
Use initial condition y(0) = 0
0 = _C1 + _C2
Compute derivative of the solution
d
y(t) = _C1 et + 3_C2 e3t + 2
dt
Use the initial condition dtd y(t)
=0
{t=0}
◦
◦
•
0 = _C1 + 3_C2 + 2
Solve for _C1 and _C2
{_C1 = 1, _C2 = −1}
Substitute constant values into general solution and simplify
y(t) = et − e3t + 2t
Solution to the IVP
y(t) = et − e3t + 2t
3 Mathematica. Time used: 0.009 (sec). Leaf size: 19
ode=D[y[t],{t,2}]-4*D[y[t],t]+3*y[t]==6*t-8;
ic={y[0]==0,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → 2t + et − e3t
chapter 2 . book solved problems
485
3 Sympy. Time used: 0.103 (sec). Leaf size: 14
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-6*t + 3*y(t) - 4*Derivative(y(t), t) + Derivative(y(t), (t, 2)) + 8,0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
y(t) = 2t − e3t + et
chapter 2 . book solved problems
2.6.10
486
Problem 10
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 487
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
Internal problem ID [8631]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 10
Date solved : Tuesday, November 11, 2025 at 12:04:18 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 +
y
t2
=
25
50
y(0) = −25
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) +
Y (s)
1
=
25
25s3
Substituting given initial conditions into Eq (1) gives
s2 Y (s) + 25s +
Y (s)
1
=
25
25s3
Solving the above equation for Y (s) results in
Y (s) = −
25s2 − 1
s3
Applying partial fractions decomposition results in
Y (s) = −
25
1
+ 3
s
s
The inverse Laplace of each term above is now found, which gives
25
L
−
= −25
s
1
t2
−1
L
=
s3
2
−1
(1)
chapter 2 . book solved problems
487
Adding the above results and simplifying gives
y = −25 +
t2
2
Solving for c1 and c2 from the given initial conditions results in
y = −25 +
t2
2
y = −25 +
t2
2
Simplifying the solution gives
Figure 2.18: Solutions plot
3 Maple. Time used: 0.090 (sec). Leaf size: 11
ode:=diff(diff(y(t),t),t)+1/25*y(t) = 1/50*t^2;
ic:=[y(0) = -25, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
t2
y = − 25
2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
chapter 2 . book solved problems
Let’s
solve
d d
t2
y(t) + y(t)
= 50
, y(0) = −25,
dt dt
25
•
488
d
y(t)
dt
=0
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
1
r2 + 25
=0
Use quadratic
formula to solve for r
q
•
•
0±
•
•
•
•
•
◦
◦
◦
◦
−4
25
r=
2
Roots of the characteristic polynomial
r = − 5I , 5I
1st solution of the homogeneous ODE
y1 (t) = cos 5t
2nd solution of the homogeneous ODE
y2 (t) = sin 5t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 cos 5t + C 2 sin 5t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
t2
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 50
Wronskian of solutions
of the homogeneous
equation
"
#
t
t
cos 5 sin 5
W (y1 (t) , y2 (t)) =
sin t
cos 5t
− 55
5
Compute Wronskian
W (y1 (t) , y2 (t)) = 15
Substitute functions
into equation
for yp (t)
R
R
cos t
◦
•
◦
◦
◦
sin t t2 dt
sin t
cos t t2 dt
5
5
yp (t) = − 5 10 5
+
10
Compute integrals
2
yp (t) = t2 − 25
Substitute particular solution into general solution to ODE
2
y(t) = C 1 cos 5t + C 2 sin 5t + t2 − 25
2
Check validity of solution y(t) = _C1 cos 5t + _C2 sin 5t + t2 − 25
Use initial condition y(0) = −25
−25 = _C1 − 25
Compute derivative of the solution
_C1 sin 5t
_C2 cos 5t
d
y(t) = −
+
+t
dt
5
5
Use the initial condition dtd y(t)
=0
{t=0}
0 = _C2
5
◦
◦
•
Solve for _C1 and _C2
{_C1 = 0, _C2 = 0}
Substitute constant values into general solution and simplify
2
y(t) = t2 − 25
Solution to the IVP
2
y(t) = t2 − 25
chapter 2 . book solved problems
489
3 Mathematica. Time used: 0.01 (sec). Leaf size: 14
ode=D[y[t],{t,2}]+4/100*y[t]==2/100*t^2;
ic={y[0]==-25,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) →
1 2
t − 50
2
3 Sympy. Time used: 0.053 (sec). Leaf size: 8
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-t**2/50 + y(t)/25 + Derivative(y(t), (t, 2)),0)
ics = {y(0): -25, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
y(t) =
t2
− 25
2
chapter 2 . book solved problems
2.6.11
490
Problem 11
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493
Internal problem ID [8632]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 11
Date solved : Tuesday, November 11, 2025 at 12:04:19 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 3y 0 +
9y
= 9t3 + 64
4
y(0) = 1
y 0 (0) =
63
2
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 3sY (s) − 3y(0) +
9Y (s)
54 64
= 4 +
4
s
s
Substituting given initial conditions into Eq (1) gives
s2 Y (s) −
69
9Y (s)
54 64
− s + 3sY (s) +
= 4 +
2
4
s
s
Solving the above equation for Y (s) results in
Y (s) =
4s5 + 138s4 + 256s3 + 216
s4 (4s2 + 12s + 9)
Applying partial fractions decomposition results in
32 24 32
1
1
Y (s) = − 3 + 4 + 2 +
+
2
s
s
s
s + 32
s + 32
(1)
chapter 2 . book solved problems
491
The inverse Laplace of each term above is now found, which gives
32
L
− 3 = −16t2
s
−1 24
L
= 4t3
s4
−1 32
L
= 32t
s2
!
3t
1
L−1
= t e− 2
2
s + 32
3t
1
−1
L
= e− 2
3
s+ 2
−1
Adding the above results and simplifying gives
3t
y = 4t3 − 16t2 + 32t + e− 2 (1 + t)
Solving for c1 and c2 from the given initial conditions results in
3t
y = 4t3 − 16t2 + 32t + e− 2 (1 + t)
Simplifying the solution gives
3t
y = 4t3 − 16t2 + 32t + e− 2 (1 + t)
Figure 2.19: Solutions plot
3 Maple. Time used: 0.109 (sec). Leaf size: 26
ode:=diff(diff(y(t),t),t)+3*diff(y(t),t)+9/4*y(t) = 9*t^3+64;
ic:=[y(0) = 1, D(y)(0) = 63/2];
dsolve([ode,op(ic)],y(t),method='laplace');
3t
y = 4t3 − 16t2 + 32t + e− 2 (1 + t)
Maple trace
chapter 2 . book solved problems
492
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) + 3 dtd y(t) + 9y(t)
= 9t3 + 64, y(0) = 1,
dt dt
4
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
d
y(t)
dt
{t=0}
= 63
2
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 3r + 94 = 0
Factor the characteristic polynomial
(2r+3)2
=0
4
Root of the characteristic polynomial
r = − 32
1st solution of the homogeneous ODE
3t
y1 (t) = e− 2
Repeated root, multiply y1 (t) by t to ensure linear independence
3t
y2 (t) = t e− 2
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
3t
3t
y(t) = C 1 e− 2 + C 2 t e− 2 + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing functioni
h variation ofRparameters to find yp here
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 9t3 + 64
Wronskian of solutions
of the homogeneous#equation
"
3t
− 3t
e 2
t e− 2
W (y1 (t) , y2 (t)) =
− 3t
− 3t
3t
− 3 e 2 2 e− 2 − 3t e2 2
Compute Wronskian
W (y1 (t) , y2 (t)) = e−3t
Substitute functions
R 3t into equation Rfor yp (t)
3t
− 3t
yp (t) = e 2 t e 2 (9t3 + 64) dt − (9t4 + 64t) e 2 dt
Compute integrals
yp (t) = 4t3 − 16t2 + 32t
Substitute particular solution into general solution to ODE
3t
3t
y(t) = C 1 e− 2 + C 2 t e− 2 + 4t3 − 16t2 + 32t
3t
3t
Check validity of solution y(t) = _C1e− 2 + _C2te− 2 + 4t3 − 16t2 + 32t
Use initial condition y(0) = 1
1 = _C1
Compute derivative of the solution
3t
3t
3_C1 e− 2
3_C2t e− 2
d
− 3t
2 −
y(t)
=
−
+
_C2
e
+ 12t2 − 32t + 32
dt
2
2
Use the initial condition dtd y(t)
= 63
2
{t=0}
chapter 2 . book solved problems
493
63
= − 3_C1
+ _C2 + 32
2
2
◦
◦
•
Solve for _C1 and _C2
{_C1 = 1, _C2 = 1}
Substitute constant values into general solution and simplify
3t
3t
y(t) = e− 2 + t e− 2 + 4t3 − 16t2 + 32t
Solution to the IVP
3t
3t
y(t) = e− 2 + t e− 2 + 4t3 − 16t2 + 32t
3 Mathematica. Time used: 0.011 (sec). Leaf size: 28
ode=D[y[t],{t,2}]+3*D[y[t],t]+225/100*y[t]==9*t^3+64;
ic={y[0]==1,Derivative[1][y][0] ==315/10};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → 4t t2 − 4t + 8 + e−3t/2 (t + 1)
3 Sympy. Time used: 0.132 (sec). Leaf size: 26
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-9*t**3 + 9*y(t)/4 + 3*Derivative(y(t), t) + Derivative(y(t), (t, 2)) 64,0)
ics = {y(0): 1, Subs(Derivative(y(t), t), t, 0): 63/2}
dsolve(ode,func=y(t),ics=ics)
3t
y(t) = 4t3 − 16t2 + 32t + (t + 1) e− 2
chapter 2 . book solved problems
2.6.12
494
Problem 12
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496
Internal problem ID [8633]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 12
Date solved : Tuesday, November 11, 2025 at 12:04:19 PM
CAS classification : [[_2nd_order, _missing_x]]
y 00 − 2y 0 − 3y = 0
y(4) = −3
y 0 (4) = −17
Using Laplace transform method.
Since second initial condition is not given at zero, then let
y 0 (0) = c2
Since first initial condition is not given at zero, then let
y(0) = c1
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) − 2sY (s) + 2y(0) − 3Y (s) = 0
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − c2 − sc1 − 2sY (s) + 2c1 − 3Y (s) = 0
Solving the above equation for Y (s) results in
Y (s) =
sc1 − 2c1 + c2
s2 − 2s − 3
Applying partial fractions decomposition results in
c1
3c1
+ c42
− c42
Y (s) = 4
+ 4
s−3
s+1
(1)
chapter 2 . book solved problems
495
The inverse Laplace of each term above is now found, which gives
−1
L
L
−1
c1
+ c42
s−3
− c42
s+1
4
3c1
4
=
(c1 + c2 ) e3t
4
(3c1 − c2 ) e−t
=
4
Adding the above results and simplifying gives
y=
et (2c1 cosh (2t) + sinh (2t) (−c1 + c2 ))
2
Solving for initial conditions the solution is
y = e−4+t ((7 sinh (8) − 3 cosh (8)) cosh (2t) + (3 sinh (8) − 7 cosh (8)) sinh (2t))
Solving for c1 and c2 from the given initial conditions results in
y = e−4+t ((7 sinh (8) − 3 cosh (8)) cosh (2t) + (3 sinh (8) − 7 cosh (8)) sinh (2t))
Simplifying the solution gives
y = e−4+t ((7 sinh (8) − 3 cosh (8)) cosh (2t) + (3 sinh (8) − 7 cosh (8)) sinh (2t))
Figure 2.20: Solutions plot
3 Maple. Time used: 0.105 (sec). Leaf size: 21
ode:=diff(diff(y(t),t),t)-2*diff(y(t),t)-3*y(t) = 0;
ic:=[y(4) = -3, D(y)(4) = -17];
dsolve([ode,op(ic)],y(t),method='laplace');
y = 2 e−t+4 − 5 e3t−12
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful
chapter 2 . book solved problems
496
Maple step by step
Let’s
solve
d d
y(t) − 2 dtd y(t) − 3y(t) = 0, y(4) = −3,
dt dt
•
•
•
•
•
•
•
•
◦
◦
◦
d
y(t)
dt
= −17
{t=4}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of ODE
r2 − 2r − 3 = 0
Factor the characteristic polynomial
(r + 1) (r − 3) = 0
Roots of the characteristic polynomial
r = (−1, 3)
1st solution of the ODE
y1 (t) = e−t
2nd solution of the ODE
y2 (t) = e3t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t)
Substitute in solutions
y(t) = C 1 e−t + C 2 e3t
Check validity of solution y(t) = _C1e−t + _C2e3t
Use initial condition y(4) = −3
−3 = _C1 e−4 + _C2 e12
Compute derivative of the solution
d
y(t) = −_C1 e−t + 3_C2 e3t
dt
Use the initial condition dtd y(t)
= −17
{t=4}
−4
◦
◦
•
12
−17 = −_C1 e + 3_C2 e
Solve for _C1 and _C2
2
_C1 = e−4
, _C2 = − e512
Substitute constant values into general solution and simplify
y(t) = 2 e4−t − 5 e−12+3t
Solution to the IVP
y(t) = 2 e4−t − 5 e−12+3t
3 Mathematica. Time used: 0.009 (sec). Leaf size: 24
ode=D[y[t],{t,2}]-2*D[y[t],t]-3*y[t]==0;
ic={y[4]==-3,Derivative[1][y][4]==-17};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → 2e4−t − 5e3(t−4)
3 Sympy. Time used: 0.103 (sec). Leaf size: 20
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-3*y(t) - 2*Derivative(y(t), t) + Derivative(y(t), (t, 2)),0)
ics = {y(4): -3, Subs(Derivative(y(t), t), t, 4): -17}
dsolve(ode,func=y(t),ics=ics)
y(t) = −
5e3t
+ 2e4 e−t
e12
chapter 2 . book solved problems
2.6.13
497
Problem 13
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 498
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 499
Internal problem ID [8634]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 13
Date solved : Tuesday, November 11, 2025 at 12:04:19 PM
CAS classification : [_quadrature]
y 0 − 6y = 0
y(−1) = 4
Using Laplace transform method.
Since initial condition is not at zero, then let
y(0) = c1
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
The given ode now becomes an algebraic equation in the Laplace domain
sY (s) − y(0) − 6Y (s) = 0
(1)
Replacing initial condition gives
sY (s) − c1 − 6Y (s) = 0
Solving for Y (s) gives
Y (s) =
c1
s−6
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
c1
−1
=L
s−6
= c1 e6t
The constant c1 is determined from the given initial condition y(−1) = 4 using the solution
found above. This results in
4 = c1 e−6
Solving gives
c1 = 4 e 6
chapter 2 . book solved problems
498
Hence the solution now becomes
y = 4 e6 e6t
(b) Slope field y 0 − 6y = 0
(a) Solution plot
3 Maple. Time used: 0.082 (sec). Leaf size: 12
ode:=diff(y(t),t)-6*y(t) = 0;
ic:=[y(-1) = 4];
dsolve([ode,op(ic)],y(t),method='laplace');
y = 4 e6t+6
Maple trace
Methods for first order ODEs:
--- Trying classification methods --trying a quadrature
trying 1st order linear
<- 1st order linear successful
Maple step by step
•
•
•
Let’s solve
d
y(t)
−
6y(t)
=
0,
y(−1)
=
4
dt
Highest derivative means the order of the ODE is 1
d
y(t)
dt
Solve for the highest derivative
d
y(t) = 6y(t)
dt
Separate variables
d
y(t)
dt
=6
Integrate both sides with respect to t
R dtd y(t)
R
dt
=
6dt + C 1
y(t)
Evaluate integral
ln (y(t)) = 6t + C 1
Solve for y(t)
y(t) = e6t+C 1
Redefine the integration constant(s)
y(t) = C 1 e6t
Use initial condition y(−1) = 4
y(t)
•
•
•
•
•
chapter 2 . book solved problems
•
•
•
499
4 = C 1 e−6
Solve for _C1
4
C 1 = e−6
4
Substitute _C1 = e−6
into general solution and simplify
6+6t
y(t) = 4 e
Solution to the IVP
y(t) = 4 e6+6t
3 Mathematica. Time used: 0.015 (sec). Leaf size: 14
ode=D[y[t],t]-6*y[t]==0;
ic={y[-1]==4};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → 4e6t+6
3 Sympy. Time used: 0.061 (sec). Leaf size: 12
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-6*y(t) + Derivative(y(t), t),0)
ics = {y(-1): 4}
dsolve(ode,func=y(t),ics=ics)
y(t) = 4e6 e6t
chapter 2 . book solved problems
2.6.14
500
Problem 14
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
Internal problem ID [8635]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 14
Date solved : Tuesday, November 11, 2025 at 12:04:20 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 + 2y 0 + 5y = 50t − 100
y(2) = −4
y 0 (2) = 14
Using Laplace transform method.
Since second initial condition is not given at zero, then let
y 0 (0) = c2
Since first initial condition is not given at zero, then let
y(0) = c1
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 2sY (s) − 2y(0) + 5Y (s) =
50 100
−
s2
s
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − c2 − sc1 + 2sY (s) − 2c1 + 5Y (s) =
50 100
−
s2
s
Solving the above equation for Y (s) results in
Y (s) =
s3 c1 + 2c1 s2 + c2 s2 − 100s + 50
s2 (s2 + 2s + 5)
Applying partial fractions decomposition results in
24 10 (−1 + 2i) − c81 − c82 − 74 + 3c81 − c82 + 41
4
Y (s) = − + 2 +
s
s
s
+
1
−
2i
(−1 − 2i) − c81 − c82 − 74 + 3c81 − c82 + 41
4
+
s + 1 + 2i
(1)
chapter 2 . book solved problems
501
The inverse Laplace of each term above is now found, which gives
24
L
−
= −24
s
−1 10
L
= 10t
s2
!
(−1 + 2i) − c81 − c82 − 74 + 3c81 − c82 + 41
e(−1+2i)t (48 − 14i − ic2 + (2 − i) c1 )
4
=
s + 1 − 2i
4
−1
L−1
L−1
!
(−1 − 2i) − c81 − c82 − 74 + 3c81 − c82 + 41
e(−1−2i)t (48 + 14i + ic2 + (2 + i) c1 )
4
=
s + 1 + 2i
4
Adding the above results and simplifying gives
y = −24 + 10t +
(2 cos (2t) (c1 + 24) + sin (2t) (c1 + c2 + 14)) e−t
2
Solving for initial conditions the solution is
y = 2 e2−t (sin (2t) cos (4) − cos (2t) sin (4)) + 10t − 24
Solving for c1 and c2 from the given initial conditions results in
y = 2 e2−t (sin (2t) cos (4) − cos (2t) sin (4)) + 10t − 24
Simplifying the solution gives
y = 2 e2−t (sin (2t) cos (4) − cos (2t) sin (4)) + 10t − 24
Figure 2.22: Solutions plot
3 Maple. Time used: 0.102 (sec). Leaf size: 23
ode:=diff(diff(y(t),t),t)+2*diff(y(t),t)+5*y(t) = 50*t-100;
ic:=[y(2) = -4, D(y)(2) = 14];
dsolve([ode,op(ic)],y(t),method='laplace');
y = −24 + 2 e−t+2 sin (2t − 4) + 10t
Maple trace
chapter 2 . book solved problems
502
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) + 2 dtd y(t) + 5y(t) = 50t − 100, y(2) = −4,
dt dt
•
d
y(t)
dt
= 14
{t=2}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 2r + 5 = 0
Use quadratic
formula to solve for r
√
•
•
(−2)±
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
−16
r=
2
Roots of the characteristic polynomial
r = (−1 − 2 I, −1 + 2 I)
1st solution of the homogeneous ODE
y1 (t) = e−t cos (2t)
2nd solution of the homogeneous ODE
y2 (t) = e−t sin (2t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−t cos (2t) + C 2 e−t sin (2t) + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function i
h variation ofRparameters to find yp here
R
(t)
(t)
yp (t) = −y1 (t) W (yy21(t)f
dt + y2 (t) W (yy11(t)f
dt, f (t) = 50t − 100
(t),y2 (t))
(t),y2 (t))
Wronskian of solutions of the homogeneous equation
e−t cos (2t)
e−t sin (2t)
W (y1 (t) , y2 (t)) =
−e−t cos (2t) − 2 e−t sin (2t) −e−t sin (2t) + 2 e−t cos (2t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 2 e−2t
Substitute functions into equation for yp (t)
R
R
yp (t) = −25 e−t et sin (2t) (−2 + t) dt cos (2t) − et cos (2t) (−2 + t) dt sin (2t)
Compute integrals
yp (t) = 10t − 24
Substitute particular solution into general solution to ODE
y(t) = C 1 e−t cos (2t) + C 2 e−t sin (2t) + 10t − 24
Check validity of solution y(t) = _C1e−t cos (2t) + _C2e−t sin (2t) + 10t − 24
Use initial condition y(2) = −4
−4 = _C1 e−2 cos (4) + _C2 e−2 sin (4) − 4
Compute derivative of the solution
d
y(t) = −_C1 e−t cos (2t) − 2_C1 e−t sin (2t) − _C2 e−t sin (2t) + 2_C2 e−t cos (2t) + 10
dt
Use the initial condition dtd y(t)
= 14
{t=2}
14 = −_C1 e−2 cos (4) − 2_C1 e−2 sin (4) − _C2 e−2 sin (4) + 2_C2 e−2 cos (4) + 10
chapter 2 . book solved problems
◦
Solve
for _C1 and _C2
_C1 = − −2 2 sin(4)
2
e
◦
•
cos(4) +sin(4)2
, _C2 =
503
2 cos(4)
e−2 cos(4)2 +sin(4)2
Substitute constant values into general solution and simplify
y(t) = 2 e2−t (cos (4) sin (2t) − sin (4) cos (2t)) + 10t − 24
Solution to the IVP
y(t) = 2 e2−t (cos (4) sin (2t) − sin (4) cos (2t)) + 10t − 24
3 Mathematica. Time used: 0.013 (sec). Leaf size: 25
ode=D[y[t],{t,2}]+2*D[y[t],t]+5*y[t]==50*t-100;
ic={y[2]==-4,Derivative[1][y][2]==14};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → 10t − 2e2−t sin(4 − 2t) − 24
3 Sympy. Time used: 0.130 (sec). Leaf size: 58
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-50*t + 5*y(t) + 2*Derivative(y(t), t) + Derivative(y(t), (t, 2)) +
100,0)
ics = {y(2): -4, Subs(Derivative(y(t), t), t, 2): 14}
dsolve(ode,func=y(t),ics=ics)
y(t) = 10t +
2e2 sin (2t) cos (4) 2e2 sin (4) cos (2t)
−
cos2 (4) + sin2 (4) cos2 (4) + sin2 (4)
e−t − 24
chapter 2 . book solved problems
2.6.15
504
Problem 15
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507
Internal problem ID [8636]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.2, page 216
Problem number : 15
Date solved : Tuesday, November 11, 2025 at 12:04:20 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 + 3y 0 − 4y = 6 e2t−3
3
y
=4
2
0 3
y
=5
2
Using Laplace transform method.
Since second initial condition is not given at zero, then let
y 0 (0) = c2
Since first initial condition is not given at zero, then let
y(0) = c1
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 3sY (s) − 3y(0) − 4Y (s) =
6 e−3
s−2
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − c2 − sc1 + 3sY (s) − 3c1 − 4Y (s) =
6 e−3
s−2
Solving the above equation for Y (s) results in
Y (s) =
s2 c1 + sc1 + c2 s + 6 e−3 − 6c1 − 2c2
(s − 2) (s2 + 3s − 4)
(1)
chapter 2 . book solved problems
505
Applying partial fractions decomposition results in
Y (s) =
−3
4c1
+ c52 − 6 e5
5
s−1
+
−3
c1
− c52 + e 5
5
s+4
+
e−3
s−2
The inverse Laplace of each term above is now found, which gives
!
4c1
c2
6 e−3
+
−
et (4c1 + c2 − 6 e−3 )
5
5
L−1 5
=
s−1
5
L−1
−3
c1
− c52 + e 5
5
!
=
s+4
L
−1
e−3
s−2
(c1 − c2 + e−3 ) e−4t
5
= e2t−3
Adding the above results and simplifying gives
y = e2t−3 +
et (4c1 + c2 − 6 e−3 ) (c1 − c2 + e−3 ) e−4t
+
5
5
Solving for initial conditions the solution is
3
y = e2t−3 + 3 e− 2 +t
Solving for c1 and c2 from the given initial conditions results in
3
y = e2t−3 + 3 e− 2 +t
Simplifying the solution gives
3
y = e2t−3 + 3 e− 2 +t
Figure 2.23: Solutions plot
3 Maple. Time used: 0.105 (sec). Leaf size: 17
ode:=diff(diff(y(t),t),t)+3*diff(y(t),t)-4*y(t) = 6*exp(2*t-3);
ic:=[y(3/2) = 4, D(y)(3/2) = 5];
dsolve([ode,op(ic)],y(t),method='laplace');
3
y = e2t−3 + 3 et− 2
Maple trace
chapter 2 . book solved problems
506
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
d
3
2t−3
y(t)
+
3
y(t)
−
4y(t)
=
6
e
,
y
= 4,
dt dt
dt
2
•
•
•
•
•
•
•
•
◦
◦
◦
◦
•
◦
◦
◦
=5
t= 32
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 3r − 4 = 0
Factor the characteristic polynomial
(r + 4) (r − 1) = 0
Roots of the characteristic polynomial
r = (−4, 1)
1st solution of the homogeneous ODE
y1 (t) = e−4t
2nd solution of the homogeneous ODE
y2 (t) = et
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−4t + C 2 et + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 6 e2t−3
Wronskian of solutions of the homogeneous equation
e−4t
et
W (y1 (t) , y2 (t)) =
−4 e−4t et
Compute Wronskian
W (y1 (t) , y2 (t)) = 5 e−3t
SubstituteR functionsRinto equation
for yp (t)
6
◦
d
y(t)
dt
e−3+t dte5t − e−3+6t dt e−4t
yp (t) =
5
Compute integrals
yp (t) = e2t−3
Substitute particular solution into general solution to ODE
y(t) = C 1 e−4t + C 2 et + e2t−3
Check validity of solution y(t) = _C1e−4t + _C2et + e2t−3
Use initial condition y 32 = 4
3
4 = _C1 e−6 + _C2 e 2 + 1
Compute derivative of the solution
d
y(t) = −4_C1 e−4t + _C2 et + 2 e2t−3
dt
Use the initial condition dtd y(t) =5
−6
5 = −4_C1 e
3
2
+ _C2 e + 2
t= 32
chapter 2 . book solved problems
◦
507
Solve
for _C1 and _C2
n
o
3
_C1 = 0, _C2 = 3
e2
◦
•
Substitute constant values into general solution and simplify
3
y(t) = 3 e− 2 +t + e2t−3
Solution to the IVP
3
y(t) = 3 e− 2 +t + e2t−3
3 Mathematica. Time used: 0.048 (sec). Leaf size: 22
ode=D[y[t],{t,2}]+3*D[y[t],t]-4*y[t]==6*Exp[2*t-3];
ic={y[15/10]==4,Derivative[1][y][15/10 ]==5};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
3
y(t) → 3et− 2 + e2t−3
3 Sympy. Time used: 0.147 (sec). Leaf size: 19
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-4*y(t) - 6*exp(2*t - 3) + 3*Derivative(y(t), t) + Derivative(y(t), (t,
2)),0)
ics = {y(3/2): 4, Subs(Derivative(y(t), t), t, 3/2): 5}
dsolve(ode,func=y(t),ics=ics)
y(t) =
3et
3
e2
+ e2t−3
chapter 2 . book solved problems
2.7
508
Chapter 6. Laplace Transforms. Problem set 6.3,
page 224
Local contents
2.7.1 Problem 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.2 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.3 Problem 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.4 Problem 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.5 Problem 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.6 Problem 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.7 Problem 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.8 Problem 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.9 Problem 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.10 Problem 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
509
512
516
520
524
528
532
536
540
543
chapter 2 . book solved problems
2.7.1
509
Problem 18
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511
Internal problem ID [8637]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 18
Date solved : Tuesday, November 11, 2025 at 12:04:20 PM
CAS classification : [[_2nd_order, _missing_x]]
9y 00 − 6y 0 + y = 0
y(0) = 3
y 0 (0) = 1
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
9s2 Y (s) − 9y 0 (0) − 9sy(0) − 6sY (s) + 6y(0) + Y (s) = 0
Substituting given initial conditions into Eq (1) gives
9s2 Y (s) + 9 − 27s − 6sY (s) + Y (s) = 0
Solving the above equation for Y (s) results in
Y (s) =
9
3s − 1
Taking inverse Laplace transform gives
L
−1
9
3s − 1
t
= 3 e3
Solving for c1 and c2 from the given initial conditions results in
t
y = 3 e3
Simplifying the solution gives
t
y = 3 e3
(1)
chapter 2 . book solved problems
510
Figure 2.24: Solutions plot
3 Maple. Time used: 0.093 (sec). Leaf size: 10
ode:=9*diff(diff(y(t),t),t)-6*diff(y(t),t)+y(t) = 0;
ic:=[y(0) = 3, D(y)(0) = 1];
dsolve([ode,op(ic)],y(t),method='laplace');
t
y = 3 e3
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful
Maple step by step
Let’s
solve
9 dtd dtd y(t) − 6 dtd y(t) + y(t) = 0, y(0) = 3,
•
•
d
y(t)
dt
=1
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Isolate 2nd derivative
d
2 y(t)
d d
y(t) = dt3 − y(t)
dt dt
9
•
Group terms with y(t) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
2 d y(t)
d d
y(t) − dt3 + y(t)
=0
dt dt
9
•
•
•
•
•
Characteristic polynomial of ODE
r2 − 23 r + 19 = 0
Factor the characteristic polynomial
(3r−1)2
=0
9
Root of the characteristic polynomial
r = 13
1st solution of the ODE
t
y1 (t) = e 3
Repeated root, multiply y1 (t) by t to ensure linear independence
t
y2 (t) = t e 3
chapter 2 . book solved problems
•
•
◦
◦
◦
511
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t)
Substitute in solutions
t
t
y(t) = C 1 e 3 + C 2 t e 3
t
t
Check validity of solution y(t) = _C1e 3 + _C2te 3
Use initial condition y(0) = 3
3 = _C1
Compute derivative of the solution
t
t
_C1 e 3 + _C2 e 3t + _C2t e 3
d
y(t)
=
dt
3
3
d
Use the initial condition dt y(t)
=1
{t=0}
1 = _C1 + _C2
3
◦
◦
•
Solve for _C1 and _C2
{_C1 = 3, _C2 = 0}
Substitute constant values into general solution and simplify
t
y(t) = 3 e 3
Solution to the IVP
t
y(t) = 3 e 3
3 Mathematica. Time used: 0.01 (sec). Leaf size: 14
ode=9*D[y[t],{t,2}]-6*D[y[t],t]+y[t]==0;
ic={y[0]==3,Derivative[1][y][0] ==1};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → 3et/3
3 Sympy. Time used: 0.098 (sec). Leaf size: 8
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(y(t) - 6*Derivative(y(t), t) + 9*Derivative(y(t), (t, 2)),0)
ics = {y(0): 3, Subs(Derivative(y(t), t), t, 0): 1}
dsolve(ode,func=y(t),ics=ics)
t
y(t) = 3e 3
chapter 2 . book solved problems
2.7.2
512
Problem 19
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515
Internal problem ID [8638]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 19
Date solved : Tuesday, November 11, 2025 at 12:04:21 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 6y 0 + 8y = e−3t − e−5t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 6sY (s) − 6y(0) + 8Y (s) =
2
(s + 3) (s + 5)
Substituting given initial conditions into Eq (1) gives
s2 Y (s) + 6sY (s) + 8Y (s) =
2
(s + 3) (s + 5)
Solving the above equation for Y (s) results in
Y (s) =
2
(s + 3) (s + 5) (s2 + 6s + 8)
Applying partial fractions decomposition results in
Y (s) =
1
1
1
1
−
−
+
3s + 6 3 (s + 5) s + 3 s + 4
(1)
chapter 2 . book solved problems
513
The inverse Laplace of each term above is now found, which gives
L
−1
1
3s + 6
=
e−2t
3
1
e−5t
L
−
=−
3 (s + 5)
3
1
−1
L
−
= −e−3t
s+3
1
−1
L
= e−4t
s+4
−1
Adding the above results and simplifying gives
y=
e−2t e−5t
−
+ e−4t − e−3t
3
3
Solving for c1 and c2 from the given initial conditions results in
y=
e−2t e−5t
−
+ e−4t − e−3t
3
3
Simplifying the solution gives
3
(et − 1) e−5t
y=
3
Figure 2.25: Solutions plot
3 Maple. Time used: 0.092 (sec). Leaf size: 16
ode:=diff(diff(y(t),t),t)+6*diff(y(t),t)+8*y(t) = exp(-3*t)-exp(-5*t);
ic:=[y(0) = 0, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
3
(et − 1) e−5t
y=
3
Maple trace
chapter 2 . book solved problems
514
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
<- double symmetry of the form [xi=0, eta=F(x)] successful
Maple step by step
Let’s
solve
d d
y(t) + 6 dtd y(t) + 8y(t) = e−3t − e−5t , y(0) = 0,
dt dt
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
d
y(t)
dt
=0
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 6r + 8 = 0
Factor the characteristic polynomial
(r + 4) (r + 2) = 0
Roots of the characteristic polynomial
r = (−4, −2)
1st solution of the homogeneous ODE
y1 (t) = e−4t
2nd solution of the homogeneous ODE
y2 (t) = e−2t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−4t + C 2 e−2t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function i
h variation ofRparameters to find yp here
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = e−3t − e−5t
Wronskian of solutions of the homogeneous equation
e−4t
e−2t
W (y1 (t) , y2 (t)) =
−4 e−4t −2 e−2t
Compute Wronskian
W (y1 (t) , y2 (t)) = 2 e−6t
Substitute functions into equation for yp (t)
R
R
yp (t) = −e−4t sinh (t) dt + e−2t sinh (t) e−2t dt
Compute integrals
−5t
yp (t) = −e−3t − e 3
Substitute particular solution into general solution to ODE
−5t
y(t) = C 1 e−4t + C 2 e−2t − e−3t − e 3
−5t
Check validity of solution y(t) = _C1e−4t + _C2e−2t − e−3t − e 3
Use initial condition y(0) = 0
0 = _C1 + _C2 − 43
Compute derivative of the solution
−5t
d
y(t) = −4_C1 e−4t − 2_C2 e−2t + 3 e−3t + 5 e3
dt
Use the initial condition dtd y(t)
=0
{t=0}
◦
◦
0 = −4_C1 − 2_C2 + 14
3
Solve for _C1 and _C2
_C1 = 1, _C2 = 13
Substitute constant values into general solution and simplify
chapter 2 . book solved problems
515
3
et −1 e−5t
•
y(t) =
3
Solution to the IVP
y(t) =
3
et −1 e−5t
3
3 Mathematica. Time used: 0.164 (sec). Leaf size: 21
ode=D[y[t],{t,2}]+6*D[y[t],t]+8*y[t]==Exp[-3*t]-Exp[-5*t];
ic={y[0]==0,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
3
1
y(t) → e−5t et − 1
3
3 Sympy. Time used: 0.219 (sec). Leaf size: 31
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(8*y(t) + 6*Derivative(y(t), t) + Derivative(y(t), (t, 2)) - exp(-3*t) +
exp(-5*t),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
y(t) =
−3t
1
e
− e−t + e−2t −
3
3
e−2t
chapter 2 . book solved problems
2.7.3
516
Problem 20
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 519
Internal problem ID [8639]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 20
Date solved : Tuesday, November 11, 2025 at 12:04:21 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]
y 00 + 10y 0 + 24y = 144t2
y(0) =
19
12
y 0 (0) = −5
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 10sY (s) − 10y(0) + 24Y (s) =
Substituting given initial conditions into Eq (1) gives
s2 Y (s) −
65 19s
288
−
+ 10sY (s) + 24Y (s) = 3
6
12
s
Solving the above equation for Y (s) results in
Y (s) =
19s2 − 60s + 144
12s3
Applying partial fractions decomposition results in
5
19
12
Y (s) = − 2 +
+ 3
s
12s s
The inverse Laplace of each term above is now found, which gives
5
−1
L
− 2 = −5t
s
19
19
−1
L
=
12s
12
−1 12
L
= 6t2
3
s
288
s3
(1)
chapter 2 . book solved problems
517
Adding the above results and simplifying gives
y = 6t2 − 5t +
19
12
Solving for c1 and c2 from the given initial conditions results in
y = 6t2 − 5t +
19
12
y = 6t2 − 5t +
19
12
Simplifying the solution gives
Figure 2.26: Solutions plot
3 Maple. Time used: 0.093 (sec). Leaf size: 14
ode:=diff(diff(y(t),t),t)+10*diff(y(t),t)+24*y(t) = 144*t^2;
ic:=[y(0) = 19/12, D(y)(0) = -5];
dsolve([ode,op(ic)],y(t),method='laplace');
y = 6t2 − 5t +
19
12
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
chapter 2 . book solved problems
518
Let’s
solve
d d
y(t) + 10 dtd y(t) + 24y(t) = 144t2 , y(0) = 19
,
dt dt
12
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
d
y(t)
dt
= −5
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 10r + 24 = 0
Factor the characteristic polynomial
(r + 6) (r + 4) = 0
Roots of the characteristic polynomial
r = (−6, −4)
1st solution of the homogeneous ODE
y1 (t) = e−6t
2nd solution of the homogeneous ODE
y2 (t) = e−4t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−6t + C 2 e−4t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 144t2
Wronskian of solutions of the homogeneous equation
e−6t
e−4t
W (y1 (t) , y2 (t)) =
−6 e−6t −4 e−4t
Compute Wronskian
W (y1 (t) , y2 (t)) = 2 e−10t
Substitute functions into equation for yp (t)
R
R
yp (t) = −72 e−6t t2 e6t dt + 72 e−4t t2 e4t dt
Compute integrals
yp (t) = 6t2 − 5t + 19
12
Substitute particular solution into general solution to ODE
y(t) = C 1 e−6t + C 2 e−4t + 6t2 − 5t + 19
12
Check validity of solution y(t) = _C1e−6t + _C2e−4t + 6t2 − 5t + 19
12
Use initial condition y(0) = 19
12
19
19
=
_C1
+
_C2
+
12
12
Compute derivative of the solution
d
y(t) = −6_C1 e−6t − 4_C2 e−4t + 12t − 5
dt
Use the initial condition dtd y(t)
= −5
{t=0}
◦
◦
•
−5 = −6_C1 − 4_C2 − 5
Solve for _C1 and _C2
{_C1 = 0, _C2 = 0}
Substitute constant values into general solution and simplify
y(t) = 6t2 − 5t + 19
12
Solution to the IVP
y(t) = 6t2 − 5t + 19
12
chapter 2 . book solved problems
519
3 Mathematica. Time used: 0.011 (sec). Leaf size: 17
ode=D[y[t],{t,2}]+10*D[y[t],t]+24*y[t]==144*t^2;
ic={y[0]==19/12,Derivative[1][y][0] ==-5};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → 6t2 − 5t +
19
12
3 Sympy. Time used: 0.130 (sec). Leaf size: 14
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-144*t**2 + 24*y(t) + 10*Derivative(y(t), t) + Derivative(y(t), (t, 2))
,0)
ics = {y(0): 19/12, Subs(Derivative(y(t), t), t, 0): -5}
dsolve(ode,func=y(t),ics=ics)
y(t) = 6t2 − 5t +
19
12
chapter 2 . book solved problems
2.7.4
520
Problem 21
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523
Internal problem ID [8640]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 21
Date solved : Tuesday, November 11, 2025 at 12:04:21 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
00
y + 9y =
8 sin (t) 0 < t < π
0
π<t
y(0) = 0
y 0 (0) = 4
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 9Y (s) =
8 + 8 e−πs
s2 + 1
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 4 + 9Y (s) =
8 + 8 e−πs
s2 + 1
Solving the above equation for Y (s) results in
Y (s) =
4s2 + 8 e−πs + 12
(s2 + 1) (s2 + 9)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
2
−πs
+ 12
−1 4s + 8 e
=L
(s2 + 1) (s2 + 9)
= sin (t) + sin (3t) −
4 Heaviside (t − π) sin (t)3
3
(1)
chapter 2 . book solved problems
521
Solving for c1 and c2 from the given initial conditions results in
y = sin (t) + sin (3t) −
4 Heaviside (t − π) sin (t)3
3
Converting the above solution to piecewise it becomes
(
y=
sin (t) + sin (3t)
t<π
4 sin(t)3
sin (t) + sin (3t) − 3
π≤t
Simplifying the solution gives
(
y=4
2
sin (t) cos (t)
sin(3t)
3
t<π
π≤t
!
Figure 2.27: Solutions plot
3 Maple. Time used: 0.302 (sec). Leaf size: 26
ode:=diff(diff(y(t),t),t)+9*y(t) = piecewise(0 < t and t < Pi,8*sin(t),Pi < t,0);
ic:=[y(0) = 0, D(y)(0) = 4];
dsolve([ode,op(ic)],y(t),method='laplace');
(
y=4
sin (t) cos (t)2 t < π
sin(3t)
π≤t
3
!
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
chapter 2 . book solved problems
522
Maple step by step
Let’s solve
8 sin (t) 0 < t < π
d d
y(t) + 9y(t) =
, y(0) = 0,
dt dt
0
π<t
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 9 = 0
Use quadratic
formula to solve for r
√
•
•
•
0±
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
−36
d
y(t)
dt
=4
{t=0}
r=
2
Roots of the characteristic polynomial
r = (−3 I, 3 I)
1st solution of the homogeneous ODE
y1 (t) = cos (3t)
2nd solution of the homogeneous ODE
y2 (t) = sin (3t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 cos (3t) + C 2 sin (3t) + yp (t)
Find a particular solution yp (t) of the ODE
Use variation of parameters to find yp here f (t) is the forcing function
R y2 (t)f (t)
R y1 (t)f (t)
8 sin (t) 0 < t < π
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) =
0
π<t
Wronskian of solutions of the homogeneous equation
cos (3t)
sin (3t)
W (y1 (t) , y2 (t)) =
−3 sin (3t) 3 cos (3t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 3
Substitute functionsinto
 equation for yp (t) 




0
t
≤
0
0
t
≤
0
R 
R  8 cos(3t) sin(t)


sin(t)
yp (t) = − cos (3t)  8 sin(3t)
dt
+
sin
(3t)
t
<
π
t < π  dt


3
3




0
π≤t
0
π≤t
Compute
 integrals

0
t≤0

4 sin(t)3
yp (t) =
t≤π
3


0
π<t
Substitute particular solution into 
general solution to ODE

0
t≤0

4 sin(t)3
y(t) = C 1 cos (3t) + C 2 sin (3t) +
t≤π
3


0
π<t


0
t≤0

4 sin(t)3
Check validity of solution y(t) = _C1 cos (3t) + _C2 sin (3t) +
t≤π
3


0
π<t
Use initial condition y(0) = 0
0 = _C1
Compute derivative of the solution

0
t≤0

2
d
y(t) = −3_C1 sin (3t) + 3_C2 cos (3t) +
4 sin (t) cos (t) t ≤ π
dt

0
π<t
Use the initial condition dtd y(t)
=4
{t=0}
4 = 3_C2
chapter 2 . book solved problems
◦
◦
•
523
Solve for _C1 and _C2
_C1 = 0, _C2 = 43
Substitute constant
 values into general solution and simplify

0
t≤0

4 sin(3t)
4 sin(t)3
y(t) = 3 +
t≤π
3


0
π<t
Solution to the IVP


0
t≤0

4 sin(3t)
4 sin(t)3
y(t) = 3 +
t≤π
3


0
π<t
3 Mathematica. Time used: 0.025 (sec). Leaf size: 30
ode=D[y[t],{t,2}]+9*y[t]==Piecewise[{{8*Sin[t],0<t<Pi},{0,t>Pi}}];
ic={y[0]==0,Derivative[1][y][0] ==4};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → {
4
sin(3t)
3
t>π∨t≤0
sin(t) + sin(3t)
True
3 Sympy. Time used: 0.278 (sec). Leaf size: 24
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Piecewise((8*sin(t), (t <= pi) & (t > 0)), (0, t > pi)) + 9*y(t) +
Derivative(y(t), (t, 2)),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 4}
dsolve(ode,func=y(t),ics=ics)
y(t) = C2 cos (3t) +



sin (t) for t ≤ π ∧ t > 0
0
for t > π


NaN
otherwise
+
4 sin (3t)
3
chapter 2 . book solved problems
2.7.5
524
Problem 22
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527
Internal problem ID [8641]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 22
Date solved : Tuesday, November 11, 2025 at 12:04:22 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
00
0
y + 3y + 2y =
4t 0 < t < 1
8
1<t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 3sY (s) − 3y(0) + 2Y (s) =
4(s − 1) e−s + 4
s2
Substituting given initial conditions into Eq (1) gives
s2 Y (s) + 3sY (s) + 2Y (s) =
4(s − 1) e−s + 4
s2
Solving the above equation for Y (s) results in
Y (s) =
4 e−s s − 4 e−s + 4
s2 (s2 + 3s + 2)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−s
−s
+4
−1 4 e s − 4 e
=L
s2 (s2 + 3s + 2)
= 2t Heaviside (1 − t) + 3 e−2t+2 − 8 e1−t + 7 Heaviside (t − 1) − e−2t + 4 e−t − 3
(1)
chapter 2 . book solved problems
525
Solving for c1 and c2 from the given initial conditions results in
y = 2t Heaviside (1 − t) + 3 e−2t+2 − 8 e1−t + 7 Heaviside (t − 1) − e−2t + 4 e−t − 3
Converting the above solution to piecewise it becomes
−3 − e−2t + 4 e−t + 2t
t<1
−2
−1
y=
1 − e + 4e
t=1

−2t
−t
−2t+2
1−t
4 − e + 4e + 3e
− 8e
1<t


Simplifying the solution gives
−3 − e−2t + 4 e−t + 2t
t<1
−2
−1
y=
1 − e + 4e
t=1

−2t
−t
−2t+2
1−t
4 − e + 4e + 3e
− 8e
1<t


Figure 2.28: Solutions plot
3 Maple. Time used: 0.229 (sec). Leaf size: 71
ode:=diff(diff(y(t),t),t)+3*diff(y(t),t)+2*y(t) = piecewise(0 < t and t < 1,4*t,1
< t,8);
ic:=[y(0) = 0, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
−3 − e−2t + 4 e−t + 2t
t<1
−2
−1
y=
1 − e + 4e
t=1

−2t
−t
2−2t
1−t
4 − e + 4e + 3e
− 8e
1<t


Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
<- double symmetry of the form [xi=0, eta=F(x)] successful
chapter 2 . book solved problems
526
Maple step by step
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
Let’s solve
4t 0 < t < 1
d d
d
d
y(t) + 3 dt y(t) + 2y(t) =
, y(0) = 0, dt y(t)
=0
dt dt
8
1<t
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 3r + 2 = 0
Factor the characteristic polynomial
(r + 2) (r + 1) = 0
Roots of the characteristic polynomial
r = (−2, −1)
1st solution of the homogeneous ODE
y1 (t) = e−2t
2nd solution of the homogeneous ODE
y2 (t) = e−t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−2t + C 2 e−t + yp (t)
Find a particular solution yp (t) of the ODE
Use variation of parameters to find yp here f (t) is the forcing function
R y2 (t)f (t)
R y1 (t)f (t)
4t 0 < t < 1
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) =
8
1<t
Wronskian of solutions of the homogeneous equation
e−2t
e−t
W (y1 (t) , y2 (t)) =
−2 e−2t −e−t
Compute Wronskian
W (y1 (t) , y2 (t)) = e−3t
Substitute functionsinto
for yp (t)
 equation 




0 t≤0
0 t≤0








R 2t 
R t
 t t<1 
 t t<1 
−2t
−t
yp (t) = −4 e
e 
dt
+
4
e
e


 dt
0 t=1 
0 t=1 






 2 1<t
 2 1<t
Compute integrals

0
t≤0

−2t
−t
yp (t) =
−e − 3 + 2t + 4 e
t≤1

−2t
2−2t
1−t
−t
−e + 3 e
+ 4 − 8e + 4e
1<t
Substitute particular solution into general solution to ODE

0
t≤0

−2t
−t
−2t
−t
y(t) = C 1 e + C 2 e +
−e − 3 + 2t + 4 e
t≤1

−2t
2−2t
1−t
−t
−e + 3 e
+ 4 − 8e + 4e
1<t

0
t≤0

−2t
−t
−2t
−t
Check validity of solution y(t) = _C1e + _C2e +
−e − 3 + 2t + 4e
t≤1

−2t
2−2t
1−t
−t
−e + 3e
+ 4 − 8e + 4e
1<
Use initial condition y(0) = 0
0 = _C1 + _C2
Compute derivative of the solution

0
t≤0

d
−2t
−t
−2t
−t
y(t) = −2_C1 e − _C2 e +
2e + 2 − 4e
t≤1
dt

−2t
2−2t
1−t
−t
2e − 6e
+ 8e − 4e
1<t
Use the initial condition dtd y(t)
=0
{t=0}
0 = −2_C1 − _C2
chapter 2 . book solved problems
◦
◦
•
Solve for _C1 and _C2
{_C1 = 0, _C2 = 0}
Substitute constant values into general solution and simplify

0
t≤0

−2t
−t
y(t) =
−e − 3 + 2t + 4 e
t≤1

−2t
2−2t
1−t
−t
−e + 3 e
+ 4 − 8e + 4e
1<t
Solution to the IVP

0
t≤0

−2t
−t
y(t) =
−e − 3 + 2t + 4 e
t≤1

−2t
2−2t
1−t
−t
−e + 3 e
+ 4 − 8e + 4e
1<t
3 Mathematica. Time used: 0.027 (sec). Leaf size: 70
ode=D[y[t],{t,2}]+3*D[y[t],t]+2*y[t]==Piecewise[{{4*t,0<t<1},{8,t>1}}];
ic={y[0]==0,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → {
0
t≤0
−t
2t − e + 4e − 3
0<t≤1
−2t
2
t
2t
t+1
e (−1 + 3e + 4e + 4e − 8e )
True
−2t
7 Sympy
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Piecewise((4*t, (t <= 1) & (t > 0)), (8, t > 1)) + 2*y(t) + 3*
Derivative(y(t), t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
527
chapter 2 . book solved problems
2.7.6
528
Problem 23
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531
Internal problem ID [8642]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 23
Date solved : Tuesday, November 11, 2025 at 12:04:23 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
00
0
y + y − 2y =
3 sin (t) − cos (t) 0 < t < 2π
3 sin (2t) − cos (2t)
2π < t
y(0) = 1
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
−2πs
3 − s + 3 e s(s+2)(s−1)
2 +4
s Y (s) − y (0) − sy(0) + sY (s) − y(0) − 2Y (s) =
2
s +1
0
2
Substituting given initial conditions into Eq (1) gives
−2πs
3 − s + 3 e s(s+2)(s−1)
2 +4
s Y (s) − 1 − s + sY (s) − 2Y (s) =
2
s +1
2
Solving the above equation for Y (s) results in
Y (s) =
s4 − s3 + 3 e−2πs s + 6s2 − 3 e−2πs − 4s + 8
(s − 1) (s2 + 1) (s2 + 4)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
4
3
−2πs
s + 6s2 − 3 e−2πs − 4s + 8
−1 s − s + 3 e
=L
(s − 1) (s2 + 1) (s2 + 4)
= et − sin (t) +
Heaviside (t − 2π) (2 sin (t) − sin (2t))
2
(1)
chapter 2 . book solved problems
529
Solving for c1 and c2 from the given initial conditions results in
y = et − sin (t) +
Heaviside (t − 2π) (2 sin (t) − sin (2t))
2
Converting the above solution to piecewise it becomes
(
y=
et − sin (t) t < 2π
et − sin(2t)
2π ≤ t
2
Simplifying the solution gives
(
y = et −
sin (t) t < 2π
sin(2t)
2π ≤ t
2
!
Figure 2.29: Solutions plot
3 Maple. Time used: 0.439 (sec). Leaf size: 28
ode:=diff(diff(y(t),t),t)+diff(y(t),t)-2*y(t) = piecewise(0 < t and t < 2*Pi,3*
sin(t)-cos(t),2*Pi < t,3*sin(2*t)-cos(2*t));
ic:=[y(0) = 1, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
(
y = et −
sin (t) t < 2π
sin(2t)
2π ≤ t
2
!
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
chapter 2 . book solved problems
530
Maple step by step
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
Let’s solve
3 sin (t) − cos (t) 0 < t < 2π
d d
d
d
y(t) + dt y(t) − 2y(t) =
, y(0) = 1, dt y(t)
=0
dt dt
3 sin (2t) − cos (2t)
2π < t
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + r − 2 = 0
Factor the characteristic polynomial
(r + 2) (r − 1) = 0
Roots of the characteristic polynomial
r = (−2, 1)
1st solution of the homogeneous ODE
y1 (t) = e−2t
2nd solution of the homogeneous ODE
y2 (t) = et
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−2t + C 2 et + yp (t)
Find a particular solution yp (t) of the ODE
Use variation of parameters to find yp here f (t) is the forcing function
R y2 (t)f (t)
R y1 (t)f (t)
3 sin (t) − cos (t) 0 < t < 2π
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) =
3 sin (2t) − cos (2t)
2π < t
Wronskian of solutions of the homogeneous equation
e−2t
et
W (y1 (t) , y2 (t)) =
−2 e−2t et
Compute Wronskian
W (y1 (t) , y2 (t)) = 3 e−t
Substitute
functions
into equation for yp (t) 









0
t≤0 
0
t≤0 















R

 3 sin (t) − cos (t)


 3 sin (t) − cos (t)


t
<
2π
t
<
2π
R
 −t 
 3t

 2t  −2t
 e 
dte − 
e dte







0
t = 2π 
0
t = 2π 


















 3 sin (2t) − cos (2t) 2π < t
 3 sin (2t) − cos (2t) 2π < t
yp (t) =
3
Compute
integrals




0
t
≤
0



 t

−2t


e
−
3
sin
(t)
−
e
t
≤
2π





3
sin(2t)

t
−2t
 e −
−
e
2π
<
t
2
yp (t) =
3
Substitute particular solution
into general solution to ODE




0
t
≤
0



 t

−2t


e
−
3
sin
(t)
−
e
t
≤
2π





3
sin(2t)

t
−2t
 e −
−
e
2π
<
t
2
y(t) = C 1 e−2t + C 2 et +
3




0
t
≤
0



 t

−2t

t ≤ 2π 
 e − 3 sin (t) − e





 et − 3 sin(2t) − e−2t
2π
<
t
2
Check validity of solution y(t) = _C1e−2t + _C2et +
3
Use initial condition y(0) = 1
1 = _C1 + _C2
Compute derivative of the solution



0
t≤0 






et − 3 cos (t) + 2 e−2t t ≤ 2π 




 t
−2t
e
−
3
cos
(2t)
+
2
e
2π
<
t
d
y(t) = −2_C1 e−2t + _C2 et +
dt
3
chapter 2 . book solved problems
◦
◦
◦
•
Use the initial condition
d
y(t)
dt
531
=0
{t=0}
0 = −2_C1 + _C2
Solve for _C1 and _C2
_C1 = 13 , _C2 = 23
Substitute
values into general solution and simplify
 constant
−2t
t
e
2
e

t≤0
 3 + 3
t
y(t) =
e − sin (t) t ≤ 2π

 et − sin(2t) 2π < t
2
Solution
the IVP
to −2t
e
2 et

t≤0
 3 + 3
t
y(t) =
e − sin (t) t ≤ 2π

 et − sin(2t) 2π < t
2
3 Mathematica. Time used: 0.03 (sec). Leaf size: 55
ode=D[y[t],{t,2}]+D[y[t],t]-2*y[t]==Piecewise[{{3*Sin[t]-Cos[t],0<t<2*Pi},{3*Sin
[2*t]-Cos[2*t],t>2*Pi}}];
ic={y[0]==1,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → {
t
e−2t
+ 2e3
3
t
t≤0
e − sin(t)
0 < t ≤ 2π
t
e − cos(t) sin(t)
True
7 Sympy
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Piecewise((3*sin(t) - cos(t), (t > 0) & (t < 2*pi)), (3*sin(2*t) - cos
(2*t), t > 2*pi)) - 2*y(t) + Derivative(y(t), t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 1, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
chapter 2 . book solved problems
2.7.7
532
Problem 24
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535
Internal problem ID [8643]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 24
Date solved : Tuesday, November 11, 2025 at 12:04:24 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
00
0
y + 3y + 2y =
1 0<t<1
0
1<t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 3sY (s) − 3y(0) + 2Y (s) =
1 − e−s
s
Substituting given initial conditions into Eq (1) gives
s2 Y (s) + 3sY (s) + 2Y (s) =
1 − e−s
s
Solving the above equation for Y (s) results in
Y (s) = −
−1 + e−s
s (s2 + 3s + 2)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−1 + e−s
−1
=L
−
s (s2 + 3s + 2)
=
Heaviside (1 − t) e−2t
(−e−2t+2 + 2 e1−t ) Heaviside (t − 1)
+
− e−t +
2
2
2
(1)
chapter 2 . book solved problems
533
Solving for c1 and c2 from the given initial conditions results in
y=
Heaviside (1 − t) e−2t
(−e−2t+2 + 2 e1−t ) Heaviside (t − 1)
+
− e−t +
2
2
2
Converting the above solution to piecewise it becomes

e−2t

− e−t + 12
t<1

2
e−2
−1
y=
−e +1
t=1
2

 e−2t − e−t − e−2t+2 + e1−t 1 < t
2
2
Simplifying the solution gives




y=

2
(e−t − 1)
t<1
e−2 − 2 e−1 + 2
t=1 
1+t
2
t
−2t
−(−2 e + e + 2 e − 1) e
1<t
2
Figure 2.30: Solutions plot
3 Maple. Time used: 0.154 (sec). Leaf size: 52
ode:=diff(diff(y(t),t),t)+3*diff(y(t),t)+2*y(t) = piecewise(0 < t and t < 1,1,1 <
t,0);
ic:=[y(0) = 0, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');




y=

2
(e−t − 1)
t<1
−2 e−1 + e−2 + 2
t=1 
2
1+t
t
−2t
−(e − 2 e + 2 e − 1) e
1<t
2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
<- double symmetry of the form [xi=0, eta=F(x)] successful
chapter 2 . book solved problems
534
Maple step by step
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
Let’s solve
1 0<t<1
d d
d
d
y(t) + 3 dt y(t) + 2y(t) =
, y(0) = 0, dt y(t)
=0
dt dt
0
1<t
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 3r + 2 = 0
Factor the characteristic polynomial
(r + 2) (r + 1) = 0
Roots of the characteristic polynomial
r = (−2, −1)
1st solution of the homogeneous ODE
y1 (t) = e−2t
2nd solution of the homogeneous ODE
y2 (t) = e−t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−2t + C 2 e−t + yp (t)
Find a particular solution yp (t) of the ODE
Use variation of parameters to find yp here f (t) is the forcing function
R y2 (t)f (t)
R y1 (t)f (t)
1 0<t<1
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) =
0
1<t
Wronskian of solutions of the homogeneous equation
e−2t
e−t
W (y1 (t) , y2 (t)) =
−2 e−2t −e−t
Compute Wronskian
W (y1 (t) , y2 (t)) = e−3t
Substitute functions into equation for yp (t)
 


 
0 t≤0
0 t≤0


R
R
yp (t) =   et t < 1  dtet −  e2t t < 1  dt e−2t


0 1≤t
0 1≤t
Compute integrals



0
t
≤
0




2

e−2t 

(et − 1)
t≤1 





1+t
t
2
1 + 2e − 2e − e 1 < t
yp (t) =
2
Substitute particular solution into general solution to ODE



0
t
≤
0




2

e−2t 

(et − 1)
t≤1 





1+t
t
2
1
+
2
e
−
2
e
−
e
1
<
t
y(t) = C 1 e−2t + C 2 e−t +
2



0
t
≤
0




2

e−2t 

(et − 1)
t≤1 





1+t
t
2
1
+
2e
−
2e
−
e
1
<
t
Check validity of solution y(t) = _C1e−2t + _C2e−t +
2
Use initial condition y(0) = 0
0 = _C1 + _C2
Compute derivative of the solution


0




t
−2t
2(e − 1) et

 e 


0
t
≤
0



2 e1+t − 2 et
2
d
−2t
−t
−2t 
t
+
y(t)
=
−2_C1
e
−
_C2
e
−
e
(e
−
1)
t
≤
1
dt
2

1 + 2 e1+t − 2 et − e2 1 < t
chapter 2 . book solved problems
◦
◦
◦
•
Use the initial condition
d
y(t)
dt
535
=0
{t=0}
0 = −2_C1 − _C2
Solve for _C1 and _C2
{_C1 = 0, _C2 = 0}
Substitute constant values into general solution and simplify



0
t≤0 



2

e−2t 

(et − 1)
t≤1 





1+t
t
2
1 + 2e − 2e − e 1 < t
y(t) =
2
Solution to the IVP



0
t≤0 



2

e−2t 

(et − 1)
t≤1 





1+t
t
2
1 + 2e − 2e − e 1 < t
y(t) =
2
3 Mathematica. Time used: 0.026 (sec). Leaf size: 57
ode=D[y[t],{t,2}]+3*D[y[t],t]+2*y[t]==Piecewise[{{1,0<t<1},{0,t>1}}];
ic={y[0]==0,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
t≤0
0<t≤1
t
(−1 − e + 2e )
True
0
y(t) → {
2
1 −2t
e (−1 + et )
2
−2t
1
(−1 + e)e
2
7 Sympy
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Piecewise((1, (t > 0) & (t < 1)), (0, t > 1)) + 2*y(t) + 3*Derivative(y
(t), t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
chapter 2 . book solved problems
2.7.8
536
Problem 25
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539
Internal problem ID [8644]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 25
Date solved : Tuesday, November 11, 2025 at 12:04:24 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
00
y +y =
t 0<t<1
0
1<t
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
−(1 + s) e−s + 1
s Y (s) − y (0) − sy(0) + Y (s) =
s2
2
0
(1)
Substituting given initial conditions into Eq (1) gives
s2 Y (s) + Y (s) =
−(1 + s) e−s + 1
s2
Solving the above equation for Y (s) results in
Y (s) = −
e−s s + e−s − 1
s2 (s2 + 1)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−s
e s + e−s − 1
−1
=L
− 2 2
s (s + 1)
= − sin (t) + t − Heaviside (t − 1) 2 sin
t 1
−
2 2
2
!
− sin (t − 1) + t − 1
chapter 2 . book solved problems
537
Solving for c1 and c2 from the given initial conditions results in
y = − sin (t) + t − Heaviside (t − 1) 2 sin
t 1
−
2 2
!
2
− sin (t − 1) + t − 1
Converting the above solution to piecewise it becomes
(
y=
− sin (t) + t
t<1
t
1 2
− sin (t) − 2 sin 2 − 2 + sin (t − 1) + 1 1 ≤ t
Simplifying the solution gives
y = − sin (t) +
t
t<1
cos (t − 1) + sin (t − 1) 1 ≤ t
Figure 2.31: Solutions plot
3 Maple. Time used: 0.375 (sec). Leaf size: 26
ode:=diff(diff(y(t),t),t)+y(t) = piecewise(0 < t and t < 1,t,1 < t,0);
ic:=[y(0) = 0, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
y = − sin (t) +
t
t<1
cos (t − 1) + sin (t − 1) 1 ≤ t
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
chapter 2 . book solved problems
538
Maple step by step
Let’s solve
t 0<t<1
d d
d
y(t) + y(t) =
, y(0) = 0, dt y(t)
=0
dt dt
0
1<t
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 1 = 0
Use quadratic
formula to solve for r
√ •
•
•
0±
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
−4
r=
2
Roots of the characteristic polynomial
r = (−I, I)
1st solution of the homogeneous ODE
y1 (t) = cos (t)
2nd solution of the homogeneous ODE
y2 (t) = sin (t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 cos (t) + C 2 sin (t) + yp (t)
Find a particular solution yp (t) of the ODE
Use variation of parameters to find yp here f (t) is the forcing function
R y2 (t)f (t)
R y1 (t)f (t)
t 0<t<1
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) =
0
1<t
Wronskian of solutions of the homogeneous equation
cos (t) sin (t)
W (y1 (t) , y2 (t)) =
− sin (t) cos (t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 1
Substitute functions into equation for yp (t)




0
t≤0
0
t≤0
R 
R 
yp (t) = − cos (t)  sin (t) t t < 1  dt + sin (t)  cos (t) t t < 1  dt


0
1≤t
0
1≤t
Compute integrals

0
t≤0

yp (t) =
t − sin (t)
t≤1

(− sin (1) + cos (1)) cos (t) + (−1 + sin (1) + cos (1)) sin (t) 1 < t
Substitute particular solution into general solution to ODE

0
t≤

y(t) = C 1 cos (t) + C 2 sin (t) +
t − sin (t)
t≤

(− sin (1) + cos (1)) cos (t) + (−1 + sin (1) + cos (1)) sin (t) 1 <

0

Check validity of solution y(t) = _C1 cos (t) + _C2 sin (t) +
t − sin (t)

(− sin (1) + cos (1)) cos (t) + (−1 +
Use initial condition y(0) = 0
0 = _C1
Compute derivative of the solution

0

d
y(t) = −_C1 sin (t) + _C2 cos (t) +
1 − cos (t)
dt

−(− sin (1) + cos (1)) sin (t) + (−1 + sin (1) + cos (1)) cos
Use the initial condition dtd y(t)
=0
{t=0}
◦
0 = _C2
Solve for _C1 and _C2
chapter 2 . book solved problems
◦
•
539
{_C1 = 0, _C2 = 0}
Substitute constant values into general solution and simplify

0

y(t) =
t − sin (t)

(− sin (1) + cos (1)) cos (t) + (−1 + sin (1) + cos (1)) sin (t)
Solution to the IVP

0

y(t) =
t − sin (t)

(− sin (1) + cos (1)) cos (t) + (−1 + sin (1) + cos (1)) sin (t)
t≤0
t≤1
1<t
t≤0
t≤1
1<t
3 Mathematica. Time used: 0.022 (sec). Leaf size: 44
ode=D[y[t],{t,2}]+y[t]==Piecewise[{{t,0<t<1},{0,t>1}}];
ic={y[0]==0,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → {
t − sin(t)
0<t≤1
cos(1 − t) − sin(1 − t) − sin(t)
t>1
3 Sympy. Time used: 0.223 (sec). Leaf size: 12
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Piecewise((t, (t > 0) & (t < 1)), (0, t > 1)) + y(t) + Derivative(y(t),
(t, 2)),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)
y(t) = C2 cos (t) +



t
for t > 0 ∧ t < 1
0
for t > 1


NaN otherwise
chapter 2 . book solved problems
2.7.9
540
Problem 26
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542
7Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 542
Internal problem ID [8645]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 26
Date solved : Tuesday, November 11, 2025 at 12:04:25 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
00
0
y + 2y + 5y =
10 sin (t) 0 < t < 2π
0
2π < t
y(π) = 1
y 0 (π) = 2 e−π − 2
Using Laplace transform method.
Since second initial condition is not given at zero, then let
y 0 (0) = c2
Since first initial condition is not given at zero, then let
y(0) = c1
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 2sY (s) − 2y(0) + 5Y (s) =
10 − 10 e−2πs
s2 + 1
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − c2 − sc1 + 2sY (s) − 2c1 + 5Y (s) =
10 − 10 e−2πs
s2 + 1
Solving the above equation for Y (s) results in
Y (s) = −
−c1 s3 − 2c1 s2 − c2 s2 − sc1 + 10 e−2πs − 2c1 − c2 − 10
(s2 + 1) (s2 + 2s + 5)
(1)
chapter 2 . book solved problems
541
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−c1 s3 − 2c1 s2 − c2 s2 − sc1 + 10 e−2πs − 2c1 − c2 − 10
−1
=L
−
(s2 + 1) (s2 + 2s + 5)
(−2 cos (2t) + sin (2t)) Heaviside (t − 2π) e2π−t
=
2
(2 cos (2t) (1 + c1 ) + sin (2t) (−1 + c1 + c2 )) e−t
+
2
+ (− cos (t) + 2 sin (t)) Heaviside (2π − t)
Solving for initial conditions the solution is
y = −2 cos (t)2 + sin (t) cos (t) + 1 Heaviside (t − 2π) e2π−t
+ (cos (t) − 2 sin (t)) Heaviside (t − 2π) + 2 e−t sin (t) cos (t) − cos (t) + 2 sin (t)
Solving for c1 and c2 from the given initial conditions results in
y = −2 cos (t)2 + sin (t) cos (t) + 1 Heaviside (t − 2π) e2π−t
+ (cos (t) − 2 sin (t)) Heaviside (t − 2π) + 2 e−t sin (t) cos (t) − cos (t) + 2 sin (t)
Converting the above solution to piecewise it becomes
y=
2 e−t sin (t) cos (t) − cos (t) + 2 sin (t)
t < 2π
2π−t
2
−t
2 e sin (t) cos (t) + −2 cos (t) + sin (t) cos (t) + 1 e
2π ≤ t
Simplifying the solution gives
(
y=
2 e−t sin (t)
cos (t) − cos (t) + 2 sin (t) t < 2π
e2π +2 e−t sin(2t)
− e−t e2π cos (2t)
2
Figure 2.32: Solutions plot
2π ≤ t
chapter 2 . book solved problems
542
3 Maple. Time used: 0.396 (sec). Leaf size: 70
ode:=diff(diff(y(t),t),t)+2*diff(y(t),t)+5*y(t) = piecewise(0 < t and t < 2*Pi
,10*sin(t),2*Pi < t,0);
ic:=[y(Pi) = 1, D(y)(Pi) = 2*exp(-Pi)-2];
dsolve([ode,op(ic)],y(t),method='laplace');

−t

 sin (2t) e − cos (t) + 2 sin (t) t < 2π
−2
t = 2π
y=

 e−t e2π +2 sin(2t) − e−t e2π cos (2t) 2π < t
2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
3 Mathematica. Time used: 0.035 (sec). Leaf size: 94
ode=D[y[t],{t,2}]+2*D[y[t],t]+5*y[t]==Piecewise[{{10*Sin[t],0<t<2*Pi},{0,t>2*Pi
}}];
ic={y[Pi]==1,Derivative[1][y][Pi]==2*Exp[-Pi]-2};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → {
1 −t
e (3 sin(2t) − 2 cos(2t))
2
−t
t≤0
− cos(t) + 2 sin(t) + e sin(2t)
0 < t ≤ 2π
1 −t
2π
2π
e ((2 + e ) sin(2t) − 2e cos(2t))
True
2
7 Sympy
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Piecewise((10*sin(t), (t > 0) & (t < 2*pi)), (0, t > 2*pi)) + 5*y(t) +
2*Derivative(y(t), t) + Derivative(y(t), (t, 2)),0)
ics = {y(pi): 1, Subs(Derivative(y(t), t), t, pi): -2 + 2*exp(-pi)}
dsolve(ode,func=y(t),ics=ics)
chapter 2 . book solved problems
2.7.10
543
Problem 27
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546
Internal problem ID [8646]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.3, page 224
Problem number : 27
Date solved : Tuesday, November 11, 2025 at 12:04:26 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
00
y + 4y =
8t2 0 < t < 5
0
5<t
y(1) = 1 + cos (2)
y 0 (1) = 4 − 2 sin (2)
Using Laplace transform method.
Since second initial condition is not given at zero, then let
y 0 (0) = c2
Since first initial condition is not given at zero, then let
y(0) = c1
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 4Y (s) =
16 − 8(25s2 + 10s + 2) e−5s
s3
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − c2 − sc1 + 4Y (s) =
16 − 8(25s2 + 10s + 2) e−5s
s3
Solving the above equation for Y (s) results in
Y (s) = −
−s4 c1 − c2 s3 + 200 e−5s s2 + 80 e−5s s + 16 e−5s − 16
s3 (s2 + 4)
(1)
chapter 2 . book solved problems
544
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−s4 c1 − c2 s3 + 200 e−5s s2 + 80 e−5s s + 16 e−5s − 16
−1
=L
−
s3 (s2 + 4)
c2 sin (2t)
+ 2t2 + cos (2t) (1 + c1 )
2
− Heaviside (t − 5) 100 sin (t − 5)2 + 2t2 + cos (−10 + 2t) − 10 sin (−10 + 2t) − 51
= −1 +
Solving for initial conditions the solution is
y = −2 Heaviside (t − 5) t2 + 2t2 + Heaviside (t − 5) − 1
+ 49 Heaviside (t − 5) cos (−10 + 2t) + 10 Heaviside (t − 5) sin (−10 + 2t) + cos (2t)
Solving for c1 and c2 from the given initial conditions results in
y = −2 Heaviside (t − 5) t2 + 2t2 + Heaviside (t − 5) − 1
+ 49 Heaviside (t − 5) cos (−10 + 2t) + 10 Heaviside (t − 5) sin (−10 + 2t) + cos (2t)
Converting the above solution to piecewise it becomes
y=
−1 + 2t2 + cos (2t)
t<5
cos (2t) + 49 cos (−10 + 2t) + 10 sin (−10 + 2t) 5 ≤ t
Simplifying the solution gives
y = cos (2t) +
2t2 − 1
t<5
49 cos (−10 + 2t) + 10 sin (−10 + 2t) 5 ≤ t
Figure 2.33: Solutions plot
3 Maple. Time used: 0.524 (sec). Leaf size: 40
ode:=diff(diff(y(t),t),t)+4*y(t) = piecewise(0 < t and t < 5,8*t^2,5 < t,0);
ic:=[y(1) = 1+cos(2), D(y)(1) = 4-2*sin(2)];
dsolve([ode,op(ic)],y(t),method='laplace');
y = cos (2t) +
Maple trace
2t2 − 1
t<5
49 cos (2t − 10) + 10 sin (2t − 10) 5 ≤ t
chapter 2 . book solved problems
545
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s solve
2
8t 0 < t < 5
d d
y(t) + 4y(t) =
, y(1) = 1 + cos (2) ,
dt dt
0
5<t
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 4 = 0
Use quadratic
formula to solve for r
√
•
•
•
0±
•
•
•
•
•
◦
◦
◦
◦
◦
d
y(t)
dt
= 4 − 2 sin (2)
{t=1}
−16
r=
2
Roots of the characteristic polynomial
r = (−2 I, 2 I)
1st solution of the homogeneous ODE
y1 (t) = cos (2t)
2nd solution of the homogeneous ODE
y2 (t) = sin (2t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 cos (2t) + C 2 sin (2t) + yp (t)
Find a particular solution yp (t) of the ODE
Use variation of parameters to find yp here f (t) is the forcing function
2
R y2 (t)f (t)
R y1 (t)f (t)
8t 0 < t < 5
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) =
0
5<t
Wronskian of solutions of the homogeneous equation
cos (2t)
sin (2t)
W (y1 (t) , y2 (t)) =
−2 sin (2t) 2 cos (2t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 2
Substitute functions into equation for yp (t)




0
t≤0
0
t≤0

R 
R
yp (t) = − cos (2t)  4 sin (2t) t2 t < 5  dt + sin (2t)  4 cos (2t) t2 t < 5  dt


0
5≤t
0
5≤t
Compute integrals

0
t≤0


2
2t + cos (2t) − 1 yp (t) =
t≤5

 (10 cos (10) + 49 sin (10)) sin (2t) + 49 cos (2t) cos (10) − 10 sin(10) + 1
5<t
49
•
49
Substitute particular solution into general solution to ODE

0


2
2t + cos (2t) − 1 y(t) = C 1 cos (2t) + C 2 sin (2t) +

 (10 cos (10) + 49 sin (10)) sin (2t) + 49 cos (2t) cos (10) − 10 s
chapter 2 . book solved problems
546
Check validity of solution y(t) = _C1 cos (2t) + _C2 sin (2t) +



2t2 +

 (10 cos (10) + 49 sin (10)) sin (2t
◦
◦
◦
Use initial condition y(1) = 1 + cos (2)
1 + cos (2) = _C1 cos (2) + _C2 sin (2) + 1 + cos (2)
Compute derivative of the solution

0


d
4t − 2 sin (2t)
y(t) = −2_C1 sin (2t) + 2_C2 cos (2t) +
dt

 2(10 cos (10) + 49 sin (10)) cos (2t) − 98 sin (2t) cos
Use the initial condition dtd y(t)
= 4 − 2 sin (2)
{t=1}
◦
◦
•
4 − 2 sin (2) = −2_C1 sin (2) + 2_C2 cos (2) + 4 − 2 sin (2)
Solve for _C1 and _C2
{_C1 = 0, _C2 = 0}
Substitute constant values into general solution and simplify

0
t≤0


2
2t + cos (2t) − 1 y(t) =
t≤5

 (10 cos (10) + 49 sin (10)) sin (2t) + 49 cos (2t) cos (10) − 10 sin(10) + 1
5<t
49
49
Solution to the IVP

0
t≤0


2
2t + cos (2t) − 1 y(t) =
t≤5

 (10 cos (10) + 49 sin (10)) sin (2t) + 49 cos (2t) cos (10) − 10 sin(10) + 1
5<t
49
49
3 Mathematica. Time used: 0.026 (sec). Leaf size: 51
ode=D[y[t],{t,2}]+4*y[t]==Piecewise[{{8*t^2,0<t<5},{0,t>5}}];
ic={y[1]==1+Cos[2],Derivative[1][y][1]==4-2*Sin[2]};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
y(t) → {
2t2 + cos(2t) − 1
0<t≤5
49 cos(2(t − 5)) + cos(2t) − 10 sin(10 − 2t)
t>5
3 Sympy. Time used: 0.338 (sec). Leaf size: 17
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Piecewise((8*t**2, (t > 0) & (t < 5)), (0, t > 5)) + 4*y(t) +
Derivative(y(t), (t, 2)),0)
ics = {y(1): cos(2) + 1, Subs(Derivative(y(t), t), t, 1): 4 - 2*sin(2)}
dsolve(ode,func=y(t),ics=ics)
y(t) =

2


2t − 1 for t > 0 ∧ t < 5
0


NaN
for t > 5
otherwise
+ cos (2t)
chapter 2 . book solved problems
2.8
547
Chapter 6. Laplace Transforms. Problem set 6.4,
page 230
Local contents
2.8.1 Problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.2 Problem 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.3 Problem 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.4 Problem 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.5 Problem 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.6 Problem 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.7 Problem 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.8 Problem 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.9 Problem 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8.10 Problem 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
548
552
556
560
564
568
572
576
580
584
chapter 2 . book solved problems
2.8.1
548
Problem 3
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 549
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 550
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551
Internal problem ID [8647]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 3
Date solved : Tuesday, November 11, 2025 at 12:04:27 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 4y = δ(t − π)
y(0) = 8
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 4Y (s) = e−πs
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 8s + 4Y (s) = e−πs
Solving the above equation for Y (s) results in
Y (s) =
e−πs + 8s
s2 + 4
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−πs
+ 8s
−1 e
=L
s2 + 4
=
Heaviside (t − π) sin (2t)
+ 8 cos (2t)
2
Solving for c1 and c2 from the given initial conditions results in
y=
Heaviside (t − π) sin (2t)
+ 8 cos (2t)
2
(1)
chapter 2 . book solved problems
549
Converting the above solution to piecewise it becomes
(
y=
8 cos (2t)
t<π
sin(2t)
8 cos (2t) + 2
π≤t
Simplifying the solution gives
(
y = 8 cos (2t) +
0
sin(2t)
2
t<π
π≤t
!
Figure 2.34: Solutions plot
3 Maple. Time used: 0.203 (sec). Leaf size: 23
ode:=diff(diff(y(t),t),t)+4*y(t) = Dirac(t-Pi);
ic:=[y(0) = 8, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
y=
Heaviside (t − π) sin (2t)
+ 8 cos (2t)
2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) + 4y(t) = Dirac(t − π) , y(0) = 8,
dt dt
d
y(t)
dt
=0
{t=0}
chapter 2 . book solved problems
•
550
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 4 = 0
Use quadratic
formula to solve for r
√
•
•
0±
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
−16
r=
2
Roots of the characteristic polynomial
r = (−2 I, 2 I)
1st solution of the homogeneous ODE
y1 (t) = cos (2t)
2nd solution of the homogeneous ODE
y2 (t) = sin (2t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 cos (2t) + C 2 sin (2t) + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = Dirac(t − π)
Wronskian of solutions of the homogeneous equation
cos (2t)
sin (2t)
W (y1 (t) , y2 (t)) =
−2 sin (2t) 2 cos (2t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 2
Substitute functions
into equation for yp (t)
R
sin(2t) Dirac(t−π)dt
yp (t) =
2
Compute integrals
yp (t) = sin(2t)H eaviside(t−π)
2
Substitute particular solution into general solution to ODE
y(t) = C 1 cos (2t) + C 2 sin (2t) + sin(2t)H eaviside(t−π)
2
Check validity of solution y(t) = _C1 cos (2t) + _C2 sin (2t) + sin(2t)H eaviside(t−π)
2
Use initial condition y(0) = 8
8 = _C1
Compute derivative of the solution
d
y(t) = −2_C1 sin (2t) + 2_C2 cos (2t) + cos (2t) H eaviside(t − π) + sin(2t)Dirac(t−π)
dt
2
d
Use the initial condition dt y(t)
=0
{t=0}
◦
◦
•
0 = 2_C2
Solve for _C1 and _C2
{_C1 = 8, _C2 = 0}
Substitute constant values into general solution and simplify
y(t) = 8 cos (2t) + sin(2t)H eaviside(t−π)
2
Solution to the IVP
y(t) = 8 cos (2t) + sin(2t)H eaviside(t−π)
2
3 Mathematica. Time used: 0.015 (sec). Leaf size: 59
ode=D[y[t],{t,2}]+4*y[t]==DiracDelta[t-Pi];
ic={y[0]==8,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Z 0
y(t) → − sin(2t)
1
1
δ(π − K[1])dK[1] + sin(2t)
2
Z t
1
δ(π − K[1])dK[1] + 8 cos(2t)
1 2
chapter 2 . book solved problems
551
3 Sympy. Time used: 0.744 (sec). Leaf size: 68
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Dirac(t - pi) + 4*y(t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 8, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)

R
R0

Dirac (t − π) cos (2t) dt
Dirac (t − π) cos (2t) dt 
−
sin (2t)
2
2


R
R0
Dirac (t − π) sin (2t) dt
Dirac (t − π) sin (2t) dt
+ −
+
+ 8 cos (2t)
2
2
y(t) = 
chapter 2 . book solved problems
2.8.2
552
Problem 4
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555
Internal problem ID [8648]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 4
Date solved : Tuesday, November 11, 2025 at 12:04:28 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 16y = 4δ(t − 3π)
y(0) = 2
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 16Y (s) = 4 e−3πs
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 2s + 16Y (s) = 4 e−3πs
Solving the above equation for Y (s) results in
Y (s) =
4 e−3πs + 2s
s2 + 16
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−3πs
+ 2s
−1 4 e
=L
s2 + 16
= Heaviside (t − 3π) sin (4t) + 2 cos (4t)
Solving for c1 and c2 from the given initial conditions results in
y = Heaviside (t − 3π) sin (4t) + 2 cos (4t)
(1)
chapter 2 . book solved problems
553
Converting the above solution to piecewise it becomes
y=
2 cos (4t)
t < 3π
2 cos (4t) + sin (4t) 3π ≤ t
Simplifying the solution gives
y = 2 cos (4t) +
0
t < 3π
sin (4t) 3π ≤ t
Figure 2.35: Solutions plot
3 Maple. Time used: 0.206 (sec). Leaf size: 22
ode:=diff(diff(y(t),t),t)+16*y(t) = 4*Dirac(t-3*Pi);
ic:=[y(0) = 2, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
y = Heaviside (t − 3π) sin (4t) + 2 cos (4t)
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) + 16y(t) = 4Dirac(t − 3π) , y(0) = 2,
dt dt
•
d
y(t)
dt
Highest derivative means the order of the ODE is 2
=0
{t=0}
chapter 2 . book solved problems
554
d d
y(t)
dt dt
•
Characteristic polynomial of homogeneous ODE
r2 + 16 = 0
Use quadratic
formula to solve for r
√
•
0±
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
−64
r=
2
Roots of the characteristic polynomial
r = (−4 I, 4 I)
1st solution of the homogeneous ODE
y1 (t) = cos (4t)
2nd solution of the homogeneous ODE
y2 (t) = sin (4t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 cos (4t) + C 2 sin (4t) + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R
(t)
y1 (t)f (t)
yp (t) = −y1 (t) W (yy21(t)f
dt
+
y
(t)
dt,
f
(t)
=
4Dirac(t
−
3π)
2
(t),y2 (t))
W (y1 (t),y2 (t))
Wronskian of solutions of the homogeneous equation
cos (4t)
sin (4t)
W (y1 (t) , y2 (t)) =
−4 sin (4t) 4 cos (4t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 4
Substitute functions into equation for yp (t)
R
yp (t) = sin (4t) Dirac(t − 3π) dt
Compute integrals
yp (t) = sin (4t) H eaviside(t − 3π)
Substitute particular solution into general solution to ODE
y(t) = C 1 cos (4t) + C 2 sin (4t) + sin (4t) H eaviside(t − 3π)
Check validity of solution y(t) = _C1 cos (4t) + _C2 sin (4t) + sin (4t) H eaviside(t − 3π)
Use initial condition y(0) = 2
2 = _C1
Compute derivative of the solution
d
y(t) = −4_C1 sin (4t) + 4_C2 cos (4t) + 4 cos (4t) H eaviside(t − 3π) + sin (4t) Dirac(t − 3π)
dt
Use the initial condition dtd y(t)
=0
{t=0}
◦
◦
•
0 = 4_C2
Solve for _C1 and _C2
{_C1 = 2, _C2 = 0}
Substitute constant values into general solution and simplify
y(t) = 2 cos (4t) + sin (4t) H eaviside(t − 3π)
Solution to the IVP
y(t) = 2 cos (4t) + sin (4t) H eaviside(t − 3π)
3 Mathematica. Time used: 0.018 (sec). Leaf size: 51
ode=D[y[t],{t,2}]+16*y[t]==4*DiracDelta[t-3*Pi];
ic={y[0]==2,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Z t
Z 0
y(t) → − sin(4t)
δ(K[1] − 3π)dK[1] + sin(4t)
1
δ(K[1] − 3π)dK[1] + 2 cos(4t)
1
chapter 2 . book solved problems
555
3 Sympy. Time used: 0.592 (sec). Leaf size: 68
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-4*Dirac(t - 3*pi) + 16*y(t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 2, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)

Dirac (t − 3π) cos (4t) dt −
y(t) = 

Z0
Z
Dirac (t − 3π) cos (4t) dt sin (4t)

Z0
Z
+ −
Dirac (t − 3π) sin (4t) dt +

Dirac (t − 3π) sin (4t) dt + 2 cos (4t)
chapter 2 . book solved problems
2.8.3
556
Problem 5
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 559
Internal problem ID [8649]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 5
Date solved : Tuesday, November 11, 2025 at 12:04:28 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + y = δ(t − π) − δ(t − 2π)
y(0) = 0
y 0 (0) = 1
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + Y (s) = e−πs − e−2πs
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 1 + Y (s) = e−πs − e−2πs
Solving the above equation for Y (s) results in
Y (s) =
e−πs − e−2πs + 1
s2 + 1
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−πs
− e−2πs + 1
−1 e
=L
s2 + 1
= sin (t) (− Heaviside (t − π) + Heaviside (2π − t))
Solving for c1 and c2 from the given initial conditions results in
y = sin (t) (− Heaviside (t − π) + Heaviside (2π − t))
(1)
chapter 2 . book solved problems
557
Converting the above solution to piecewise it becomes

t<π
 sin (t)
y=
0
t ≤ 2π

− sin (t) 2π < t
Simplifying the solution gives


 1 t<π
y = sin (t) 
0 t ≤ 2π 

−1 2π < t
Figure 2.36: Solutions plot
3 Maple. Time used: 0.304 (sec). Leaf size: 25
ode:=diff(diff(y(t),t),t)+y(t) = Dirac(t-Pi)-Dirac(t-2*Pi);
ic:=[y(0) = 0, D(y)(0) = 1];
dsolve([ode,op(ic)],y(t),method='laplace');
y = sin (t) (1 − Heaviside (t − 2π) − Heaviside (t − π))
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s solve
chapter 2 . book solved problems
•
d d
y(t) + y(t) = Dirac(t − π) − Dirac(t − 2π) , y(0) = 0,
dt dt
558
d
y(t)
dt
=1
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 1 = 0
Use quadratic
formula to solve for r
√ •
•
0±
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
−4
r=
2
Roots of the characteristic polynomial
r = (−I, I)
1st solution of the homogeneous ODE
y1 (t) = cos (t)
2nd solution of the homogeneous ODE
y2 (t) = sin (t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 cos (t) + C 2 sin (t) + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = Dirac(t − π) − Dirac(t − 2π)
Wronskian of solutions of the homogeneous equation
cos (t) sin (t)
W (y1 (t) , y2 (t)) =
− sin (t) cos (t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 1
Substitute functions into equation for yp (t)
R
yp (t) = sin (t) (−Dirac(t − 2π) − Dirac(t − π)) dt
Compute integrals
yp (t) = − sin (t) (H eaviside(t − 2π) + H eaviside(t − π))
Substitute particular solution into general solution to ODE
y(t) = C 1 cos (t) + C 2 sin (t) − sin (t) (H eaviside(t − 2π) + H eaviside(t − π))
Check validity of solution y(t) = _C1 cos (t) + _C2 sin (t) − sin (t) (H eaviside(t − 2π) + H eavisi
Use initial condition y(0) = 0
0 = _C1
Compute derivative of the solution
d
y(t) = −_C1 sin (t) + _C2 cos (t) − cos (t) (H eaviside(t − 2π) + H eaviside(t − π)) − sin (t) (D
dt
Use the initial condition dtd y(t)
=1
{t=0}
◦
◦
•
1 = _C2
Solve for _C1 and _C2
{_C1 = 0, _C2 = 1}
Substitute constant values into general solution and simplify
y(t) = sin (t) (1 − H eaviside(t − 2π) − H eaviside(t − π))
Solution to the IVP
y(t) = sin (t) (1 − H eaviside(t − 2π) − H eaviside(t − π))
3 Mathematica. Time used: 0.022 (sec). Leaf size: 73
ode=D[y[t],{t,2}]+y[t]==DiracDelta[t-Pi]-DiracDelta[t-2*Pi];
ic={y[0]==0,Derivative[1][y][0] ==1};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Z 0
Z t
y(t) → sin(t) −
− cos(K[1])(δ(K[1] − 2π) − δ(K[1] − π))dK[1] + sin(t)
− cos(K[1])(δ(K[1] − 2π)
1
1
chapter 2 . book solved problems
559
3 Sympy. Time used: 1.178 (sec). Leaf size: 99
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(Dirac(t - 2*pi) - Dirac(t - pi) + y(t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 1}
dsolve(ode,func=y(t),ics=ics)

Z0
Z
Dirac (t − 2π) sin (t) dt −
y(t) = 
Z
Dirac (t − 2π) sin (t) dt −

Z0
+

Dirac (t − π) sin (t) dt cos (t)
Z0
Z
+ −
Dirac (t − π) sin (t) dt
Dirac (t − 2π) cos (t) dt +
Dirac (t − 2π) cos (t) dt
Z
Z0
+
Dirac (t − π) cos (t) dt −

Dirac (t − π) cos (t) dt + 1 sin (t)
chapter 2 . book solved problems
2.8.4
560
Problem 6
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 561
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563
Internal problem ID [8650]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 6
Date solved : Tuesday, November 11, 2025 at 12:04:29 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 4y 0 + 5y = δ(t − 1)
y(0) = 0
y 0 (0) = 3
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 4sY (s) − 4y(0) + 5Y (s) = e−s
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 3 + 4sY (s) + 5Y (s) = e−s
Solving the above equation for Y (s) results in
Y (s) =
e−s + 3
s2 + 4s + 5
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−s
e +3
−1
=L
s2 + 4s + 5
= Heaviside (t − 1) e−2t+2 sin (t − 1) + 3 e−2t sin (t)
Solving for c1 and c2 from the given initial conditions results in
y = Heaviside (t − 1) e−2t+2 sin (t − 1) + 3 e−2t sin (t)
(1)
chapter 2 . book solved problems
561
Converting the above solution to piecewise it becomes
y=
3 e−2t sin (t)
t<1
−2t
−2t+2
3 e sin (t) + e
sin (t − 1) 1 ≤ t
Simplifying the solution gives
y=e
−2t
3 sin (t) +
0
t<1
2
e sin (t − 1) 1 ≤ t
Figure 2.37: Solutions plot
3 Maple. Time used: 0.276 (sec). Leaf size: 25
ode:=diff(diff(y(t),t),t)+4*diff(y(t),t)+5*y(t) = Dirac(t-1);
ic:=[y(0) = 0, D(y)(0) = 3];
dsolve([ode,op(ic)],y(t),method='laplace');
y = e−2t Heaviside (t − 1) e2 sin (t − 1) + 3 sin (t)
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) + 4 dtd y(t) + 5y(t) = Dirac(t − 1) , y(0) = 0,
dt dt
•
Highest derivative means the order of the ODE is 2
d
y(t)
dt
=3
{t=0}
chapter 2 . book solved problems
562
d d
y(t)
dt dt
•
Characteristic polynomial of homogeneous ODE
r2 + 4r + 5 = 0
Use quadratic
formula to solve for r
√ •
(−4)±
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
−4
r=
2
Roots of the characteristic polynomial
r = (−2 − I, −2 + I)
1st solution of the homogeneous ODE
y1 (t) = e−2t cos (t)
2nd solution of the homogeneous ODE
y2 (t) = e−2t sin (t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−2t cos (t) + C 2 e−2t sin (t) + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R
(t)
y1 (t)f (t)
yp (t) = −y1 (t) W (yy21(t)f
dt
+
y
(t)
dt,
f
(t)
=
Dirac(t
−
1)
2
(t),y2 (t))
W (y1 (t),y2 (t))
Wronskian of solutions of the homogeneous equation
e−2t cos (t)
e−2t sin (t)
W (y1 (t) , y2 (t)) =
−2 e−2t cos (t) − e−2t sin (t) −2 e−2t sin (t) + e−2t cos (t)
Compute Wronskian
W (y1 (t) , y2 (t)) = e−4t
Substitute functions into equation for yp (t)
R
yp (t) = Dirac(t − 1) dt(− cos (t) sin (1) + cos (1) sin (t)) e2−2t
Compute integrals
yp (t) = H eaviside(t − 1) (− cos (t) sin (1) + cos (1) sin (t)) e2−2t
Substitute particular solution into general solution to ODE
y(t) = C 1 e−2t cos (t) + C 2 e−2t sin (t) + H eaviside(t − 1) (− cos (t) sin (1) + cos (1) sin (t)) e2−2t
Check validity of solution y(t) = _C1e−2t cos (t) + _C2e−2t sin (t) + H eaviside(t − 1) (− cos (t) si
Use initial condition y(0) = 0
0 = _C1
Compute derivative of the solution
d
y(t) = −2_C1 e−2t cos (t) − _C1 e−2t sin (t) − 2_C2 e−2t sin (t) + _C2 e−2t cos (t) + Dirac(t − 1)
dt
Use the initial condition dtd y(t)
=3
{t=0}
◦
◦
•
3 = −2_C1 + _C2
Solve for _C1 and _C2
{_C1 = 0, _C2 = 3}
Substitute constant values into general solution and simplify
y(t) = e−2t (3 sin (t) + H eaviside(t − 1) (− cos (t) sin (1) + cos (1) sin (t)) e2 )
Solution to the IVP
y(t) = e−2t (3 sin (t) + H eaviside(t − 1) (− cos (t) sin (1) + cos (1) sin (t)) e2 )
3 Mathematica. Time used: 0.028 (sec). Leaf size: 105
ode=D[y[t],{t,2}]+4*D[y[t],t]+5*y[t]==DiracDelta[t-1];
ic={y[0]==0,Derivative[1][y][0] ==3};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
−2t
y(t) → −e
Z 0
Z t
Z 0
2
2
sin(t)
e cos(1)δ(K[1] − 1)dK[1] − sin(t)
e cos(1)δ(K[1] − 1)dK[1] + cos(t)
−e2 δ(
1
1
1
chapter 2 . book solved problems
563
3 Sympy. Time used: 1.593 (sec). Leaf size: 76
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Dirac(t - 1) + 5*y(t) + 4*Derivative(y(t), t) + Derivative(y(t), (t, 2)
),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 3}
dsolve(ode,func=y(t),ics=ics)

Z
y(t) = −
Dirac (t − 1)e2t sin (t) dt +
Z0

Z
+
Dirac (t − 1)e2t cos (t) dt −

Dirac (t − 1)e2t sin (t) dt cos (t)
Z0


Dirac (t − 1)e2t cos (t) dt + 3 sin (t) e−2t
chapter 2 . book solved problems
2.8.5
564
Problem 7
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567
Internal problem ID [8651]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 7
Date solved : Tuesday, November 11, 2025 at 12:04:29 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
00
0
−t
4y + 24y + 37y = 17 e
1
+δ t−
2
y(0) = 1
y 0 (0) = 1
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
4s2 Y (s) − 4y 0 (0) − 4sy(0) + 24sY (s) − 24y(0) + 37Y (s) =
s
17
+ e− 2
1+s
Substituting given initial conditions into Eq (1) gives
4s2 Y (s) − 28 − 4s + 24sY (s) + 37Y (s) =
s
17
+ e− 2
1+s
Solving the above equation for Y (s) results in
s
s
e− 2 s + 4s2 + e− 2 + 32s + 45
Y (s) =
(1 + s) (4s2 + 24s + 37)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−s
−s
2
2 s + 4s + e 2 + 32s + 45
−1 e
=L
(1 + s) (4s2 + 24s + 37)
3
Heaviside t − 12 e−3t+ 2 sin 2t − 14
t
−3t
=
+ 4 e sin
+ e−t
2
2
(1)
chapter 2 . book solved problems
565
Solving for c1 and c2 from the given initial conditions results in
3
Heaviside t − 12 e−3t+ 2 sin
y=
2
t
− 14
2
−3t
+ 4e
t
sin
+ e−t
2
Converting the above solution to piecewise it becomes
(
y=
4 e−3t sin 2t + e−t
t < 12
3
e−3t+ 2 sin 2t − 14
1
4 e−3t sin 2t + e−t +
≤t
2
2
Simplifying the solution gives
y = e−3t
t
2t
e + 4 sin
+
2
(
t < 12
0
3
e 2 sin 2t − 14
2
!!
1
≤t
2
Figure 2.38: Solutions plot
3 Maple. Time used: 0.274 (sec). Leaf size: 36
ode:=4*diff(diff(y(t),t),t)+24*diff(y(t),t)+37*y(t) = 17*exp(-t)+Dirac(t-1/2);
ic:=[y(0) = 1, D(y)(0) = 1];
dsolve([ode,op(ic)],y(t),method='laplace');
y=
3
e 2 Heaviside t − 12 sin − 14 + 2t + 2 e2t + 8 sin
−3t
t
e
2
2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
chapter 2 . book solved problems
566
Maple step by step
Let’s
solve
4 dtd dtd y(t) + 24 dtd y(t) + 37y(t) = 17 e−t + Dirac t − 12 , y(0) = 1,
•
d
y(t)
dt
=1
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Isolate 2nd derivative
•
1
−t
Dirac t− 2
d d
y(t) = −6 dtd y(t) − 37y(t)
+ 17 4e +
dt dt
4
4
•
Group terms with y(t) on the lhs of the ODE
and the rest on the rhs of the ODE; ODE is linear
1
−t
Dirac t− 2
d d
y(t) + 6 dtd y(t) + 37y(t)
= 17 4e +
dt dt
4
4
•
Characteristic polynomial of homogeneous ODE
r2 + 6r + 37
=0
4
Use quadratic
formula to solve for r
√ •
(−6)±
•
•
•
•
•
◦
◦
◦
◦
−1
r=
2
Roots of the characteristic polynomial
r = −3 − 2I , −3 + 2I
1st solution of the homogeneous ODE
y1 (t) = e−3t cos 2t
2nd solution of the homogeneous ODE
y2 (t) = e−3t sin 2t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−3t cos 2t + C 2 e−3t sin 2t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
−t
Dirac t− 12
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 17 4e +
4
Wronskian of solutions
of the homogeneous equation
"
#
t
−3t
e−3t cos 2t
e
sin
e−3t sin t 2 e−3t cos t W (y1 (t) , y2 (t)) =
t
t
−3t
−3t
2
2
−3 e cos 2 −
−3 e sin 2 +
2
2
Compute Wronskian
−6t
W (y1 (t) , y2 (t)) = e 2
Substitute functions
into equation for yp (t)
e−3t
◦
R
3
R
3
sin 14 e 2 Dirac t− 12 +17 e2t sin 2t dt cos 2t − cos 14 e 2 Dirac t− 12 +17 e2t cos 2t dt sin 2t
yp (t) = −
Compute integrals
H eaviside t− 1
•
◦
◦
◦
◦
◦
2
3
cos 1 sin t −sin 1 cos t e 2 −3t
2
4
2
4
2
yp (t) =
+ e−t
2
Substitute particular solution into general solution to ODE
H eaviside t− 12 cos 14 sin 2t −sin 14 cos 2t e 32 −3t
t
t
−3t
−3t
y(t) = C 1 e cos 2 + C 2 e sin 2 +
+ e−t
2
H eaviside t− 1 cos 1 sin t −sin
2
4
2
Check validity of solution y(t) = _C1e−3t cos 2t + _C2e−3t sin 2t +
2
Use initial condition y(0) = 1
1 = _C1 + 1
Compute derivative of the solution
_C1 e−3t sin 2t
d
y(t) = −3_C1 e−3t cos
dt
t
2
Use the initial condition
d
y(t)
dt
−
2
− 3_C2 e−3t sin
t
2
+
_C2 e−3t cos 2t
=1
{t=0}
1 = −3_C1 − 1 + _C2
2
Solve for _C1 and _C2
{_C1 = 0, _C2 = 4}
Substitute constant values into general solution and simplify
H eaviside t− 12 cos 14 sin 2t −sin 14 cos 2t e 32 −3t
y(t) = 4 e−3t sin 2t +
+ e−t
2
2
+
Dirac t− 12
cos 14
chapter 2 . book solved problems
•
567
Solution to the IVP
H eaviside t− 1 cos 1 sin t −sin 1 cos t e 32 −3t
t
−3t
2
4
2
4
2
y(t) = 4 e sin 2 +
+ e−t
2
3 Mathematica. Time used: 0.12 (sec). Leaf size: 179
ode=4*D[y[t],{t,2}]+24*D[y[t],t]+27*y[t]==17*Exp[-t]+DiracDelta[t-1/2];
ic={y[0]==1,Derivative[1][y][0] ==1};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Z t
Z 0
1 −9t/2
1 7K[1]
1 K[2]
K[1]
3t
2
y(t) → − e
−6
− e
2e δ(2K[1] − 1) + 17 dK[1] + 6e
e 2 2eK[2] δ(2K[2] − 1) +
6
12
1
1 12
3 Sympy. Time used: 32.888 (sec). Leaf size: 143
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Dirac(t - 1/2) + 37*y(t) + 24*Derivative(y(t), t) + 4*Derivative(y(t),
(t, 2)) - 17*exp(-t),0)
ics = {y(0): 1, Subs(Derivative(y(t), t), t, 0): 1}
dsolve(ode,func=y(t),ics=ics)

R0
2t
 17e sin
y(t) = 
2
t
2
dt
R
−
R0
+

R0
t
2
Dirac t − 12 et + 17 e2t sin
2
dt

Dirac t − 12 e3t sin
2
t
2
dt

+ 1 cos
t
2
Dirac t − 12 et + 17 e2t cos 2t dt
+
2


3t
R0
1
t
Dirac t − 2 e cos 2 dt
t  −3t

−
+ 8 sin
e
2
2
17e2t cos

+ −
2
t
2
dt
R
chapter 2 . book solved problems
2.8.6
568
Problem 8
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 570
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 571
Internal problem ID [8652]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 8
Date solved : Tuesday, November 11, 2025 at 12:04:30 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 3y 0 + 2y = 10 sin (t) + 10δ(t − 1)
y(0) = 1
y 0 (0) = −1
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 3sY (s) − 3y(0) + 2Y (s) =
10
s2 + 1
+ 10 e−s
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 2 − s + 3sY (s) + 2Y (s) =
10
s2 + 1
+ 10 e−s
Solving the above equation for Y (s) results in
Y (s) =
10 e−s s2 + s3 + 2s2 + 10 e−s + s + 12
(s2 + 1) (s2 + 3s + 2)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−s 2
3
2
−s
+ s + 12
−1 10 e s + s + 2s + 10 e
=L
(s2 + 1) (s2 + 3s + 2)
= −3 cos (t) − 2 e−2t + 6 e−t + sin (t) + 10 −e−2t+2 + e1−t Heaviside (t − 1)
Solving for c1 and c2 from the given initial conditions results in
y = −3 cos (t) − 2 e−2t + 6 e−t + sin (t) + 10 −e−2t+2 + e1−t Heaviside (t − 1)
(1)
chapter 2 . book solved problems
569
Converting the above solution to piecewise it becomes
y=
−3 cos (t) − 2 e−2t + 6 e−t + sin (t)
t<1
−2t
−t
1−t
−2t+2
−3 cos (t) − 2 e + 6 e + sin (t) + 10 e − 10 e
1≤t
Simplifying the solution gives
−t
y = −3 cos (t) + sin (t) + 6 e
−2t
− 2e
+
0
t<1
1−t
−2t+2
10 e − 10 e
1≤t
Figure 2.39: Solutions plot
3 Maple. Time used: 0.256 (sec). Leaf size: 47
ode:=diff(diff(y(t),t),t)+3*diff(y(t),t)+2*y(t) = 10*sin(t)+10*Dirac(t-1);
ic:=[y(0) = 1, D(y)(0) = -1];
dsolve([ode,op(ic)],y(t),method='laplace');
y = −10 Heaviside (t − 1) e2−2t + 10 Heaviside (t − 1) e1−t − 3 cos (t) + sin (t) − 2 e−2t + 6 e−t
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
<- double symmetry of the form [xi=0, eta=F(x)] successful
Maple step by step
Let’s
solve
d d
y(t) + 3 dtd y(t) + 2y(t) = 10 sin (t) + 10Dirac(t − 1) , y(0) = 1,
dt dt
•
•
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 3r + 2 = 0
d
y(t)
dt
= −1
{t=0}
chapter 2 . book solved problems
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
570
Factor the characteristic polynomial
(r + 2) (r + 1) = 0
Roots of the characteristic polynomial
r = (−2, −1)
1st solution of the homogeneous ODE
y1 (t) = e−2t
2nd solution of the homogeneous ODE
y2 (t) = e−t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−2t + C 2 e−t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = 10 sin (t) + 10Dirac(t − 1)
Wronskian of solutions of the homogeneous equation
e−2t
e−t
W (y1 (t) , y2 (t)) =
−2 e−2t −e−t
Compute Wronskian
W (y1 (t) , y2 (t)) = e−3t
Substitute functions into equation for yp (t)
R
R
yp (t) = −10 e−2t (e2 Dirac(t − 1) + sin (t) e2t ) dt + 10 e−t et (sin (t) + Dirac(t − 1)) dt
Compute integrals
yp (t) = 10 e1−t H eaviside(t − 1) − 3 cos (t) + sin (t) − 10 e2−2t H eaviside(t − 1)
Substitute particular solution into general solution to ODE
y(t) = C 1 e−2t + C 2 e−t + 10 e1−t H eaviside(t − 1) − 3 cos (t) + sin (t) − 10 e2−2t H eaviside(t − 1)
Check validity of solution y(t) = _C1e−2t + _C2e−t + 10e1−t H eaviside(t − 1) − 3 cos (t) + sin (t)
Use initial condition y(0) = 1
1 = _C1 + _C2 − 3
Compute derivative of the solution
d
y(t) = −2_C1 e−2t − _C2 e−t − 10 e1−t H eaviside(t − 1) + 10 e1−t Dirac(t − 1) + 3 sin (t) + cos (
dt
Use the initial condition dtd y(t)
= −1
{t=0}
◦
◦
•
−1 = −2_C1 − _C2 + 1
Solve for _C1 and _C2
{_C1 = −2, _C2 = 6}
Substitute constant values into general solution and simplify
y(t) = −2 e−2t + 6 e−t + 10 e1−t H eaviside(t − 1) − 3 cos (t) + sin (t) − 10 e2−2t H eaviside(t − 1)
Solution to the IVP
y(t) = −2 e−2t + 6 e−t + 10 e1−t H eaviside(t − 1) − 3 cos (t) + sin (t) − 10 e2−2t H eaviside(t − 1)
3 Mathematica. Time used: 0.087 (sec). Leaf size: 121
ode=D[y[t],{t,2}]+3*D[y[t],t]+2*y[t]==10*(Sin[t]+DiracDelta[t-1]);
ic={y[0]==1,Derivative[1][y][0] ==-1};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
−2t
y(t) → −e
Z t
Z 0
2K[1]
t
−
−10e
(δ(K[1] − 1) + sin(K[1]))dK[1] + e
10eK[2] (δ(K[2] − 1) + sin(K[2]))dK[2
1
1
chapter 2 . book solved problems
571
3 Sympy. Time used: 1.327 (sec). Leaf size: 90
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-10*Dirac(t - 1) + 2*y(t) - 10*sin(t) + 3*Derivative(y(t), t) +
Derivative(y(t), (t, 2)),0)
ics = {y(0): 1, Subs(Derivative(y(t), t), t, 0): -1}
dsolve(ode,func=y(t),ics=ics)

Z
y(t) = −10
2t
Z0
(Dirac (t − 1) + sin (t)) e dt + 10
Z0
+ 10
Dirac (t − 1)e2t dt

e sin (t) dt e−t + 10
2t
Z0
− 10
Z
(Dirac (t − 1) + sin (t)) et dt
Dirac (t − 1)et dt − 10
Z0

et sin (t) dt + 1 e−t
chapter 2 . book solved problems
2.8.7
572
Problem 9
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573
7Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575
Internal problem ID [8653]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 9
Date solved : Tuesday, November 11, 2025 at 12:04:31 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 4y 0 + 5y = (1 − Heaviside (t − 10)) et − e10 δ(t − 10)
y(0) = 0
y 0 (0) = 1
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
−s e10−10s + 1
s Y (s) − y (0) − sy(0) + 4sY (s) − 4y(0) + 5Y (s) =
s−1
2
0
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 1 + 4sY (s) + 5Y (s) =
−s e10−10s + 1
s−1
Solving the above equation for Y (s) results in
Y (s) = −
s(e10−10s − 1)
(s − 1) (s2 + 4s + 5)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
s(e10−10s − 1)
−1
=L
−
(s − 1) (s2 + 4s + 5)
=
et Heaviside (−t + 10) e30−2t Heaviside (t − 10) (cos (t − 10) − 7 sin (t − 10))
+
10
10
(− cos (t) + 7 sin (t)) e−2t
+
10
(1)
chapter 2 . book solved problems
573
Solving for c1 and c2 from the given initial conditions results in
y=
et Heaviside (−t + 10) e30−2t Heaviside (t − 10) (cos (t − 10) − 7 sin (t − 10))
+
10
10
(− cos (t) + 7 sin (t)) e−2t
+
10
Converting the above solution to piecewise it becomes
y=





(− cos(t)+7 sin(t))e−2t
et
+ 10
10
10
(− cos(10)+7 sin(10))e−20
+ e5
10
(− cos(t)+7 sin(t))e−2t
e30−2t (cos(t−10)−7 sin(t−10))
+
10
10
t < 10
t = 10
10 < t
Simplifying the solution gives




y=−

(−e3t + cos (t) − 7 sin (t)) e−2t
t < 10
(−2 e30 + cos (10) − 7 sin (10)) e−20
t = 10 
−2t
30
30
e (−e cos (t − 10) + 7 e sin (t − 10) + cos (t) − 7 sin (t)) 10 < t
10
Figure 2.40: Solutions plot
3 Maple. Time used: 0.353 (sec). Leaf size: 66
ode:=diff(diff(y(t),t),t)+4*diff(y(t),t)+5*y(t) = (1-Heaviside(t-10))*exp(t)-exp
(10)*Dirac(t-10);
ic:=[y(0) = 0, D(y)(0) = 1];
dsolve([ode,op(ic)],y(t),method='laplace');
y=
sin(10)
(cos (10) + 7 sin (10)) cos (t) − 7 cos (10) − 7
sin (t) Heaviside (t − 10) e30−2t
10
et Heaviside (t − 10) (− cos (t) + 7 sin (t)) e−2t
et
−
+
+
10
10
10
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
chapter 2 . book solved problems
574
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) + 4 dtd y(t) + 5y(t) = (1 − H eaviside(t − 10)) et − e10 Dirac(t − 10) , y(0) = 0,
dt dt
•
d
y(t)
dt
{t=
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Isolate 2nd derivative
d d
y(t) = −5y(t) − et H eaviside(t − 10) − e10 Dirac(t − 10) − 4 dtd y(t) + et
dt dt
Group terms with y(t) on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear
d d
y(t) + 4 dtd y(t) + 5y(t) = −et H eaviside(t − 10) − e10 Dirac(t − 10) + et
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 4r + 5 = 0
Use quadratic
formula to solve for r
√ •
•
•
•
(−4)±
•
•
•
•
•
◦
◦
◦
◦
◦
−4
r=
2
Roots of the characteristic polynomial
r = (−2 − I, −2 + I)
1st solution of the homogeneous ODE
y1 (t) = e−2t cos (t)
2nd solution of the homogeneous ODE
y2 (t) = e−2t sin (t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−2t cos (t) + C 2 e−2t sin (t) + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = −et H eaviside(t − 10) − e10 Dirac(t
Wronskian of solutions of the homogeneous equation
e−2t cos (t)
e−2t sin (t)
W (y1 (t) , y2 (t)) =
−2 e−2t cos (t) − e−2t sin (t) −2 e−2t sin (t) + e−2t cos (t)
Compute Wronskian
W (y1 (t) , y2 (t)) = e−4t
Substitute functions into equation for yp (t)
R
R
yp (t) = e−2t (− cos (t) e3t (−1 + H eaviside(t − 10)) − cos (10) e30 Dirac(t − 10)) dt sin (t) − (−
Compute
integrals
(cos(t)−7 sin(t)) cos(10)+7 sin(10) cos(t)+
sin(t)
H eaviside(t−10)e30−2t
7
yp (t) =
10
Substitute particular solution into general
solution to ODE
•
−2t
y(t) = C 1 e
◦
◦
cos (t) + C 2 e
−2t
sin (t) +
t
− e (−1+H eaviside(t−10))
10
sin(t)
(cos(t)−7 sin(t)) cos(10)+7 sin(10) cos(t)+ 7
H eaviside(t−10)e30−2t
10
Check validity of solution y(t) = _C1e−2t cos (t) + _C2e−2t sin (t) +
Use initial condition y(0) = 0
1
0 = _C1 + 10
Compute derivative of the solution
d
y(t) = −2_C1 e−2t cos (t) − _C1 e−2t sin (t) − 2_C2 e−2t sin (t) + _C2 e−2t cos (t) +
dt
◦
Use the initial condition
d
y(t)
dt
=1
{t=0}
−
(cos(t)−7 sin(t)) cos(10)+7 sin(10) cos
10
(− sin(t)−7 cos(
chapter 2 . book solved problems
◦
◦
575
1
1 = −2_C1 + 10
+ _C2
Solve for _C1 and _C2
1
7
_C1 = − 10
, _C2 = 10
Substitute constant values into general solution and simplify
−e3t +(cos(t)(cos(10)+7 sin(10))+sin(t)(−7 cos(10)+sin(10)))e30 H eaviside(t−10)−cos(t)+7 sin(t)+e3t e−2t
10
•
y(t) =
Solution to the IVP
y(t) =
−e3t +(cos(t)(cos(10)+7 sin(10))+sin(t)(−7 cos(10)+sin(10)))e30 H eaviside(t−10)−cos(t)+7 sin(t)+e3t e−2t
10
7 Mathematica
ode=D[y[t],{t,2}]+4*D[y[t],t]+5*y[t]==(1-UnitStep[t-10])*Exp[t]-Exp[10]*
DiracDelta[t-10];
ic={y[0]==0,Derivative[1][y][0] ==1};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Timed out
3 Sympy. Time used: 4.251 (sec). Leaf size: 224
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq((Heaviside(t - 10) - 1)*exp(t) + Dirac(t - 10)*exp(10) + 5*y(t) + 4*
Derivative(y(t), t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 1}
dsolve(ode,func=y(t),ics=ics)

y(t) = −
(e3t sin (t) + 3e3t cos (t) − e30 sin (10) − 3e30 cos (10)) θ(t − 10) e3t sin (t)
+
10
10
Z
Z0
3e3t cos (t)
10
2t
10
+
−e
Dirac (t − 10)e cos (t) dt + e
Dirac (t − 10)e2t cos (t) dt
10


+
7
(3e3t sin (t) − e3t cos (t) + e30 cos (10) − 3e30 sin (10)) θ(t − 10)
sin (t) + 
10
10
Z
3e3t sin (t) e3t cos (t)
10
−
+
+e
Dirac (t − 10)e2t sin (t) dt
10
10


Z0
1
− e10 Dirac (t − 10)e2t sin (t) dt −  cos (t) e−2t
10
chapter 2 . book solved problems
2.8.8
576
Problem 10
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577
7Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 579
Internal problem ID [8654]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 10
Date solved : Tuesday, November 11, 2025 at 12:04:33 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
π
y 00 + 5y 0 + 6y = δ t −
+ Heaviside (t − π) cos (t)
2
y(0) = 0
y 0 (0) = 0
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
πs
s2 Y (s) − y 0 (0) − sy(0) + 5sY (s) − 5y(0) + 6Y (s) = e− 2 −
e−πs s
s2 + 1
Substituting given initial conditions into Eq (1) gives
πs
s2 Y (s) + 5sY (s) + 6Y (s) = e− 2 −
e−πs s
s2 + 1
Solving the above equation for Y (s) results in
πs
πs
e− 2 s2 − e−πs s + e− 2
Y (s) = 2
(s + 1) (s2 + 5s + 6)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
− πs 2
πs −πs
2 s − e
s + e− 2
−1 e
=L
(s2 + 1) (s2 + 5s + 6)
=
Heaviside (t − π) (4 e−2t+2π − 3 e−3t+3π + sin (t) + cos (t))
10
π −3t+ 3π
−2t+π
2 + e
+ Heaviside t −
−e
2
(1)
chapter 2 . book solved problems
577
Solving for c1 and c2 from the given initial conditions results in
y=
Heaviside (t − π) (4 e−2t+2π − 3 e−3t+3π + sin (t) + cos (t))
10
π −3t+ 3π
−2t+π
2 + e
+ Heaviside t −
−e
2
Converting the above solution to piecewise it becomes



0
t < π2
3π
y=
−e−3t+ 2 + e−2t+π
t<π

 − 3 e−3t+3π + 2 e−2t+2π + sin(t) + cos(t) − e−3t+ 3π2 + e−2t+π π ≤ t
10
5
10
10
Simplifying the solution gives



0
t < π2
3π
y=
−e−3t+ 2 + e−2t+π
t<π

 − 3 e−3t+3π + 2 e−2t+2π + sin(t) + cos(t) − e−3t+ 3π2 + e−2t+π π ≤ t
10
5
10
10
Figure 2.41: Solutions plot
3 Maple. Time used: 0.294 (sec). Leaf size: 62
ode:=diff(diff(y(t),t),t)+5*diff(y(t),t)+6*y(t) = Dirac(t-1/2*Pi)+cos(t)*
Heaviside(t-Pi);
ic:=[y(0) = 0, D(y)(0) = 0];
dsolve([ode,op(ic)],y(t),method='laplace');
y=
Heaviside (t − π) (cos (t) − 3 e−3t+3π + 4 e−2t+2π + sin (t))
10
5π
5t
π
t π
+ 2 Heaviside t −
sinh
−
e 4 −2
2
2 4
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
chapter 2 . book solved problems
578
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) + 5 dtd y(t) + 6y(t) = Dirac t − π2 + H eaviside(t − π) cos (t) , y(0) = 0,
dt dt
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
d
y(t)
dt
=
{t=0}
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 5r + 6 = 0
Factor the characteristic polynomial
(r + 3) (r + 2) = 0
Roots of the characteristic polynomial
r = (−3, −2)
1st solution of the homogeneous ODE
y1 (t) = e−3t
2nd solution of the homogeneous ODE
y2 (t) = e−2t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−3t + C 2 e−2t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = Dirac t − π2 + H eaviside(t − π) c
Wronskian of solutions of the homogeneous equation
e−3t
e−2t
W (y1 (t) , y2 (t)) =
−3 e−3t −2 e−2t
Compute Wronskian
W (y1 (t) , y2 (t)) = e−5t
Substitute functions
for yp (t)
3π into equation
R
R π
yp (t) = −e−3t
e 2 Dirac t − π2 + H eaviside(t − π) cos (t) e3t dt + e−2t
e Dirac t − π2 + H
Compute integrals
3π−3t H eaviside(t−π)
3π
yp (t) = eπ−2t H eaviside t − π2 − 3 e
− e 2 −3t H eaviside t − π2 + H eaviside(t−π)(sin(
10
10
Substitute particular solution into general solution to ODE
3π−3t H eaviside(t−π)
3π
y(t) = C 1 e−3t + C 2 e−2t + eπ−2t H eaviside t − π2 − 3 e
− e 2 −3t H eaviside t − π2
10
3π−3t H eaviside(t−
Check validity of solution y(t) = _C1e−3t + _C2e−2t + eπ−2t H eaviside t − π2 − 3e
10
Use initial condition y(0) = 0
0 = _C1 + _C2
Compute derivative of the solution
9 e3π−3t H eavisid
d
π
π
−3t
−2t
π−2t
π−2t
y(t)
=
−3_C1
e
−
2_C2
e
−
2
e
H
eaviside
t
−
+
e
Dirac
t
−
+
dt
2
2
10
d
Use the initial condition dt y(t)
=0
{t=0}
◦
◦
•
0 = −3_C1 − 2_C2
Solve for _C1 and _C2
{_C1 = 0, _C2 = 0}
Substitute constant values into general solution and simplify
3π−3t H eaviside(t−π)
3π
y(t) = eπ−2t H eaviside t − π2 − 3 e
− e 2 −3t H eaviside t − π2 + H eaviside(t−π)(sin(t)
10
10
Solution to the IVP
chapter 2 . book solved problems
579
3π−3t H eaviside(t−π)
3π
y(t) = eπ−2t H eaviside t − π2 − 3 e
− e 2 −3t H eaviside t − π2 + H eaviside(t−π)(sin(t)
10
10
7 Mathematica
ode=D[y[t],{t,2}]+5*D[y[t],t]+6*y[t]==DiracDelta[t-1/2*Pi]+UnitStep[t-Pi]*Cos[t];
ic={y[0]==0,Derivative[1][y][0] ==0};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Timed out
3 Sympy. Time used: 1.654 (sec). Leaf size: 119
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Dirac(t - pi/2) + 6*y(t) - cos(t)*Heaviside(t - pi) + 5*Derivative(y(t),
t) + Derivative(y(t), (t, 2)),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 0}
dsolve(ode,func=y(t),ics=ics)

Z0
π
π 3t
3t


y(t) =
−
Dirac t −
+ cos (t)θ(t − π) e dt + Dirac t −
e dt
2
2

Z0
Z π
3t
−t
+ e cos (t)θ(t − π) dt e +
Dirac t −
+ cos (t)θ(t − π) e2t dt
2

Z0
Z0
π 2t
− Dirac t −
e dt − e2t cos (t)θ(t − π) dt e−2t
2
Z chapter 2 . book solved problems
2.8.9
580
Problem 11
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583
Internal problem ID [8655]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 11
Date solved : Tuesday, November 11, 2025 at 12:04:34 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 5y 0 + 6y = Heaviside (t − 1) + δ(t − 2)
y(0) = 0
y 0 (0) = 1
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 5sY (s) − 5y(0) + 6Y (s) =
e−s
+ e−2s
s
(1)
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 1 + 5sY (s) + 6Y (s) =
e−s
+ e−2s
s
Solving the above equation for Y (s) results in
Y (s) =
e−2s s + e−s + s
s (s2 + 5s + 6)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
−2s
s + e−s + s
−1 e
=L
s (s2 + 5s + 6)
= e−2t − e−3t +
Heaviside (t − 1) (1 + 2 e−3t+3 − 3 e−2t+2 )
+ Heaviside (t − 2) e−2t+4 − e−3t+6
6
chapter 2 . book solved problems
581
Solving for c1 and c2 from the given initial conditions results in
y = e−2t −e−3t +
Heaviside (t − 1) (1 + 2 e−3t+3 − 3 e−2t+2 )
+Heaviside (t−2) e−2t+4 −e−3t+6
6
Converting the above solution to piecewise it becomes


e−2t − e−3t
t<1

1
e−3t+3
e−2t+2
−2t
−3t
y=
e −e + 6 + 3 − 2
t<2

 e−2t − e−3t + 1 + e−3t+3 − e−2t+2 + e−2t+4 − e−3t+6 2 ≤ t
6
3
2
Simplifying the solution gives


e−2t − e−3t
t<1

1
e−3t+3
e−2t+2
−2t
−3t
e − e + 6 + 3 − 2
t<2
y=

 e3t −6 e6 +6 e4+t +2 e3 −3 et+2 +6 et −6 e−3t 2 ≤ t
6
Figure 2.42: Solutions plot
3 Maple. Time used: 0.149 (sec). Leaf size: 68
ode:=diff(diff(y(t),t),t)+5*diff(y(t),t)+6*y(t) = Heaviside(t-1)+Dirac(t-2);
ic:=[y(0) = 0, D(y)(0) = 1];
dsolve([ode,op(ic)],y(t),method='laplace');
y = e−2t − e−3t + Heaviside (t − 2) e4−2t − Heaviside (t − 2) e6−3t
Heaviside (t − 1) e2−2t Heaviside (t − 1) e3−3t Heaviside (t − 1)
−
+
+
2
3
6
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
chapter 2 . book solved problems
582
Maple step by step
Let’s
solve
d d
y(t) + 5 dtd y(t) + 6y(t) = H eaviside(t − 1) + Dirac(t − 2) , y(0) = 0,
dt dt
•
•
•
•
•
•
•
•
◦
◦
◦
◦
◦
•
◦
◦
◦
◦
=1
{t=0}
Wronskian of solutions of the homogeneous equation
e−3t
e−2t
W (y1 (t) , y2 (t)) =
−3 e−3t −2 e−2t
Compute Wronskian
W (y1 (t) , y2 (t)) = e−5t
Substitute functions into equation for yp (t)
R
R
yp (t) = −e−3t (e6 Dirac(t − 2) + H eaviside(t − 1) e3t ) dt + e−2t (e4 Dirac(t − 2) + H eaviside(
Compute integrals
3−3t
2−2t
yp (t) = e H eaviside(t−1)
+ H eaviside(t−1)
− e6−3t H eaviside(t − 2) − e H eaviside(t−1)
+ e4−2t H eavis
3
6
2
Substitute particular solution into general solution to ODE
3−3t
H eaviside(t−1)
e2−2t H eavisid
6−3t
y(t) = C 1 e−3t + C 2 e−2t + e H eaviside(t−1)
+
−
e
H
eaviside(t
−
2)
−
3
6
2
3−3t
H eaviside(t−1)
6−3t
Check validity of solution y(t) = _C1e−3t + _C2e−2t + e H eaviside(t−1)
+
−
e
H
e
3
6
Use initial condition y(0) = 0
0 = _C1 + _C2
Compute derivative of the solution
3−3t
d
y(t) = −3_C1 e−3t − 2_C2 e−2t − e3−3t H eaviside(t − 1) + e Dirac(t−1)
+ Dirac(t−1)
+ 3 e6−3t H e
dt
3
6
Use the initial condition dtd y(t)
=1
1 = −3_C1 − 2_C2
Solve for _C1 and _C2
{_C1 = −1, _C2 = 1}
Substitute
constant values into general solution and simplify
3t
2+t
e3 + e 2 − 3 e 2
•
Highest derivative means the order of the ODE is 2
d d
y(t)
dt dt
Characteristic polynomial of homogeneous ODE
r2 + 5r + 6 = 0
Factor the characteristic polynomial
(r + 3) (r + 2) = 0
Roots of the characteristic polynomial
r = (−3, −2)
1st solution of the homogeneous ODE
y1 (t) = e−3t
2nd solution of the homogeneous ODE
y2 (t) = e−2t
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−3t + C 2 e−2t + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R y1 (t)f (t)
y2 (t)f (t)
yp (t) = −y1 (t) W (y1 (t),y2 (t)) dt + y2 (t) W (y1 (t),y2 (t)) dt, f (t) = H eaviside(t − 1) + Dirac(t − 2)
{t=0}
◦
d
y(t)
dt
y(t) =
Solutionto the IVP 3t
y(t) =
2+t
e3 + e 2 − 3 e 2
H eaviside(t−1)+ −3 e6 +3 e4+t H eaviside(t−2)+3 et −3 e−3t
3
H eaviside(t−1)+ −3 e6 +3 e4+t H eaviside(t−2)+3 et −3 e−3t
3
chapter 2 . book solved problems
583
3 Mathematica. Time used: 0.335 (sec). Leaf size: 306
ode=D[y[t],{t,2}]+5*D[y[t],t]+6*y[t]==UnitStep[t-1]+DiracDelta[t-2];
ic={y[0]==0,Derivative[1][y][0] ==1};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Z 0
Z t
Z t
−3t
t
4
3K[3]
t
y(t) → θ(1 − t) e
e
e δ(K[2] − 2)dK[2] −
−e
(δ(K[3] − 2) + 1)dK[3] − e
e2K[4] (δ(K[4]
1
1
1
3 Sympy. Time used: 1.072 (sec). Leaf size: 102
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-Dirac(t - 2) + 6*y(t) - Heaviside(t - 1) + 5*Derivative(y(t), t) +
Derivative(y(t), (t, 2)),0)
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 1}
dsolve(ode,func=y(t),ics=ics)

Z
y(t) = −
3t
(Dirac (t − 2) + θ(t − 1)) e dt +
Z0
Dirac (t − 2)e dt +

− 1 e−t +
Z
Z0
3t
(Dirac (t − 2) + θ(t − 1)) e2t dt −
Z0
−
Z0
e3t θ(t − 1) dt
Dirac (t − 2)e2t dt

e2t θ(t − 1) dt + 1 e−2t
chapter 2 . book solved problems
2.8.10
584
Problem 12
Local contents
3Maple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585
3Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586
3Sympy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587
Internal problem ID [8656]
Book : ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section : Chapter 6. Laplace Transforms. Problem set 6.4, page 230
Problem number : 12
Date solved : Tuesday, November 11, 2025 at 12:04:35 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]
y 00 + 2y 0 + 5y = 25t − 100δ(t − π)
y(0) = −2
y 0 (0) = 5
Using Laplace transform method.
Solving using the Laplace transform method. Let
L(y) = Y (s)
Taking the Laplace transform of the ode and using the relations that
L(y 0 ) = sY (s) − y(0)
L(y 00 ) = s2 Y (s) − y 0 (0) − sy(0)
The given ode now becomes an algebraic equation in the Laplace domain.
s2 Y (s) − y 0 (0) − sy(0) + 2sY (s) − 2y(0) + 5Y (s) =
25
− 100 e−πs
s2
Substituting given initial conditions into Eq (1) gives
s2 Y (s) − 1 + 2s + 2sY (s) + 5Y (s) =
25
− 100 e−πs
s2
Solving the above equation for Y (s) results in
Y (s) = −
100 e−πs s2 + 2s3 − s2 − 25
s2 (s2 + 2s + 5)
Taking the inverse Laplace transform gives
y = L−1 (Y (s))
100 e−πs s2 + 2s3 − s2 − 25
−1
=L
−
s2 (s2 + 2s + 5)
= −50 Heaviside (t − π) e−t+π sin (2t) + 5t − 2
Solving for c1 and c2 from the given initial conditions results in
y = −50 Heaviside (t − π) e−t+π sin (2t) + 5t − 2
(1)
chapter 2 . book solved problems
585
Converting the above solution to piecewise it becomes
y=
−2 + 5t
t<π
−t+π
−2 + 5t − 50 e
sin (2t) π ≤ t
Simplifying the solution gives
y = −2 + 5t −
0
t<π
−t+π
50 e
sin (2t) π ≤ t
Figure 2.43: Solutions plot
3 Maple. Time used: 0.225 (sec). Leaf size: 27
ode:=diff(diff(y(t),t),t)+2*diff(y(t),t)+5*y(t) = 25*t-100*Dirac(t-Pi);
ic:=[y(0) = -2, D(y)(0) = 5];
dsolve([ode,op(ic)],y(t),method='laplace');
y = −50 sin (2t) Heaviside (t − π) e−t+π + 5t − 2
Maple trace
Methods for second order ODEs:
--- Trying classification methods --trying a quadrature
trying high order exact linear fully integrable
trying differential order: 2; linear nonhomogeneous with symmetry [0,1]
trying a double symmetry of the form [xi=0, eta=F(x)]
-> Try solving first the homogeneous part of the ODE
checking if the LODE has constant coefficients
<- constant coefficients successful
<- solving first the homogeneous part of the ODE successful
Maple step by step
Let’s
solve
d d
y(t) + 2 dtd y(t) + 5y(t) = 25t − 100Dirac(t − π) , y(0) = −2,
dt dt
•
Highest derivative means the order of the ODE is 2
d
y(t)
dt
=5
{t=0}
chapter 2 . book solved problems
586
d d
y(t)
dt dt
•
Characteristic polynomial of homogeneous ODE
r2 + 2r + 5 = 0
Use quadratic
formula to solve for r
√
•
(−2)±
•
•
•
•
•
◦
◦
◦
◦
−16
r=
2
Roots of the characteristic polynomial
r = (−1 − 2 I, −1 + 2 I)
1st solution of the homogeneous ODE
y1 (t) = e−t cos (2t)
2nd solution of the homogeneous ODE
y2 (t) = e−t sin (2t)
General solution of the ODE
y(t) = C 1 y1 (t) + C 2 y2 (t) + yp (t)
Substitute in solutions of the homogeneous ODE
y(t) = C 1 e−t cos (2t) + C 2 e−t sin (2t) + yp (t)
Find a particular solution yp (t) of the ODE
Use
f (t) is the forcing function
h variation ofRparameters to find yp here
i
R
(t)
y1 (t)f (t)
yp (t) = −y1 (t) W (yy21(t)f
dt
+
y
(t)
dt,
f
(t)
=
25t
−
100Dirac(t
−
π)
2
(t),y2 (t))
W (y1 (t),y2 (t))
Wronskian of solutions of the homogeneous equation
e−t cos (2t)
e−t sin (2t)
W (y1 (t) , y2 (t)) =
−e−t cos (2t) − 2 e−t sin (2t) −e−t sin (2t) + 2 e−t cos (2t)
Compute Wronskian
W (y1 (t) , y2 (t)) = 2 e−2t
Substitute functions
into equation
for yp (t)
R
R
25 e−t
◦
•
◦
◦
◦
et sin(2t)tdt cos(2t)−
−4 eπ Dirac(t−π)+et cos(2t)t dt sin(2t)
yp (t) = −
2
Compute integrals
yp (t) = 5t − 2 − 50 eπ−t sin (2t) H eaviside(t − π)
Substitute particular solution into general solution to ODE
y(t) = C 1 e−t cos (2t) + C 2 e−t sin (2t) + 5t − 2 − 50 eπ−t sin (2t) H eaviside(t − π)
Check validity of solution y(t) = _C1e−t cos (2t) + _C2e−t sin (2t) + 5t − 2 − 50eπ−t sin (2t) H eav
Use initial condition y(0) = −2
−2 = _C1 − 2
Compute derivative of the solution
d
y(t) = −_C1 e−t cos (2t) − 2_C1 e−t sin (2t) − _C2 e−t sin (2t) + 2_C2 e−t cos (2t) + 5 + 50 eπ−t
dt
Use the initial condition dtd y(t)
=5
{t=0}
◦
◦
•
5 = −_C1 + 5 + 2_C2
Solve for _C1 and _C2
{_C1 = 0, _C2 = 0}
Substitute constant values into general solution and simplify
y(t) = 5t − 2 − 50 eπ−t sin (2t) H eaviside(t − π)
Solution to the IVP
y(t) = 5t − 2 − 50 eπ−t sin (2t) H eaviside(t − π)
3 Mathematica. Time used: 0.189 (sec). Leaf size: 163
ode=D[y[t],{t,2}]+2*D[y[t],t]+5*y[t]==25*t-100*DiracDelta[t-Pi];
ic={y[0]==-2,Derivative[1][y][0] ==5};
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
Z 0
Z t
1 −t
25 K[1]
25 K[1]
π
y(t) → − e
2 sin(2t)
e
cos(2K[1])K[1] − 50e δ(π − K[1]) dK[1] − 2 sin(2t)
e
cos
2
2
2
1
1
chapter 2 . book solved problems
587
3 Sympy. Time used: 9.656 (sec). Leaf size: 139
from sympy import *
t = symbols("t")
y = Function("y")
ode = Eq(-25*t + 100*Dirac(t - pi) + 5*y(t) + 2*Derivative(y(t), t) + Derivative(
y(t), (t, 2)),0)
ics = {y(0): -2, Subs(Derivative(y(t), t), t, 0): 5}
dsolve(ode,func=y(t),ics=ics)

y(t) =
 25
R0
R
tet sin (2t) dt 25 (t − 4 Dirac (t − π)) et sin (2t) dt
−
2
2
R0
+
t

25 (−4 Dirac (t − π)e sin (2t)) dt
− 2 cos (2t)
2
R0 t
R
25
te
cos
(2t)
dt
25
(t − 4 Dirac (t − π)) et cos (2t) dt
+ −
+
2
2


R0
t
25 (−4 Dirac (t − π)e cos (2t)) dt 3 
−
+
sin (2t) e−t
2
2
