SECOND ORDER DIFFERENTIAL EQUATIONS

UNIT - V
SECOND ORDER DIFFERENTIAL
EQUATIONS
5.1. Solutionof secondorderdifferentialequationswithconstant 2
d y dy
coefficients in the form a + b + cy = 0.
S imple P roblems
2
dx dx
5.2 Solutionofsecondorderdifferentialequationsintheform 2
d y dy
a + b + cy = f(
x)
. Where a,b and c are constants and
2
dx dx
f(x ) = e mx . Simple problems.
5.3. Solutionofsecondorderdifferentialequationsintheform 2
d y dy
a + b + cy = f(
x).
W here a,b and c are constants a nd f(x ) =
2
dx dx
sinmx or cosmx. Simple problems
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
2
d y dy
a + b + cy = f(
x).
(1)
2
dx dx
W here a,b and c are real numbers and f(x ) is a function of x .
W e use differential operators D y, D 2 y in (1), w e get
2
d
( aD + bD + c)
y = f(
x)
where D=
dx
Now, w e put f(x ) = 0 in (1), w e get
2
d y dy
a + b + cy = 0
(3)
2
dx dx
297
(2)

The solution of (3) is called complementary function (CF) of (1).
T o solv e (3), we assume a trial solution y = e px for some value of
2
dy px d y 2 px
p. Then = pe and = p e .
dx
2
dx
S ubstituting these v alues in (3), we get
2
px
px
ap e + bpe + ce
px 2
e [ ap + bp + c]
= 0
2
ap + bp + c = 0
px
= 0
T his equation in p is called the A ux illary E quation (A E )
Solving (4), we get two roots say p1 and p2. Then the following
three cases arise.
Case (i)
I f the roots p1 and p2 are real a nd distinct, then the solution of (3) is
Case (ii)
y = Ae +
p1x p2x
Be
I f the roots p1 and p2 are real and equal, then the solution of (3) is
Case (iii)
p1 x
y = e ( Ax + B)
If the roots 1 pand 2 p arecomplexsayp1 = α + i β and p2 = α -iβ, then
the solution of (3) is
y
x
= α
e
[ A cos βx
+ Bsinβx]
I n a ll cases, A a nd B are arbitrary consta nts.
298
(4)

5.1 WORKED EXAMPLES
PART – A
1. If roots of the auxillary equation are
of the differential equation?
1
± i
2
3
, what is the solution
2
Solution:
1
H ere, the roots are complex and α = ,
2
.
β =
3
,
2
. . The solution of differential equation is
1
x
2
y = e [ A cos
3
x + Bsin
2
3
x]
2
2. Find the solution of (D 2 – 81) y = 0
Solution:
The auxillary equation is p 2 –81 = 0
(p+9) (p-9) = 0
p1 = -9, p2 = 9
H ere, the roots are rea l a nd distinct
. . . The solution of differential equation is
y =Ae -9x +Be x 9
2
d y
3. Solve + 64y
= 0
2
dx
Solution:
2
d y
2
Given + 64y
= 0 ( D + 64)
y = 0
2
dx
The auxillary equation is p 2 +64=0
p = ± 8i
H ere, the roots are complex ,α=0 and β = 8
. . . T he solution is y = A cos 8x + B sin 8x
299

4. Solve (D 2 -2D-3)y=0
Solution:
The auxillary equation is p 2 -2p-3=0
(p+1) (p-3) = 0
p1 = -1, p2 = 3
H ere, the roots are rea l a nd distinct
. . . The solution is y = Ae -x + Be 3x
5. Solve (D 2 -4D-1) y =0
Solution:
The auxillary equation is p 2 -4p-1 = 0
H ere a = 1, b = -4, c = -1
P =
=
=
− b ±
b
2a
2 −
2(
1)
4ac
4 ± 16 − 4(
1)(
−1)
4 ± 20
2
4 ± 2 5
= = 2 ± 5
2
So, p1 = 2 + 5andp2= 2 − 5
H ere, the roots are rea l a nd distinct
. . . The solution is
y =Ae (2 + 5 )x + Be
(2 - 5 )x
300

2
d y dy
6. Solve − 6 + 9y
= 0
2
dx dx
Solution:
2
Given:
d y dy
− 6 + 9y
= 0
2
dx dx
2
( D − 6D
+ 9)
y = 0
The auxillary equation is p 2 –6p+9 = 0
(p-3)(p-3)=0
p1=3, p2 = 3
Here, the roots are real and equal.
.. The solution is y = e 3x [Ax+B]
7. Solve (D 2 +D+2)y =0
Solution:
The auxillary equation is p 2 +p+2 = 0
Here a = 1, b = 1, c = 2
2
− b ± b − 4ac
P =
2a
−1±
1−
4(
1)(
2)
=
2(
1)
−1±
− 7
=
2
−1±
i 7
=
2
−1
7
= ± i
2 2
1
H ere, the roots are complex , α = − , β =
2
−1
. . 7 7
. The solution is y = e [ A cos x Bsin
x
2 2
x
2
+
301
7
2

PART – B
1. Solve (D 2 π
+1) y = 0 when x = 0, y = 2 and x = , y=-2.
2
Solution:
The auxillary equation is p 2 + 1 = 0
p = ± i
H ere, 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
π
When x = , y=-2, the equation (1) becomes
2
π π
A cos +B sin = -2
2 2
0 + B = -2
B = -2
. . . The required solution is
y = 2 cosx – 2 sinx
2. Show that the solution of the equation (D 2 + 3D + 2) y = 0 if y(0)
= 1 and y 1 (0) = 0 is y = 2e -x –e -2x
Solution:
The auxillary equation is p 2 +3p+2=0
Here, the roots are real and distinct
(p+1) (p+2) = 0
p1 = -1, p2 = -2
302

. . . The solution is y = Ae -x + Be -2x
… 1
Now, y′ = -A e -x –2Be -2x
… 2
If y(0) = 1, the equation (1) becomes
A +B =1 … 3
I f y’(0) =0 , the equation (2) becomes
A+2B=0 … 4
S olv ing (3) and (4) we get A =2, B=-1
∴The required solution is
y=2e -x -e -2x
5.2. SOLUTION OF SECOND ORDER EQUATIONS IN THE FORM
2
d y dy
a + b + cy = f(
x)
WHERE A,B AND C ARE CONSTANTS
2
dx dx
mx
AND f(x) = e .
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 differentia l equation.
The Solution of Differential equation with Constant Coefficients is
y=CF+PI
Method of finding particular integral
Consider (aD 2 +bD+c)y = e mx where m is a constant.
L et f(D ) = aD 2 +bD+c
1
T hen P I is giv en by
f(
D)
Three cases arise in PI
Case (i)
If f(m) ≠
0
then PI
=
1
f(
D)
e
mx =
e
mx =
303
mx
e
f(
m)
mx
e
f(
m)

Case (ii)
Case (iii)
If f(m) = 0
If f(m) = 0
and
f '(m)
and f '(m)
≠ 0then
= 0
and
304
PI =
f ''(m)
mx
x e
f'(
m)
≠ 0 then
5.2 WORKED EXAMPLE
PART – A
PI =
2
mx
x e
f"
( m)
1. Find the complementary function of (D 2 +16)y= e x
Solution:
The auxiliary equation is p 2 +16=0 p=±4i
Here, the roots are complex , β=4
∴C F = A cos 4x + B sin 4x
2. F ind the complementary function of (D 2 -60D+800)y=e 40x
Solution:
The auxiliary equation is p 2 -60p+800=0
(p-40) (P-20) =0
P1=40, P 2=20
H ere the roots are rea l a nd distinct
∴CF = Ae 40x + Be 20x
3. F ind the particular integra l of (D 2 +1) y =1
Solution:
1 1
PI =
= e
2 2
D + 1 D + 1
1 1
= = = 1
0 + 1 1
0

4. F ind the particular integra l of (D 2 +7D+14) = 8e -x
Solution:
1
−x
PI =
8e
2
D + 7D
+ 14
−x
8e
=
2
( −1)
+ 7(
−1)
+ 14
=
−x
8e
8
−x
= e
5. F ind the particular integra l of (D 2 -2D-3)y = e -x
Solution:
1 −x
PI =
e
2
D − 2D
− 3
−x
x e
=
2D
− 2
Since f(
−1)
= 0
−x
x e
=
2(
−1)
− 2
−x
x e
= −
4
PART - B
1. Solve (D 2 +5D+6)y=30
Solution:
The auxiliary equation is p 2 +5p+6=0
(p+2) (P+3) =0
P1=-2, P 2=-3
Here, the roots are real and distinct
∴CF = Ae -2x +Be -3x
1
Now PI =
30
2
D + 5D
+ 6
30e°
=
2
D + 5D
+ 6
30e°
=
2
0 + 5(
0)
+ 6
30
=
6
PI = 5
∴ T he R equired solution is
Y=CF+PI =Ae -2x +Be -3x +5
305

2. Solve (D 2 +6D +5) y =2e x
Solution:
The auxiliary equation is p 2 +6p+5=0
(p+1) (P+5) =0
P1=-1, P 2=-5
H ere the roots are rea l a nd distinct
∴CF = Ae -x +Be -5x
1
x
Now PI =
2e
2
D + 6D
+ 5
x
2e
=
2
1 + 6(
1)
+ 5
x
2e
=
12
x
e
PI =
6
∴The required solution is
Y=CF+PI
x
-x
-5x
e
Ae + Be +
6
3. Solve
x
2
(D + D)
y = e 2
Solution:
The auxiliary equation is p 2 +p=0
p (p+1)=0
P1=0, P 2=-1
H ere the roots are rea l a nd distinct
∴CF = Ae 0 +Be -x =A +Be -x
x
2
1
Now PI =
e
2
D + D
306

x
2
e
=
2
1
+
2
e
=
x
2
3
4
x
2
1
2
4
PI = e
3
∴The required solution is
y=CF+PI
-x
3
= A + Be + e
4
2
4x
4. Solve (D − D − 12)
y = e
Solution:
The auxiliary equation is p 2 -p-12=0
(p-4) (p+3)=0
p1=4, p2=-3
H ere the roots are rea l a nd distinct
∴CF = Ae 4x +Be -3x
1 4x
Now PI =
e
2
D − D − 12
4x
x e
=
2D
− 1
4x
x e
=
2(
4)
−1
4x
x
2
x e
PI =
7
∴ T he required solution is
y=CF + PI
= Ae
4x
+ Be
-3x
x e
+
7
4x
Since f(
4)
307
= 0

5. Solve (D 2 -2D+1) y =e x
Solution:
6
The auxiliary equation is p 2 -2p+1=0
(p-1) (p-1) =0
p1=1, p 2=1
H ere the roots are rea l a nd equa l
∴C F = e x (A x +B)
Now PI
=
D
2
2
x
PI =
2
1
e
− 2D
+ 1
∴The required solution is
Y=CF+PI
Solution:
e
x x
= e +
2
x
( Ax + B)
e
2
x
2
x
Since
308
f ( 1)
d y dy
−2x
Solve −13
+ 12y
= 2e
+ 5e
2
dx dx
Given
2
2
d y dy
−13
+ 12y
= 2e
2
dx dx
−2x
−2x
( D −13D
+ 12)
y = 2e
+ 5e
x
+ 5e
T he aux ilary equa tion is p 2 -13p+12=0
x
=
x
0,
f'(
1)
(p-1) (p-12) =0
p1=1, p2=12
H ere the roots are rea l a nd distinct
∴CF = Ae x +Be 12x
= 0

1
−2x
Now PI1
=
2e
2
D − 13D
+ 12
−2x
2e
=
2
( −2)
− 13(
−2)
+ 12
−2x
2e
=
4 + 26 + 12
−2x
e
=
21
1
x
Now PI2
=
5e
2
D − 13D
+ 12
x
5xe
=
Sincef(
1)
= 0
2D
− 13
x
5xe
=
2(
1)
− 13
x
5xe
= −
11
∴ T he required solution is
Y =CF+PI1+PI2
−2x
x
x 12x e 5xe
= Ae + Be + −
21 11
5.3 SOLUTION OF SECOND ORDER DIFFERENTIALEQUATIONS
2
d y dy
IN THE FORM a −b + cy = f(
x)
WHERE a,b AND c ARE
2
dx dx
CONSTANTS AND f(x) = sin mx or cos mx where m is a
constant ≠ 0
INTRODUCTION
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
C onsider f(x ) =sin m x
309

Case (i)
Express f(D) as function of D 2 ,say φ (D 2 ) and then replace D 2 with –m 2
If φ(-m 2 )≠0,then
PI =
1
sin mx
f(
D)
1
= sin mx
2
φ(
D )
1
PI = sin mx
2
φ(
−m
)
Case (ii)
Sometimes we cannot form φ (D 2 ) Then we shall try to get
φ(D,D 2 ) that is a function of D and D 2 . In such cases we proceed as
follows.
For Example
1
Now PI =
sin2x
2
D + 2D
+ 3
1 2 2
=
sin2x
Replace
D by − 2
2
− 2 + 2D
+ 3
1
= Sin 2x
2D
−1
2D
+ 1
= sin2x
multiply anddivide
by 2D
+ 1
2
4D
−1
2D(sin
2x)
+ sin2x
=
2
4(
−2
) −1
4cos
2x
+ sin2x
=
−17
1
= [ 4cos
2x
+ sin2x]
−17
Now consider f(x) = cos m x
1
Case (i): PI = cos mx
2
φ(-m
)
Case(ii): S ame as sin m x method
General Solution:
The general solution is y= CF+PI
310

5.3 WORKED EXAMPLE
PART - A
1. Find the complementary function of (D 2 +49) y= cos 4x
Solution:
The auxiliary equation is p 2 +49=0
p=±7i
Here, the roots are complex ,β =7
∴ C F = A cos 7x +B sin 7x
2. Find the particular integral of (D 2 +14) y = sin 3x
Solution:
1
PI = sin 3x
2
D + 14
1
= sin 3x
2
− 3 + 14
sin 3x
=
5
3. F ind the particular integra l of (D 2 +a 2 ) y = Cos b x
Solution:
1
PI =
2 2
D + a
cos bx
1
=
2 2
− b + a
cos bx
=
2 2
a − b
cos bx
1.) Solve ( − 4)
y = sin2x
Solution:
D 2
PART - B
The auxiliary equation is − 4 = 0
p
2
p 2
= 4
p = + 2
p = 2,
p = −2
1
2
311

Here, the roots are real and distinct
∴ CF = Ae
2x
+ Be
−2x
Now PI = ( sin2x
)
D
1
2 −
4
1
= sin2x
2
− 2 − 4
sin2x
= −
8
∴ T he R equired solution is
y = CF + PI
= Ae
2x
+ Be
−2x
sin2X
−
8
2.) Solve y = −16sin4x
Solution:
D 2
The auxiliary equation is p 0
2 =
p, 0,
p2
0 = =
Here, the roots are real and equal
∴ CF = e
0
( Ax + B)
= Ax + B
1
Now PI 16 sin4x
2
D
− =
1
16 sin4x
2
4
− =
−
PI= Sin4x
∴ T he R equired solution is
y =
CF + PI
= Ax + B + Sin4x
312

2
d y
2
3.) Solve + 16y
= cos x
2
dx
Solution:
2
d y
2
Given + 16y
= cos x
2
dx
2
2
( D + 16)
y = cos x
2 1 cos 2x
( D + 16)
y = +
2 2
1 0 1
= e + cos 2x
2 2
The auxiliary equation is p 16 0
2
+ =
p = + 4i
Here, the roots are complex, β = 4
∴CF
= A cos 4x
+ BSin4x
1 0
e
PI 2
1 =
2
D + 16
0
1 e
=
2 0 + 16
1
=
32
1 cos 2x
PI 2 = .
2 2 D + 16
1 cos 2x
= .
2 2 − 2 + 16
cos 2x
=
24
∴ T he R equired solution is
y =
CF + PI
1 cos 2x
= A cos 4x
+ BSin4x
+ +
32 24
313

4.) Solve ( + 3D
+ 2)
y = sin2x
D 2
Solution:
The auxiliary equation is p 3p
2 0
2
+ + =
( p + 2)(
p + 1)
= 0
p1
= −2,
p2
= −1
Here, the roots are real and distinct
−2x
−x
∴CF
= Ae + Be
1
Now, PI =
. Sin2x
2
D + 3D
+ 2
1
=
. Sin2x
2
− 2 + 3D
+ 2
1
= . Sin2x
3D
− 2
3D
=
9D
2
+ 2
. Sin2x
− 4
3D
+ 2
= . Sin2x
− 36 − 4
3D(
sin2x
) + 2sin2x
=
− 40
6cos
2x
+ 2sin2x
=
− 40
−1
=
20
[ 3cos
2x
+ sin2x]
∴ T he R equired solution is
y = CF + PI
= Ae
−2x
+ Be
−x
1
−
20
[ 3cos
2x
+ sin2x]
314

5.) Solve ( − 2D
− 8)
y = 4cos
3x
Solution:
D 2
Solution: The auxiliary equation is − 2p
− 8 = 0
Here, the roots are real and distinct
4x
−2x
∴CF
= Ae + Be
1
Now, PI =
4cos
3x
2
D − 2D
− 8
1
=
4cos
3x
2
− 3 − 2D
− 8
1
= 4cos
3x
− 2D
−17
1
= −4
4cos
3x
2D
+ 17
2D
−17
= −4
cos 3x
2
4D
− 289
2D
= −4
( cos 3x)
315
p 2
p
−17cos
3x
− 325
− 6sin3x
−17cos
3x
= −4
325
−
− 4
= [ 6sin3x
+ 17cos
3x]
325
∴ T he R equired solution is
y = CF + PI
= Ae
4x
+ Be
−2x
4
−
325
( p − 4)(
p + 2)
1
=
[ 6sin3x + 17cos3x ]
4,
p
2
= 0
= −2

EXERCISE
PART - A
1.) If roots of the aux ilary equation are 2,7 what is the solution of the
differential equation?
2.) If roots of the aux ilary 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 ( − 1)
y = 0
D 2
d y
5.) Find the solution of − 16y
= 0
2
dx
6.) Solve ( + 9)
y = 0
D 2
7.) Find the solution of ( + 100)
y = 0
2
D 2
8.) Solve ( + 4D
− 1020)
y = 0
D 2
9.) Solve ( D − 5D
+ 2)
y = 0
3 2
10.) Solve ( D − 7D
− 6)
y = 0
3 2
d y dy
11.) Solve + = 0
2
dx dx
2
12.) Solve ( − D −1)
y = 0
D 2
13.) Solve ( + 4D
+ 4)
y = 0
D 2
2
d y dy
14.) Solve − 12 + 36y
= 0
2
dx dx
15.) Solve ( + D + 1)
y = 0
D 2
16.) Solve ( D − D + 1)
y = 0
3 2
2
x
17.) Find the Complementary function of ( D +
13D
− 90)
y = e
316

2
x
18.) Find the Particular integral of ( D − 3D
+ 2)
y = e
2
2x
19.) Find the Particular integral of ( D + D + 4)
y = 10e
2
3x
20.) Find the Particular integral of ( D − 8D
+ 15)
y = e
2
5x
21.) Find the Particular integral of ( D + 10D
+ 25)
y = e
22.) Find the Complementary integral of ( + 25)
y = cos ax
23.) Find the Particular integral of ( + 25)
y = Sinx
24.) Find the Particular integral of ( + 10)
y = sin3x
d y
25.) Find the Particular integral of − 4y
= cos 4x
2
dx
D 2
D 2
2
PART - B
1.) Solve ( 36)
y = 0 when y(
0)
= 2
D 2
2
317
D 2
+ and ( 0)
= 12
d y
dy
2.) Solve + y = 0 giv en that = 2and
y=1 when x=0
2
dx
dx
3.) Solve ( 2D
− 15)
y = 0
x=0
D 2
4.) Solve ( D − 20)
y = 0
D 2
5.) Solve ( + 7D
+ 12)
y = 3
D 2
dy d y
− given that = 0 and = 2 when
2
dx dx
dy
− giv en that y=5 and = −2
when x =0
dx
2
x
6.) Solve ( )
D + 3D
+ 2 y = 2e
2
x
7.) Solve ( D + 12D
+ 36)
y = e
2
8.) Solve ( ) 2
D + D + 4 y = e
2
2x
9.) Solve ( D − 3D
+ 2)
y = e
2
4x
10.) Solve ( )
D +
6D
+ 8 y = e
x
−
y 1
−
−
2

11.) Solve
2
d y
−
2
dx
dy
4 + 4y
= e
dx
2x
2
2 ax
12.) Solve ( )
D
+ 2aD
+ a
y = e
2
7x
13.) Solve ( )
D
+ 14D
+ 49 y = 4e
2
x
14.) Solve ( D − 2D
+ 4)
y = 5 + 3e
15.) Solve
2
d y
+
2
dx
−
dy
8 + 15y
= e
dx
−
−
−3x
+ e
2
5x
5x
16.) Solve ( D + 10D
+ 25)
y = e + e
17.) Solve ( + 16)
y = sin9x
D 2
18.) Solve ( − 25)
y = sin5x
D 2
19.) Solve ( + 100)
y = cos 2x
D 2
2
d y
20.) Solve − 2y
= cos 3x
2
dx
21.) Solve ( + 2D
− 3)
y = sinx
D 2
22.) Solve ( + D − 2)
y = Sin3x
D 2
23.) Solve ( + 4D
+ 13)
y = 4cos
3x
D 2
24.) Solve ( − 8D
+ 9)
y = 8cos
5x
D 2
25.) Solve ( − 2D
− 8)
y = 4cos
2x
1.)
D 2
2x
7x
318
−
3x
ANSWERS
PART - A
y = Ae + Be
2.)
2x
3.) y = e [ A cos x + Bsinx]
−
5.)
4x
−4x
4.)
y = A + Be
y = Ae
x
x −x
+ Be
y = Ae + Be
6.) y = A cos 3x
+ B sin3x
7.) y = A cos10x
+ B sin10x
8.) y =
Ae
30x
+ Be
−34x

9.)
11.)
2 x
x
Be 3
y = Ae +
10.) y = Ae
−x
y = A + Be
12.) y = Ae
2x
13.) y = e ( Ax + B)
−
15.)
16.)
17.)
y
= −x
2
e
A cos
3
x + Bsin
2
3
x
2
x
11 11
y = e 6 A cos x + Bsin
x
6
6
5x
−18x
CF = Ae + Be
18.)
6
xe x 3
319
3x
+ Be
1 5
+
x
2
−2
x
3
+ Be
6x
14.) y = e ( Ax + B)
e x −
20.) −
2
21.)
x −5x
e
2
22.) CF=Acos5x+Bsin5x 23.)
sinx
24
24.) Sin3x
Cos4x
25.) −
20
Part-B
2
19.)
1.) y = 2cos
6x
+ 2sin6x
2.) y = cos x + 2sin
x
3.)
5.)
1 5x
1 −3x
y = e + e
4.) y = 2e
20 2
y = Ae
y
e
−4x
6x
= −
+ Be
−3x
7.) ( ) 49
8.)
+
e
Ax + B +
1
4
x
6.)
x
2 15 15
y = e A
cos x + Bsin
x
+
2
2
−
x
2x
y =
Ae + Be + xe
2x
y = Ae
4
19
e
x
2
5x
−x
9.)
+ 3e
+ Be
−4x
−2x
1 5
−
x
2
2x
e
x
e
+
3

−4x
10.)
−4x
−2x
xe
y = Ae + Be −
2
11.)
2
2x
x 2x
y = e ( Ax + B)
+ e
2
12.)
2
−ax
x −ax
y = e ( Ax + B)
+ e
2
13.)
−7x
2 −7x
y = e ( Ax + B)
+ 2x
e
14.)
x
y = e ( A cos 3x
+ Bsin
5 3 −x
3x)
+ + e
4 7
15.)
−3x
3x
−3x
−5x
xe e
y = Ae + Be + +
2 48
16.)
5x
2 −5x
−5x
e x e
y = e ( Ax + B)
+ +
100 2
17.)
sin9x
y = A cos 4x
+ B sin 4x
−
65
18.)
5x
−5x
sin5x
y = Ae + Be −
50
19.)
cos 2x
y = A cos10x
+ Bsin10x
+
96
20.) y = Ae
2x
−
+ Be
2x
cos 3x
−
11
21.)
−3x
x 1
y = Ae + Be − ( cos x + 2sinx
)
10
1
130
x −2x
22.) y = Ae + Be − ( 3cos
3x
+ 11sin3x
)
2x
= −
1
10
23.) y e ( A cos 3x
+ Bsin3x
) + 3sin3x
+ cos 3x
1
29
24.) ( 4+
7 ) x ( 4−
7 ) x
y = Ae + Be − ( 5sin5x
+ 2cos
5x)
1
10
4x
−2x
25.) y = Ae + Be − ( sin2x
+ 3cos
2x)
320