Section 3.1
Introduction & Review of Power Series

How would we solve y ″ + xy ′ + 3 x 2 y = 0 ?

Recall: n



0 a x n n  a
0
 a x
1
 a x
2
2  a x
3
3   a x n n 

Recall: n



0 a x n n  a
0
 a x
1
 a x
2
2  a x
3
3   a x n n  n



0 a n
   n  a
0
   a
2
  
2    
3   a n
 x c
 n 


 n

0 n lim
N
 n
N  n converges on an interval J if exists for all x values on the interval.

0 a x n

Maclaurin Series:
 n



0 f n !
(0) x n

Maclaurin Series:
 n



0 f n !
(0) x n
Taylor Series:
 n



0 f ( )   n n !

Examples e x 

 n

0
1 n !
x n (interval of convergence is (–∞,∞))

Examples e x 

 n

0
1 n !
x n cos x
 n



0
 
  n
2 n !
x
2 n
(interval of convergence is (–∞,∞))
(interval of convergence is (–∞,∞))

Examples e x 

 n

0
1 n !
x n cos x
 n



0
 
  n
2 n !
x
2 n
(interval of convergence is (–∞,∞))
(interval of convergence is (–∞,∞)) sin x


 n

0

2
  n
 n
 x
2 n

1 (interval of convergence is (–∞,∞))

Examples e x 

 n

0
1 n !
x n cos x
 n



0
 
  n
2 n !
x
2 n
(interval of convergence is (–∞,∞))
(interval of convergence is (–∞,∞)) sin x


 n

0

2
  n
 n
 x
2 n

1 (interval of convergence is (–∞,∞))
1
1
 x
 n



0 x n
(interval of convergence is (–1, 1))

The Ratio Test
 b n
 b n
 lim n
 b n

1 b n

1 lim n
 b n

1 b n

1

Ex. 1 Find the interval of convergence for the following power series:

(a)
 n

0
1 n !
x n

Ex. 1 Find the interval of convergence for the following power series:

(a)
 n

0
1 n !
x n

Ex. 1 Find the interval of convergence for the following power series:
(b) n



0
 
3 n n n x n

Ex. 1 Find the interval of convergence for the following power series:
(b) n



0
 
3 n n n x n

Important Properties of Power Series:
1.

 n



0 n

 n

0 n



________________
2. c



 n

0 a x n



________________
3. d dx

 n



0 a x n n



________________

n



1
Shifting Indices: n

1
= a
1
+ a
2 x + a
3 x 2 + a
4 x 3 + a
5 x 4 + a
6 x 5 + · · · ·

n



1
Shifting Indices: n

1
= a
1
+ a
2 x + a
3 x 2 + a
4 x 3 + a
5 x 4 + a
6 x 5 + · · · ·

 n

0 a n

1 x n

n



2
Shifting Indices: n

2
= a
2
+ a
3 x + a
4 x 2 + a
5 x 3 + a
6 x 4 + a
7 x 5 + · · · ·

n



2
Shifting Indices: n

2
= a
2
+ a
3 x + a
4 x 2 + a
5 x 3 + a
6 x 4 + a
7 x 5 + · · · ·  n



0 a n

2 x n

Ex. 2 Use power series to solve the differential equation y
′ + 4 y = 0. Then determine the interval of convergence. (Note that we can use the material in chapter 2 to see that the solution should be y = Ce
–4x
)

Ex. 2 Use power series to solve the differential equation y
′ + 4 y = 0. Then determine the interval of convergence.

Ex. 2 Use power series to solve the differential equation y
′ + 4 y = 0. Then determine the interval of convergence.

Ex. 3 Use power series to solve the differential equation ( x
– 3) y
′ + 2 y = 0, but just give the first five terms in the power series. Then determine the interval of convergence.

Ex. 3 Use power series to solve the differential equation ( x
– 3) y
′ + 2 y = 0, but just give the first five terms in the power series. Then determine the interval of convergence.

Ex. 4
(a) Use power series to solve y
″ + y = 0 . Then determine the interval of convergence.

Ex. 4
(a) Use power series to solve y
″ + y = 0 . Then determine the interval of convergence.

Ex. 4
(a) Use power series to solve y
″ + y = 0 . Then determine the interval of convergence.

Ex. 4
(b) Use power series to solve the initial value problem y
″ + y = 0, y (0) = 3, y
′(0) = 5.

Section 3.2
Series Solutions near Ordinary Points

Question - Solve x 2 y
′ – y = – x
– 1
The power series method (learned in the last section) would produce the
 following answer: y x
 n

1
 n
  x n
Something's wrong with this solution though.

Definition:
A function f ( x ) is analytic at a point x = c if there is a power series expansion of f ( x ) which converges on some interval that contains x = c.
Definition:
Given the differential equation y ″ + P ( x ) y ′ + Q ( x ) y = 0, we say that x = c is an ordinary point if both P ( x ) and Q ( x ) are analytic at x = c.
Otherwise, x = c is said to be a singular point.

Ex. 1 Identify the singular points for each of the following differential equations.
(a) x 2 y ″ + y ′ + x 5 y = 0

Ex. 1 Identify the singular points for each of the following differential equations.
(b) ( x 2 – 1) y ″ + sin( x ) y ′ + x 3 y = 0

Ex. 1 Identify the singular points for each of the following differential equations.
(c) x 2 y ″ + x 3 y ′ + x 5 y = 0

Ex. 1 Identify the singular points for each of the following differential equations.
(d) x y ″ + sin( x ) y ′ + 3 x 10 y = 0

Ex. 1 Identify the singular points for each of the following differential equations.
(e) y ″ + x 1/2 y = 0

Theorem:
Suppose that x = c is an ordinary point of the differential equation y
″ +
P ( x ) y
′ +
Q ( x ) y = 0. Then this differential equation has two linearly independent solutions: y
1
 n



0 a n
   n y
2


 n

0
  n
The radius of convergence of any such series solution is at least as large as the distance from c to the nearest (real or complex) singular point.

Ex. 2 Determine if the theorem guarantees a power series (in powers of x ) solution for each of the following. If a power series solution is guaranteed then give the guaranteed radius of convergence.
(a) x 2 y
″ + y
′ + x 5 y = 0

Ex. 2 Determine if the theorem guarantees a power series (in powers of x ) solution for each of the following. If a power series solution is guaranteed then give the guaranteed radius of convergence.
(b) ( x 2 – x
– 6) y
″ + sin( x ) y
′ + y = 0

Ex. 2 Determine if the theorem guarantees a power series (in powers of x ) solution for each of the following. If a power series solution is guaranteed then give the guaranteed radius of convergence.
(c) ( x 2 + 64) y
″ + y
′ + x 5 y = 0

Ex. 2 Determine if the theorem guarantees a power series (in powers of x ) solution for each of the following. If a power series solution is guaranteed then give the guaranteed radius of convergence.
(d) ( x 2 – 6 x + 58) y
″ + 3 x 10 y = 0

y
 c
1

 n

0 a x n c
2 n



0 b x equation. If initial conditions of y (0) = 5429 and y
′(0) = 1171 are given, what are the values of c
1 and c
2
?

y
 c
1 n



0 a n
 x
  n  c
2

 n

0
   n
Suppose is the general solution to a differential equation. If initial conditions of y (44) = 5429 and y
′(44) = 1171 are given, what are the values of c
1 and c
2
?

Ex. 3
Solve the following initial value problem: xy
″ + y
′ + y = 0, y (3) = 4, y
′(3) = 5.
(Actually let's just set this one up as if we were going to solve it with power series, but let's not actually go through all the steps of solving it!)

Ex. 4 Solve the following initial value problem: y
″ + ( x
– 1) y
′ + y = 0, y (1) = 2, y
′(1) = 0.

Ex. 4 Solve the following initial value problem: y
″ + ( x
– 1) y
′ + y = 0, y (1) = 2, y
′(1) = 0.