UNIT - V
SECOND ORDER DIFFERENTIAL
EQUATIONS
5.1. Solution of second order differential equations with constant
d2 y
dy
+b
+ cy = 0. Simple Problems
2
dx
dx
5.2 Solution of second order differential equations in the form
coefficients in the form a
d2 y
dy
+ cy = f ( x ) . W here a,b and c are constants and
dx
dx
mx
f(x) = e . Simple problems.
5.3. Solution of second order differential equations in the form
a
2
+b
d2 y
dy
+b
+ cy = f (x ). W here a,b and c are constants and f(x) =
dx
dx 2
sinmx or cosmx. Simple problems
a
5.1 SECOND ORDER DIFFERENTIAL EQUATIONS
Introduction:
In the last unit, we learnt first order differential equation. In this
unit, we will learn second order differential equation.
The second order differential equation is of the form
a
d2 y
dy
+b
+ cy = f (x ).
dx
dx 2
(1)
Where a,b and c are real numbers and f(x) is a function of x.
2
We use differential operators Dy, D y in (1), we get
(aD 2 + bD + c)y = f (x ) where D=
d
dx
(2)
Now, we put f(x) = 0 in (1), we get
a
d2 y
dx
2
+b
dy
+ cy = 0
dx
(3)
297
The solution of (3) is called complementary function (CF) of (1).
px
To solve (3), we assume a trial solution y = e for some value of
p. Then
dy
d2 y
= pe px and 2 = p2epx .
dx
dx
Substituting these values in (3), we get
ap2e px + bpe px + cepx = 0
Ÿ epx [ap2 + bp + c] = 0
(4)
Ÿ ap + bp + c = 0
2
This equation in p is called the Auxillary Equation (AE)
Solving (4), we get two roots say p1 and p2. Then the following
three cases arise.
Case (i)
If the roots p1 and p2 are real and distinct, then the solution of (3) is
y = Ae p1x + Be p 2 x
Case (ii)
If the roots p1 and p2 are real and equal, then the solution of (3) is
y = ep1x ( Ax + B)
Case (iii)
If the roots p1 and p2 are complex say p1 = α + i β and p2 = α -iβ, then
the solution of (3) is
y = e αx [A cos βx + B sin βx ]
In all cases, A and B are arbitrary constants.
298
5.1 WORKED EXAMPLES
PART – A
1.
If roots of the auxillary equation are
3
1
, what is the solution
±i
2
2
of the differential equation?
Solution:
1
3
, β=
,
2
2
.
. . The solution of differential equation is
Here, the roots are complex and α =
1
x
y = e 2 [A cos
3
3
x + B sin
x]
2
2
2
2. Find the solution of (D –81) y = 0
Solution:
2
The auxillary equation is p –81 = 0
Ÿ (p+9) (p-9) = 0
Ÿ p1 = -9, p2 = 9
Here, the roots are real and distinct
.
. . The solution of differential equation is
-9x
y = Ae
3.
Solve
d2 y
dx 2
+Be 9 x
+ 64y = 0
Solution:
Given
d2 y
dx
2
+ 64y = 0 Ÿ (D 2 + 64)y = 0
2
The auxillary equation is p +64=0
Ÿ p = ± 8i
Here, the roots are complex,α=0 and β = 8
.
. . The solution is y = A cos8x + B sin 8x
299
4.
2
Solve (D -2D-3)y=0
Solution:
2
The auxillary equation is p -2p-3=0
Ÿ (p+1) (p-3) = 0
Ÿ p1 = -1, p2 = 3
Here, the roots are real and distinct
.
-x
3x
. . The solution is y = Ae + Be
5.
2
Solve (D -4D-1) y =0
Solution:
2
The auxillary equation is p -4p-1 = 0
Here a = 1, b = -4, c = -1
P=
− b ± b2 − 4ac
2a
=
4 ± 16 − 4(1)(−1)
2(1)
=
4 ± 20
2
=
4±2 5
=2± 5
2
So, p1 = 2 +
5 and p 2 = 2 − 5
Here, the roots are real and distinct
.
. . The solution is
(2 + 5 )x
(2 - 5
y = Ae
+ Be
)x
300
6.
Solve
d2 y
dy
−6
+ 9y = 0
dx
dx 2
Solution:
Given:
dy
d2 y
−6
+ 9y = 0 Ÿ (D 2 − 6D + 9) y = 0
2
dx
dx
2
The auxillary equation is p –6p+9 = 0
Ÿ (p-3)(p-3)=0
Ÿ p1=3, p2 = 3
Here, the roots are real and equal.
3x
.. The solution is y = e [Ax+B]
7.
2
Solve (D +D+2)y =0
Solution:
2
The auxillary equation is p +p+2 = 0
Here a = 1, b = 1, c = 2
P=
− b ± b2 − 4ac
2a
=
− 1 ± 1 − 4(1)(2)
2(1)
− 1± − 7
2
− 1± i 7
=
2
−1
7
=
±i
2
2
=
1
7
Here, the roots are complex, α = − , β =
2
2
−1
.
x
. . The solution is y = e 2 [A cos
301
7
7
x
x + B sin
2
2
PART – B
1.
2
Solve (D +1) y = 0 when x = 0, y = 2 and x =
π
, y=-2.
2
Solution:
2
The auxillary equation is p + 1 = 0
Ÿ p =± i
Here, the roots are complex, β = 1
.
. . The solution is
y = A cosx + B sinx
…1
When x=0, y=2, the equation (1) becomes
A cos0 + B sin0 = 2
A +0 =2
A =2
When x =
π
, y=-2, the equation (1) becomes
2
π
π
+B sin
= -2
2
2
0 + B = -2
B = -2
.
. . The required solution is
y = 2 cosx –2 sinx
A cos
2.
2
Show that the solution of the equation (D + 3D + 2) y = 0 if y(0)
1
-x
-2x
= 1 and y (0) = 0 is y = 2e –e
Solution:
2
The auxillary equation is p +3p+2=0
Ÿ (p+1) (p+2) = 0
Ÿ p1 = -1, p2 = -2
Here, the roots are real and distinct
302
.
-x
-2x
. . The solution is y = Ae + Be
-x
-2x
Now, y′ = -Ae –2Be
…1
…2
If y(0) = 1, the equation (1) becomes
A +B =1
…3
If y’(0) =0 , the equation (2) becomes
A+2B=0
…4
Solving (3) and (4) we get A=2, B=-1
∴The required solution is
-x
-2x
y=2e -e
5.2. SOLUTION OF SECOND ORDER EQUATIONS IN THE FORM
d2 y
dy
+b
+ cy = f( x ) WHERE A,B AND C ARE CONSTANTS
dx
dx 2
AND f(x) = emx .
a
Introduction:
In previous section, we find the complementary function . In this
section, we have to find the particular integral (PI) and the general
solution of a second order differential equation.
The Solution of Differential equation with Constant Coefficients is
y=CF+PI
Method of finding particular integral
2
Consider (aD +bD+c)y = e
mx
where m is a constant.
2
Let f(D) = aD +bD+c
Then PI is given by
1 mx emx
e =
f(D)
f (m)
Three cases arise in PI
Case (i)
If f(m) ≠ 0 then PI =
1 mx emx
e =
f (D)
f(m)
303
Case (ii)
If f(m) = 0 and f '(m) ≠ 0 then
PI =
x emx
f '(m)
Case (iii)
If f(m) = 0 and f '(m) = 0 and f ''(m) ≠ 0 then
PI =
x 2emx
f "(m)
5.2 WORKED EXAMPLE
PART – A
1.
2
Find the complementary function of (D +16)y= e
x
Solution:
2
2.
The auxiliary equation is p +16=0 Ÿp=±4i
Here, the roots are complex , β=4
∴CF = A cos 4x + B sin 4x
2
40x
Find the complementary function of (D -60D+800)y=e
Solution:
The auxiliary equation is
3.
2
p -60p+800=0
Ÿ(p-40) (P-20) =0
ŸP 1=40, P 2=20
Here the roots are real and distinct
40x
20x
∴CF = Ae + Be
2
Find the particular integral of (D +1) y =1
Solution:
1
1
=
e0
D2 + 1 D2 + 1
1
1
=
= =1
0 +1
1
PI =
304
2
-x
4. Find the particular integral of (D +7D+14) = 8e
Solution:
1
PI = 2
8e − x
D + 7D + 14
8e − x
8e − x
=
= e− x
8
(−1)2 + 7(−1) + 14
2
-x
5. Find the particular integral of (D -2D-3)y = e
Solution:
1
PI = 2
e− x
D − 2D − 3
=
=
x e− x
2D − 2
=
x e−x
x e−x
=−
2(−1) − 2
4
Since f (−1) = 0
PART - B
2
1. Solve (D +5D+6)y=30
Solution:
The auxiliary equation is
2
p +5p+6=0
Ÿ(p+2) (P+3) =0
ŸP 1=-2, P 2=-3
Here, the roots are real and distinct
-2x
-3x
∴CF = Ae +Be
1
Now PI = 2
30
D + 5D + 6
30e°
= 2
D + 5D + 6
30e°
= 2
0 + 5(0) + 6
30
=
6
PI = 5
∴ The Required solution is
-2x
-3x
Y=CF+PI = Ae +Be +5
305
2.
2
Solve (D +6D+5) y =2e
x
Solution:
2
The auxiliary equation is
p +6p+5=0
Ÿ(p+1) (P+5) =0
ŸP 1=-1, P 2=-5
Here the roots are real and distinct
-x
-5x
∴CF = Ae +Be
1
Now PI = 2
2e x
D + 6D + 5
=
2e x
12 + 6(1) + 5
2e x
12
ex
PI =
6
∴The required solution is
Y=CF+PI
=
Ae -x + Be -5x +
ex
6
x
3.
Solve (D2 + D)y = e 2
Solution:
2
The auxiliary equation is
p +p=0
Ÿp (p+1)=0
ŸP 1=0, P 2=-1
Here the roots are real and distinct
∴CF = Ae +Be =A+Be
0
Now PI =
-x
1
D2 + D
-x
x
2
e
306
=
=
x
2
e
2
1
§ 1·
¨ ¸ +
2
2
© ¹
x
2
e
3
4
x
4
PI = e 2
3
∴The required solution is
y=CF+PI
x
3
= A +Be -x + e 2
4
2
4. Solve (D − D − 12)y = e 4 x
Solution:
2
The auxiliary equation is p -p-12=0
Ÿ(p-4) (p+3)=0
Ÿp1=4, p2=-3
Here the roots are real and distinct
4x
-3x
∴CF = Ae +Be
1
e4 x
Now PI = 2
D − D − 12
x e4x
=
Since f(4) = 0
2D − 1
x e4 x
=
2(4) − 1
x e4 x
7
∴ The required solution is
y=CF + PI
PI =
= Ae 4x +Be -3x +
x e 4x
7
307
5.
2
Solve (D -2D+1) y =e
x
Solution:
The auxiliary equation is
2
p -2p+1=0
Ÿ(p-1) (p-1) =0
Ÿp1=1, p 2=1
Here the roots are real and equal
∴CF = e (Ax+B)
x
Now PI =
PI =
1
D − 2D + 1
2
ex
x2 x
e
2
Since f (1) = 0, f '(1) = 0
∴The required solution is
Y=CF+PI
x2 x
e
2
= e x (Ax + B) +
6
Solve
d2 y
dx
2
− 13
dy
+ 12y = 2e − 2x + 5e x
dx
Solution:
Given
d2 y
dx 2
− 13
dy
+ 12y = 2e − 2 x + 5e x
dx
Ÿ (D 2 − 13D + 12) y = 2e −2 x + 5e x
The auxilary equation is
2
p -13p+12=0
Ÿ(p-1) (p-12) =0
Ÿp1=1, p2=12
Here the roots are real and distinct
x
12x
∴CF = Ae +Be
308
Now PI1 =
=
1
D − 13D + 12
2
2e − 2x
(−2)2 − 13(−2) + 12
=
2e − 2x
4 + 26 + 12
=
e−2 x
21
Now PI2 =
2 e − 2x
1
D − 13D + 12
2
=
5xex
2D − 13
=
5xex
2(1) − 13
=−
5 ex
Sincef (1)= 0
5xex
11
∴ The required solution is
Y =CF+PI1+PI2
= Ae x + Be12x +
e −2 x 5 xe x
−
21
11
5.3 SOLUTION OF SECOND ORDER DIFFERENTIALEQUATIONS
d2 y
dy
+ cy = f( x ) WHERE a,b AND c ARE
dx
dx
CONSTANTS AND f(x) = sin mx or cos mx where m is a
constant ≠ 0
INTRODUCTION
IN THE FORM a
2
−b
In this section, we have to find the particular integral when f(x)
=sin mx or cos m x where m is a constant
Methods of finding PI
Consider f(x) =sin m x
309
Case (i)
2
2
2
2
Express f(D) as function of D ,say φ (D ) and then replace D with –m
2
If φ(-m )≠0,then
1
PI =
sin mx
f (D)
1
sin mx
=
φ(D 2 )
1
PI =
sin mx
φ(−m2 )
Case (ii)
2
Sometimes we cannot form φ (D ) Then we shall try to get
2
2
φ(D,D ) that is a function of D and D . In such cases we proceed as
follows.
For Example
1
Now PI = 2
sin 2x
D + 2D + 3
1
=
sin 2x Re place D 2 by − 22
2
− 2 + 2D + 3
1
=
Sin 2x
2D − 1
2D + 1
=
sin 2x multiply and divide by 2D + 1
4D 2 − 1
2D(sin 2x ) + sin 2x
=
4(−22 ) − 1
4 cos 2x + sin 2x
=
− 17
1
[4 cos 2x + sin 2x ]
=
− 17
Now consider f(x) = cos m x
1
Case (i):
PI =
cos m x
φ(-m2 )
Case(ii):
Same as sin m x method
General Solution:
The general solution is y= CF+PI
310
5.3 WORKED EXAMPLE
PART - A
2
1. Find the complementary function of (D +49) y= cos 4x
Solution:
2
The auxiliary equation is p +49=0
Ÿp=±7i
Here, the roots are complex ,β =7
∴ CF = A cos 7x+B sin 7x
2
2. Find the particular integral of (D +14) y = sin 3x
Solution:
1
PI = 2
sin 3 x
D + 14
1
=
sin 3 x
2
− 3 + 14
sin 3 x
=
5
3.
2
2
Find the particular integral of (D +a ) y = Cos b x
Solution:
1
cos bx
D + a2
1
cos bx
=
2
− b + a2
cos bx
= 2
a − b2
PI =
1.)
2
(
)
PART - B
Solve D 2 − 4 y = sin 2x
Solution:
The auxiliary equation is p2 − 4 = 0
Ÿ p2 = 4
Ÿ p = +2
Ÿ p1 = 2, p2 = −2
311
Here, the roots are real and distinct
∴ CF = Ae2x + Be−2 x
(sin 2x )
D −4
1
=
sin 2x
2
−2 −4
sin 2x
=−
8
Now PI =
1
2
∴ The Required solution is
y = CF + PI
sin 2X
= Ae 2 x + Be − 2x −
8
2.)
Solve D 2 y = −16 sin 4x
Solution:
The auxiliary equation is p2 = 0
Ÿ p, = 0,p2 = 0
Here, the roots are real and equal
∴ CF = e0 (Ax + B) = Ax + B
Now PI =
=
1
− 16 sin 4x
D2
1
− 16 sin 4x
− 42
PI= Sin4x
∴ The Required solution is
y = CF + PI
= Ax + B + Sin4x
312
3.)
Solve
d2 y
+ 16 y = cos 2 x
dx 2
Solution:
d2 y
+ 16 y = cos 2 x
dx 2
Ÿ D 2 + 16 y = cos 2 x
1 cos 2x
Ÿ D 2 + 16 y = +
2
2
1 0 1
= e + cos 2x
2
2
The auxiliary equation is p2 + 16 = 0
Ÿ p = + 4i
Here, the roots are complex, β = 4
∴ CF = A cos 4x + BSin4x
Given
(
(
)
)
1 0
e
PI1 = 22
D + 16
1 e0
2 0 + 16
1
=
32
1 cos 2x
PI 2 = . 2
2 D + 16
1
cos 2x
.
=
2 − 22 + 16
cos 2x
=
24
=
∴ The Required solution is
y = CF + PI
= A cos 4x + BSin4x +
1 cos 2x
+
32
24
313
4.)
(
)
Solve D 2 + 3D + 2 y = sin 2x
Solution:
The auxiliary equation is p2 + 3p + 2 = 0
Ÿ (p + 2)(p + 1) = 0
Ÿ p1 = −2, p2 = −1
Here, the roots are real and distinct
∴ CF = Ae −2 x + Be − x
Now, PI =
=
1
D + 3D + 2
2
.Sin2x
1
.Sin2x
− 2 + 3D + 2
1
.Sin2x
=
3D − 2
3D + 2
.Sin2x
=
9D 2 − 4
2
3D + 2
.Sin2x
− 36 − 4
3D(sin 2x ) + 2 sin 2x
=
− 40
6 cos 2x + 2 sin 2x
=
− 40
−1
[3 cos 2x + sin 2x ]
=
20
=
∴ The Required solution is
y = CF + PI
1
= Ae − 2 x + Be − x −
[3 cos 2x + sin 2x ]
20
314
5.)
(
)
Solve D 2 − 2D − 8 y = 4 cos 3 x
Solution:
Solution: The auxiliary equation is p2 − 2p − 8 = 0
Ÿ (p − 4)(p + 2) = 0
Ÿ p1 = 4, p2 = −2
Here, the roots are real and distinct
∴ CF = Ae 4 x + Be −2 x
Now, PI =
=
1
4 cos 3x
D 2 − 2D − 8
1
4 cos 3x
− 3 − 2D − 8
1
=
4 cos 3x
− 2D − 17
ª 1
º
= −4 «
4 cos 3x »
+
2
D
17
¬
¼
2
ª 2D − 17
º
= −4 « 2
cos 3 x »
¬ 4D − 289
¼
ª 2D(cos 3x ) − 17 cos 3 x º
= −4 «
»
− 325
¬
¼
ª − 6 sin 3 x − 17 cos 3 x º
= −4 «
»
− 325
¬
¼
−4
[6 sin 3x + 17 cos 3x ]
=
325
∴ The Required solution is
y = CF + PI
= Ae 4 x + Be − 2 x −
4
[6sin3x + 17cos3x
325
315
]
EXERCISE
PART - A
1.)
If roots of the auxilary equation are 2,7 what is the solution of the
differential equation?
2.)
If roots of the auxilary equation are 0,1 what is the solution of the
differential equation?
3.)
If roots of the auxilary equation are -2,± i, what is the solution of
the differential equation?
4.)
Find the solution of D 2 − 1 y = 0
5.)
Find the solution of
6.)
Solve D 2 + 9 y = 0
7.)
Find the solution of D 2 + 100 y = 0
(
(
)
(
9.) Solve (3D
10.) Solve (3D
8.)
)
d2 y
dx 2
(
− 16 y = 0
)
)
Solve D 2 + 4D − 1020 y = 0
11.) Solve
2
2
d2 y
dx
(
13.) Solve (D
)
− 7D − 6)y = 0
− 5D + 2 y = 0
+
2
dy
=0
dx
)
12.) Solve D 2 − D − 1 y = 0
14.) Solve
2
d2 y
dx
2
(
16.) Solve (3D
)
+ 4D + 4 y = 0
− 12
dy
+ 36 y = 0
dx
)
15.) Solve D 2 + D + 1 y = 0
2
)
− D + 1y = 0
(
)
17.) Find the Complementary function of D 2 + 13D − 90 y = e x
316
(
19.) Find the Particular integral of (D
20.) Find the Particular integral of (D
21.) Find the Particular integral of (D
)
18.) Find the Particular integral of D 2 − 3D + 2 y = e − x
)
2
+ D + 4 y = 10e 2x
2
− 8D + 15 y = e3 x
2
+ 10D + 25 y = e −5 x
)
(
)
)
22.) Find the Complementary integral of D 2 + 25 y = cos ax
(
24.) Find the Particular integral of (D
)
+ 10 )y = sin 3x
23.) Find the Particular integral of D 2 + 25 y = Sinx
25.) Find the Particular integral of
(
2
d2 y
dx 2
− 4y = cos 4x
PART - B
)
y(0) = 2 and y1(0 ) = 12
1.)
Solve D 2 + 36 y = 0 when
2.)
Solve
3.)
Solve D 2 − 2D − 15 y = 0 given that
d2 y
dx
2
+ y = 0 given that
(
)
x=0
(
5.) Solve (D
6.) Solve (D
7.) Solve (D
8.) Solve (D
9.) Solve (D
10.) Solve (D
4.)
dy
= 2 and y=1 when x=0
dx
)
d2 y
dy
= 0 and
= 2 when
dx
dx 2
Solve D 2 − D − 20 y = 0 given that y=5 and
)
2
+ 7D + 12 y = 3
2
+ 3D + 2 y = 2e x
2
+ 12D + 36 y = e x
2
+D + 4 y = e
2
2
)
)
)
x
2
)
+ 6D + 8 )y = e
− 3D + 2 y = e 2x
−4 x
317
dy
= −2 when x=0
dx
11.) Solve
d2 y
dx 2
(
13.) Solve (D
14.) Solve (D
−4
dy
+ 4y = e2x
dx
)
+ 14D + 49)y = 4e
− 2D + 4)y = 5 + 3e
12.) Solve D 2 + 2aD + a 2 y = e −ax
15.) Solve
2
2
d2 y
dx 2
(
17.) Solve (D
18.) Solve (D
19.) Solve (D
−7 x
+8
−x
dy
+ 15 y = e − 3 x + e3 x
dx
)
16.) Solve D 2 + 10D + 25 y = e5 x + e −5 x
20.) Solve
2
2
2
d2 y
dx 2
(
22.) Solve (D
23.) Solve (D
24.) Solve (D
25.) Solve (D
)
− 25)y = sin 5x
+ 100)y = cos 2x
+ 16 y = sin 9x
− 2y = cos 3x
)
21.) Solve D 2 + 2D − 3 y = sin x
)
2
+ D − 2 y = Sin3x
2
+ 4D + 13 y = 4 cos 3x
2
2
)
)
− 2D − 8)y = 4 cos 2x
− 8D + 9 y = 8 cos 5x
ANSWERS
PART - A
1.)
y = Ae 2 x + Be 7 x
2.) y = A + Be x
3.)
y = e −2x [A cos x + B sin x ]
4.) y = Ae x + Be − x
5.)
y = Ae 4 x + Be −4 x
6.) y = A cos 3x + B sin 3 x
7.)
y = A cos 10 x + B sin 10 x
8.) y = Ae30 x + Be −34 x
318
9.)
y = Ae x + Be
11.) y = A + Be
2 x
3
10.) y = Ae 3 x + Be
−x
12.) y =
−x
16.) y = e
x
§
¨
©
2 ¨ A cos
3
3 ·¸
x + B sin
x
2
2 ¸¹
§
11
11 ·¸
cos
x + B sin
x
¨
6
6 ¸¹
©
6¨ A
17.) CF = Ae5 x + Be −18 x
20.) −
§ 1− 5 ·
¨
¸x
¨ 2 ¸
¹
+ Be ©
14.) y = e 6 x (Ax + B )
13.) y = e −2 x (Ax + B)
15.) y = e
§ 1+ 5 ·
¨
¸x
¨ 2 ¸
¹
Ae ©
−2 x
3
18.)
e−x
6
19.) e2x
x 2 −5 x
e
2
sin x
23.)
24
xe3 x
2
21.)
22.) CF=Acos5x+Bsin5x
25.) −
24.) Sin3x
Cos4x
20
Part - B
1.)
y = 2 cos 6x + 2 sin 6x
3.)
y=
5.)
y = Ae − 4x + Be − 3 x +
7.)
y = e − 6 x (Ax + B) +
8.)
y=e
2.) y = cos x + 2 sin x
1 5x 1 −3x
e + e
20
2
−x
2
4.) y = 2e5 x + 3e −4x
1
4
6.) y = Ae − x + Be − 2 x +
ex
49
ª
15
15 º 4 x 2
x + B sin
x» +
e 9.)
« A cos
2
2 »¼ 19
«¬
y = Ae x + Be2x + xe2 x
319
ex
3
10.) y = Ae − 4 x + Be − 2 x −
11.) y = e2 x (Ax + B ) +
xe−4 x
2
x 2 2x
e
2
12.) y = e − ax (Ax + B ) +
x 2 − ax
e
2
13.) y = e−7 x (Ax + B) + 2x 2e −7 x
(
)
14.) y = e x A cos 3 x + B sin 3 x +
15.) y = Ae − 3 x + Be − 5 x +
5 3 −x
+ e
4 7
xe−3 x e3 x
+
2
48
e5 x x 2e −5 x
+
100
2
sin 9x
y = A cos 4x + B sin 4x −
65
sin
5
x
y = Ae5 x + Be − 5 x −
50
cos 2x
y = A cos 10x + B sin10x +
96
cos 3x
y = Ae 2x + Be − 2x −
11
1
y = Ae − 3 x + Be x −
(cos x + 2 sin x )
10
1
y = Ae x + Be − 2 x −
(3 cos 3x + 11sin 3x )
130
1
y = e − 2x (A cos 3x + B sin 3 x ) +
3 sin 3x + cos 3 x
10
1
y = Ae (4 + 7 )x + Be (4 − 7 )x −
(5 sin 5x + 2 cos 5x )
29
1
y = Ae 4x + Be − 2x −
(sin 2x + 3 cos 2x )
10
16.) y = e − 5 x (Ax + B ) +
17.)
18.)
19.)
20.)
21.)
22.)
23.)
24.)
25.)
320