Non-homogenous linear differential
equation with constant coefficients
CALENG3
Dr. Susan A. Roces
Department of Chemical Engineering
2T, AY 2022 - 2023
1
Linear differential equation of order n
II. Non-homogenous linear differential equation
with constant coefficients
Std. Form:
dny
d n −1 y
dy
a0
+ a1
+ ....... + an −1
+ an y = R (x )
n
n −1
dx
dx
dx
(a
n
n −1
D
+
a
D
+ .... + an −1 D + an
0
1
) y = R( x)
General Sol’n.: y = yc + y p
where.:
yc = complementary function which is the
general solution of the homogenous linear
d. e.
2
Complementary sol’n.: yc
(a D
0
n
+ a1 D
n −1
Auxiliary eq’n.:
+ .... + an −1 D + an
) y=0
f (m) = 0
yc
will either have the following general
solution depending on the roots of the auxiliary
equation.
yc = c1e
m1 x
+ c2e
m2 x
+ c3e
2
m3 x
+ .... + cn e mn x
yc = (c1 + c2 x + c3 x + .... + cn x
yc = c1 e
a x
sin bx + c2 e
a x
n −1
)e
m x
cos bx
3
n
yc = (c1 + c2 x + ... + cn x )e
n
ax
sin bx
ax
+ (c3 + c4 x + ... + cm x )e cos bx
where:
c1 , c2 , c3 ,...cn , = arbitrary constants
Particular solution y p :
y p = particular solution of the original
differential equation
4
Two methods in solving the particular solution ( y p )
1. Method of Undetermined Coefficients (MUC)
= R(x) itself is a solution of the homogenous
linear d. e. with constant coefficients, and
hence the roots can be determined from the
given R(x).
R (x ) = polynomial in x, exponential function,
sine function or cosine function
5
Note:
1. y p will be based on the m’ of R(x) and treat
the roots as roots from the auxiliary
equation.
2. y p will either have the following solution.
y p = Ae
m1 x
+ Be
m2 x
+ Ce
2
m3 x
y p = ( A + Bx + Cx + .... + Zx
y p = Ae
a x
sin bx + Be
a x
+ .... + Ze
n −1
)e
mn x
m x
cos bx
6
n
y p = ( A + Bx + ... + Cx )e
n
a x
+ ( D + Ex + ... + Zx )e
sin bx
a x
cos bx
where:
A, B, C , ..., Z = constants to be determined
numerically so that y p will
satisfy the original d. e.
7
Solve the particular solution using MUC given
the R(x):
1. R( x) = 3 + x − xe
2x
+ sin x
term (3 + x) is associated with : m' = 0, 0
2x
term xe is associated with : m' = 2, 2
term sin x is associated with : m' = 0 ± i
Auxiliary equation:
2
2
[
2
2
]
m (m − 2) (m − 0) + 1 = 0
2
2
2
m (m − 2) (m + 1) = 0
8
Particular Sol’n.:
y p = ( A + Bx )e
+Ee
(0) x
(0) x
+ (C + Dx)e
sin x + F e
2x
(0) x
cos x
2x
y p = A + Bx + (C + Dx)e + E sin x + F cos x
Compare:
R( x) = 3 + x − xe
2x
+ sin x
A = 3, B = 1, C = 0, D = −1, E = 1, F = 0
9
2
2. R( x) = x + x e
4x
x
+ 6 x e cos 2 x
term x is associated with : m' = 0, 0
2 4x
term x e is associated with : m' = 4, 4, 4
x
term 6 xe cos 2 x is associated with :
m' = 1 ± 2i, 1 ± 2i
Auxiliary equation:
2
[
3
]
2 2
m (m − 4) (m − 1) + 2 = 0
2
3
2
2
m (m − 4) (m − 2m + 1 + 4) = 0
2
3
2
2
m (m − 4) (m − 2m + 5) = 0
2
10
Particular Sol’n.:
y p = ( A + Bx )e
(0) x
2
+ (C + Dx + Ex )e
x
4x
x
+ ( F + Gx) e sin 2 x + ( H + Ix) e cos 2 x
2
y p = A + Bx + (C + Dx + Ex )e
x
4x
x
+ ( F + Gx) e sin 2 x + ( H + Ix) e cos 2 x
11
3. R( x) = sin 3x − 4 cos 3x + sin 2 x cos x
Recall:
1
sin A cos B = [sin( A + B) + sin( A − B)]
2
1
1
1
sin 2 x cos x = [sin 3x + sin x] = sin 3x + sin x
2
2
2
Therefore:
1
1
R( x) = sin 3x − 4 cos 3x + sin 3x + sin x
2
2
3
1
R( x) = sin 3x − 4 cos 3 x + sin x
2
2
12
3
term sin 3x − 4 cos 3x :
2
1
term sin x :
2
m' = 0 ± 3i
m' = 0 ± i
Auxiliary equation:
[(m − 0)
2
+3
2
2
] [(m − 0)
2
2
]
+1 = 0
2
(m + 9) (m + 1) = 0
Particular Sol’n.:
y p = A sin 3 x + B cos 3 x + C sin x + D cos x
A = 3 / 2, B = −4, C = 1 / 2, D = 0
13
4. R( x) = sin x cos x + cosh 2 x
Recall:
sin 2 x = 2 sin x cos x
sin 2 x
sin x cos x =
2
2x
e +e
cosh 2 x =
2
−2x
Therefore:
1
1 2x 1 −2x
R( x) = sin 2 x + e + e
2
2
2
14
1
term sin 2 x : m' = 0 ± 2 i
2
1 2x
term e : m' = 2
2
1 −2x
term e :
2
m' = − 2
Auxiliary equation:
[(m − 0)
]
+ 2 (m − 2) (m + 2) = 0
2
(m + 4) (m − 2)(m + 2) = 0
2
2
Particular Sol’n.:
2x
y p = A sin 2 x + B cos 2 x + Ce + De − 2 x
A = 1 / 2, B = 0, C = 1 / 2, D = 1 / 2
15
Solve the ff. non-homogenous d. e. using MUC:
1. y ' '+ y = cos x
2
D y + y = cos x
2
( D + 1) y = cos x
Complementary solution:
2
( D + 1) y = 0
m2 + 1 = 0
m = 0 ± i (1st) a = 0, b = 1
yc : yc = c1 e
( 0) x
sin x + c2 e
( 0) x
cos x
yc = c1 sin x + c2 cos x
16
Particular solution:
R( x) = cos x
term cos x : m = 0 ± i (2nd)
y p : yp = A x e
( 0) x
a = 0, b = 1
( 0) x
sin x + B x e
cos x
y p = A x sin x + B x cos x
y' p =A [ x cos x + sin x]+ B [ x(− sin x) + cos x]
y ' p = A x cos x + A sin x − B x sin x + B cos x
y ' ' p = A [− x sin x + cos x] + A cos x
− B [ x cos x + sin x] − B sin x
17
y ' ' p = − A x sin x + A cos x + A cos x
− B x cos x − B sin x − B sin x
y ' ' p = − A x sin x + 2 A cos x − Bx cos x − 2 B sin x
Substitute to the original d. e.:
2
D y p + y p = cos x
− A x sin x + 2 A cos x − Bx cos x − 2 B sin x
+ A x sin x + B x cos x = cos x
2 A cos x − 2 B sin x = cos x
18
2 A cos x − 2 B sin x = cos x
coeff . of cos x :
2A = 1
1
A=
2
B=0
coeff . of sin x : − 2 B = 0
1
Therefore, yp: y p = x sin x + 0
2
Gen. sol’n.:
y = yc + y p
1
y = c1 sin x + c2 cos x + x sin x
2
19
5
4
3
2
2. ( D + D − D + D − 2D) y =
2
2
x + 2 cosh 2 x − sin (1 / 2) x + 2 + 1 + cos x
Complementary solution:
5
4
3
2
( D + D − D + D − 2 D) y = 0
5
4
3
2
m + m − m + m − 2m = 0 5 roots
4
3
2
m(m + m − m + m − 2) = 0
2
m(m + 2)(m − 1)(m + 1) = 0
m = 0, − 2, 1, 0 ± i
yc :
( 0) x
−2 x
( 0) x
( 0) x
x
yc = c1 e + c2 e + c3 e + c4 e sin x + c5 e cos x
20
yc = c1 + c2 e
−2 x
x
+ c3 e + c4 sin x + c5 cos x
Particular solution:
1
R( x) = x + 2 cosh 2 x − sin x + 2 + 1 + cos x
2
2
2
Recall:
1
sin x = [1 − cos 2 x]
2
1⎡
⎛ 1 ⎞⎤ 1 1
2 1
sin x = ⎢1 − cos 2⎜ x ⎟⎥ = − cos x
2
2⎣
⎝ 2 ⎠⎦ 2 2
2
21
2
cos 2 x = 2 cos x − 1
2
2 cos x = 1 + cos 2 x
1
2
cos x = (1 + cos 2 x)
2
2x
⎛e +e
R( x) = x +2⎜⎜
2
⎝
x 1
cos = (1 + cos x)
2 2
x
1 + cos x
cos =
2
2
x
1 + cos x = 2 cos
2
2
−2x
⎞ 1
x
⎟⎟ − (1 − cos x ) + 2 + 2 cos
2
⎠ 2
1 1
x
2
2x
−2x
R( x) = x + e + e − + cos x + 2 + 2 cos
2 2
2
2
22
1 1
x
R( x) = x + e + e − + cos x + 2 + 2 cos
2 2
2
x
⎡ 3 2 ⎤ 2x −2x 1
R( x) = ⎢ + x ⎥ + e + e + cos x + 2 cos
2
2
⎣2
⎦
2
2x
−2x
m' = 0 (2nd) , 0(3rd) , 0 (4th) , 2, − 2 (2nd) , 0 ± i (2nd),
1
Note : yc : m = 0, − 2, 1, 0 ± i
0± i
2
y p : y p = Ax + Bx 2 + Cx 3 + De 2 x + E x e − 2 x
x
x
+ F x sin x + G x cos x + H sin + I cos
2
2
23
Gen. Sol’n.:
y = yc + y p
y = c1 + c2 e
−2 x
x
+ c3 e + c4 sin x + c5 cos x
2
3
+ Ax + Bx + Cx + De
2x
+ E xe
−2x
x
x
+ F x sin x + G x cos x + H sin + I cos
2
2
24
2
d I
dI
−t
3.
+2
+ 5I = 8e
2
dt
dt
dI
when : t = 0, I = 0,
=8
dt
2
D I + 2 DI + 5 I = 8e
2
( D + 2 D + 5) I = 8e
−t
−t
Complementary solution:
2
m + 2m + 5 = 0
m=
−2±
4 − 4(1)(5)
− 2 ± − 16
=
2
2(1)
25
− 2 ± 4i
m=
2
m = −1± 2i
Ic :
a = −1, b = 2
−t
I c = c1e sin 2t + c2 e
−t
cos 2t
Particular solution:
R(t ) = 8e
−t
m' = −1
−t
I p = Ae
−t
I ' p = − Ae
I ' ' p = Ae
−t
26
Substitute Ip in the original d. e.:
2
D I p + 2 DI p + 5I p = 8e
Ae
−t
−t
+ 2(− Ae ) + 5 Ae
4 Ae
Therefore, Ip:
Gen. Sol’n.:
−t
−t
−t
= 8e
−t
−t
= 8e
−t
8e
A=
=2
−t
4e
−t
I p = 2e
I = Ic + I p
27
−t
−t
−t
I = c1e sin 2t + c2e cos 2t + 2e
(1)
dI
= c1[e −t (2 cos 2t ) + sin 2t (e −t )(−1)]
dt
−t
−t
+ c2 [e (−2 sin 2t ) + cos 2t (e )(−1)]
− 2e
−t
dI
= c1e −t [2 cos 2t − sin 2t ]
dt
−t
−t
+ c2e [−2 sin 2t − cos 2t ] − 2e
( 2)
28
when : t = 0, I = 0, use eq ' n. (1) :
−0
−0
−0
0 = c1e sin 0 + c2 e cos 0 + 2e
0 = 0 + c2 + 2
c2 = − 2
dI
when : t = 0,
= 8, use eq ' n.( 2) :
dt
8 = c1e −0 [2 cos 0 − sin 0]
−0
+ c2 e [−2 sin 0 − cos 0] − 2e
8 = 2c1 − c2 − 2
P.S.:
−t
8 = 2c1 + 2 − 2
−t
−0
c1 = 4
I = 4e sin 2t − 2e cos 2t + 2e
−t
29
2. Method of Variation of Parameters
= R(x) is not a particular solution of some
homogeneous linear d. e. with constant
coefficient
Complementary sol’n.: yc
Auxiliary eq’n.:
f (m) = 0
yc
will either have the following general
solution depending on the roots of the auxiliary
equation.
m3 x
m1 x
m2 x
mn x
yc = c1e
+ c2e
+ c3e
+ .... + cn e
31
2
yc = (c1 + c2 x + c3 x + .... + cn x
yc = c1 e
a x
sin bx + c2 e
a x
n −1
)e
m x
cos bx
n
yc = (c1 + c2 x + ... + cn x )e
n
ax
sin bx
ax
+ (c3 + c4 x + ... + cm x )e cos bx
where:
c1 , c2 , c3 ,...cn , = arbitrary constants
Particular solution y p:
y p will be based on the solved yc above except
for the arbitrary constants and will either have
the following solutions:
32
y p = A( x)e
m1 x
+ B ( x )e
m2 x
+ .... + Z ( x)e
[
y p = A( x) + B( x) x + .... + Z ( x) x
y p = A( x)e
a x
sin bx + B( x)e
a x
n −1
]e
mn x
m x
cos bx
[
]e
+ [D( x) + E ( x) x + ... + Z ( x) x ] e
n
a x
sin bx
n
a x
cos bx
y p = A( x) + B( x) x + ... + C ( x) x
where:
A, B, C , ..., Z =functions of x to be determined
numerically so that y p will
satisfy the original d. e.
33
Solve the ff. non-homogenous d. e.:
1. y ' '+ y = tan x
2
D y + y = tan x
2
( D + 1) y = tan x
Complementary solution:
2
( D + 1) y = 0
m2 + 1 = 0
m = 0±i
yc : yc = c1 e
( 0) x
a = 0, b = 1
sin x + c2 e
( 0) x
cos x
yc = c1 sin x + c2 cos x
34
Particular solution y p :
y p = A( x) sin x + B ( x) cos x
y p = A sin x + B cos x
y ' p = A cos x + A' sin x − B sin x + B' cos x
Let:
A' sin x + B' cos x = 0
(1)
y ' p = A cos x − B sin x
y ' ' p = − A sin x + A' cos x − B cos x − B ' sin x
Substitute in original d. e.:
2
D y p + y p = tan x
35
− A sin x + A' cos x − B cos x − B ' sin x
+ A sin x + B cos x = tan x
A' cos x − B ' sin x = tan x
( 2)
From eq’n. (1): A' sin x + B' cos x = 0
B' cos x
A' = −
sin x
Subs’t. A’ in eq’n. (2):
B ' cos x
(cos x) − B' sin x = tan x
−
sin x
36
2
⎛ cos x
⎞
− B' ⎜⎜
+ sin x ⎟⎟ = tan x
⎝ sin x
⎠
⎛ cos 2 x + sin 2 x ⎞
⎟⎟ = − tan x
B' ⎜⎜
sin x
⎝
⎠
⎛ sin x ⎞
B ' = − sin x⎜
⎟
⎝ cos x ⎠
2
2
(1 − cos x)
dB
sin x
=−
=−
cos x
dx
cos x
37
∫
dB = ∫ (cos x − sec x)dx
B = sin x − ln sec x + tan x
From eq’n. (1):
B' cos x
A' = −
sin x
dA
⎛ sin 2 x ⎞ cos x
= − ⎜⎜ −
⎟⎟
dx
⎝ cos x ⎠ sin x
dA
= sin x
dx
38
∫ dA = ∫
sin x dx
A = − cos x
Particular solution y p :
y p = − cos x sin x + (sin x − ln sec x + tan x ) cos x
y p = − cos x sin x + cos x sin x − (cos x) ln sec x + tan x
y p = −(cos x) ln sec x + tan x
39
Gen. Sol’n:
y = yc + y p
y = c1 sin x + c2 cos x
− (cos x) ln sec x + tan x
40
2x
e
2
2. ( D − 3D + 2) y =
2x
1+ e
Complementary solution:
2
m − 3m + 2 = 0
(m − 2)(m − 1) = 0
m = 2, 1
2x
x
yc : yc = c1 e + c2 e
yp :
y p = A( x) e
yp = Ae
2x
2x
+ B( x) e
+ Be
x
x
41
yp = Ae
2x
y' p = 2 A e
A' e
+ Be
2x
2x
y' p = 2 A e
x
+ A' e
2x
+ B e x + B' e x
x
+ B' e = 0
2x
y' ' p = 2[2 A e
y' ' p = 4 A e
+ Be
2x
2x
(1)
x
+ A' e
2x
+ 2 A' e
2x
x
] + B e + B' e
x
+ Be + B' e
x
x
42
Substitute in original d. e.:
2x
e
D y p − 3Dy p + 2 y p =
2x
1+ e
2
4Ae
2x
+ 2 A' e
+ 2( A e
4Ae
2x
+ 2 A' e
2x
x
+ Be + B' e − 3(2 A e
2x
2x
+ 2Ae
x
e
x
+ Be ) =
x
x
+ 2B e =
+ Be )
2x
+ Be + B' e − 6 A e
2x
x
2x
1+ e
x
2x
e
2x
− 3B e
x
2x
1+ e
2x
43
2 A' e
2x
e
x
+ B' e =
From eq’n. (1): A' e
2x
1+ e
2x
( 2)
2x
x
+ B' e = 0
x
B' e
B'
A' = − 2 x = − x
e
e
Subs’t. in eq’n. (2):
2x
e
B' ⎞ 2 x
⎛
x
2 ⎜ − x ⎟ e + B' e =
2x
1+ e
⎝ e ⎠
44
− 2 B' e
x
x
x
+ B' e =
− B' e =
e
2x
e
2x
1+ e
2x
2x
1+ e
2x
−e
x
dB
−e
= x
2x =
dx e (1 + e ) 1+ e 2 x
x
x
e
u=e
dB
=
−
dx
∫
∫ 1+ e2 x
x
du = e dx
x
B = − arctan e
45
B'
A' = − x
e
dA
1
= − x
dx
e
⎛ −e x
⎜
⎜ 1 + e2 x
⎝
1
∫ dA = ∫ 1 + e 2 x dx
2
sec θ dθ
A= ∫
2
[1 + (tan θ ) ] tan θ
dθ
A= ∫
tan θ
By trig. Sub:
⎞
⎟
⎟
⎠
a = 1, u = e
u = a tan θ
x
e = tan θ
1+ e 2 x
θ
e
x
x
1
x
2
e dx = sec θ dθ
sec 2 θ dθ
dx =
tan θ
46
e
A = ∫ cot θ dθ = ln sin θ = ln
yp : yp = e
Gen. Sol’n:
y = c1 e
+ e
2x
2x
2x
e
ln
x
1+ e
y = yc + y p
+ c2 e
ln
x
1 + e2 x
x
2x
− e arctan e
x
x
e
x
1+ e
x
2x
− e arctan e
x
47
2
3. ( D + 4 D + 4) y = e
−2 x
ln x
Complementary solution:
2
m + 4m + 4 = 0
2
(m + 2) = 0
m = −2, − 2
yc :
yp :
yc = c1 e
−2x
y p = A( x) e
yp = Ae
+ c2 x e
−2x
−2 x
− 2x
+ B( x) [ x]e
+ B xe
−2 x
−2 x
48
yp = Ae
−2 x
y' p = −2 A e
+ B xe
−2 x
− 2B x e
A' e
−2 x
y' p = −2 Ae
+ A'e
−2 x
−2 x
−2 x
−2 x
−2 x
−2 x
−2 x
+ A' e
−2 x
−2 x
(1)
+B e
−2 x
+B e
+ B 'e
+ B' x e
=0
− 2B x e
− 2[−2 B x e
− 2B e
−2 x
+B e
+ B' x e
y' ' p = −2[−2 Ae
−2 x
−2 x
]
−2 x
+ B'xe
−2 x
]
−2 x
49
−2 x
−2 x
−2 x
y' ' p = 4 Ae − 2 A' e + 4 B x e − 2 B e
−2 x
−2 x
−2 x
−2B ' x e
− 2B e
+ B 'e
−2 x
−2 x
−2 x
y' ' p = 4 Ae − 2 A' e
−2 x
−2 x
−2 x
−2 x
+ 4B x e − 4B e − 2B ' x e + B ' e
Substitute in original d. e.:
2
D y p + 4D y p + 4 y p = e
−2 x
−2 x
−2 x
−2 x
ln x
−2 x
4 Ae − 2 A' e + 4B x e − 4B e − 2B ' x e
−2 x
−2 x
−2 x
− 8 Ae
− 8B x e
+ 4 Be
+ 4Ae
−2 x
+ 4B x e
−2 x
=e
−2 x
−2 x
+ B 'e
−2 x
ln x
50
− 2 A' e−2 x + B' e−2 x (1 − 2 x) = e−2 x ln x
( 2)
From eq’n. (1):
A' e −2 x + B' x e −2 x = 0
A' = − B' x
Sub. A’ in eq’n. (2):
− 2 ( − B ' x )e
−2 x
+ B' e
−2 x
(1 − 2 x) = e
2 B ' x + B '−2 B ' x = ln x
By parts:
∫ dB = ∫ ln xdx
u = ln x
B = x ln x − x
dx
du =
x
−2 x
ln x
∫ dv = ∫ dx
v=x
51
From eq’n. (1):
∫
A' = − B' x
A' = − (ln x ) x
dA = − ∫ x ln x dx
By parts: u = ln x
dx
du =
x
⎡ x2
⎤
1
A = − ⎢ ln x − ∫ x dx ⎥
2
⎣2
⎦
2
2
x
x
A=−
ln x +
2
4
dv
=
x
dx
∫
∫2
x
v=
2
52
Particular Sol’n.:
2
2
⎛ x
x
ln x +
y p = ⎜⎜ −
4
⎝ 2
2
2
x
−2 x ⎛ x
⎜⎜
yp = e
−
2
⎝ 4
⎞ −2 x
−2 x
⎟⎟ e
+ ( x ln x − x) x e
⎠
2
2⎞
ln x + x ln x − x ⎟⎟
⎠
2
2
2
⎡
⎛
x ⎞⎤
−2 x x − 4 x
2
⎟⎟⎥
yp = e
+ ln x⎜⎜ x −
⎢
4
2 ⎠⎦
⎝
⎣
2
2
⎡
⎛
⎞⎤
−
3
x
x
−2 x
⎟⎟⎥
yp = e
+ ln x⎜⎜
⎢
⎝ 2 ⎠⎦
⎣ 4
53
2
x −2 x ⎛
3⎞
yp =
e
⎜ ln x − ⎟
2
2⎠
⎝
2
x −2 x ⎛ 2 ln x − 3 ⎞
yp =
e
⎜
⎟
2
2
⎝
⎠
2
x −2 x
(2 ln x − 3)
yp =
e
4
Gen. Sol’n.: y = yc + y p
y = c1 e
−2x
+ c2 x e
− 2x
2
x −2 x
(2 ln x − 3)
+
e
4
54
Problem Set: Non-homogeneous Linear Differential Equations
1. y '' + y ' − 2 y = x
2
x
Ans. y = c1e + c2 e
−2 x
1 2
1
3
−
x −
x−
2
2
4
2. y '' + y ' − 2 y = sin x
x
Ans. y = c1e + c2 e
−2 x
1
−
(cos x + 3sin x)
10
3. y ''− 4y '+ 5y = e− x
1 −x
Ans. y = e (c1 cos x + c2 sin x) +
e
10
2x
4. y ''− 4y = e x cos x;
y(0) = 1, y '(0) = 2
1
5. y '' − 3y ' + 2 y =
−x
1+ e
−x ⎤ x
⎡
Ans. y = ⎣c1 + ln(1+ e )⎦ e
+ ⎡⎣c2 − e− x + ln(1+ e− x )⎤⎦ e 2 x
ex
6. y '' − 2 y ' + y =
1+ x 2
⎤
1
x⎡
2
Ans. y = e ⎢c1 + c2 x − ln(1+ x ) + x arctan x ⎥
⎣
⎦
2
7. y ''+ y '− 2 y = x + sin 2 x;
y(0) = 1, y '(0) = 0
8. y ''+ 3y '− 4y = (x 3 + x) e x
9.
y '' + 4 y = e3 x + x sin 2 x
(Do not det er min e the coefficients)
10. y '' + 3y ' + 2 y = sin(e x )
Assignment 13_Non-Homogenous Linear DE or order n
1. y '' + 4y ' + 3y = x −1
Ans. y = c1e
−x
+ c2 e
x
2. y ''− y = e + 2e
−3 x
x
7
+
−
3
9
2x
x
Ans. y = c1e + c2 e
−x
xe
+
2
x
2x
2e
+
3
3. y ''− 2 y ' = sin 4x
Ans. y = c1 + c2 e
2x
1
1
+
cos 4x −
sin 4x
40
20
4. y '' − 2 y ' + y = xe x − e x
3
x
2
x
x e
x e
Ans. y = c1e + c2 x e +
−
6
2
−2 x
5. y '' + 4y ' + 4y = e
ln x
2
⎛
⎞
x
−2 x
Ans. y = e ⎜ c1 + c2 x +
(2 ln x − 3)⎟
4
⎝
⎠
x
x
x
3
6. y '' + y = e + x ;
y(0) = 2, y '(0) = 0
3
11
1 x
3
Ans. y = cos x +
sin x + e + x − 6 x
2
2
2
7. y '' − y ' = xe x ;
y(0) = 2, y '(0) = 1
⎞
x⎛1
2
Ans. y = e ⎜ x − x + 2 ⎟
⎝2
⎠
8. y '' + 9 y = e 2 x + x 2 sin x
(Do not det er min e the coefficients)
Ans. y p = Ae 2 x + (Bx 2 + Cx + D)cos x
+ (Ex 2 + Fx + G)sin x
9. y ''+ y = sec 2 x
Ans. y = c1 sin x + c2 cos x
+ (sin x ) ln(sec x + tan x) −1
10. y '' + 2 y ' +10 y = x 2 e− x cos3x
Ans. y p = x e− x ⎡⎣(Ax 2 + Bx + C)cos3x
+ (Dx 2 + Ex + F)sin 3x ⎤⎦