For 12 weeks course only

Mathematical Methods and its Applications (Solution of assignment-3)

Solution 1 The differential equation is

y 00 + y = cosec x
Its auxiliary equation is

m2 + m = 0
m=±i
Since the roots of the auxiliary equation are complex, we have

C.F. = c1 cos x + c2 sin x

Let us consider
yp = u(x)y1 (x) + v(x)y2 (x)
here y1 (x) = cos x and y2 (x) = sin x
Then the equations connecting u0 and v 0 are

y1 u0 + y2 v 0 = 0
y10 u0 + y20 v 0 = r(x)
   
 y1 y2   cos x sin x 
Wronskian W =  0 = =1
y1 y20   − sin x cos x 
 

 0 sin x 

 cosec x cos x  1
u0 = =− = −1
W W

u = −x
Similarly,  
 cos x 0 
 
 − sin x cosec x  cot x
v0 = = = cot x
W W

v 0 = cot x
v = ln sin x

1
Particular integral is
yp = −x cos x + sin x (ln sin x)
Hence the general solution is

y = c1 cos x + c2 sin x − x cos x + sin x (ln sin x)

So option ‘d’ is correct.

Solution 2 The differential equation is

y 000 − 3y 00 + 2y 0 = 0
Its auxiliary equation is

m3 − 3m2 + 2m = 0
m = 0, 1, 2
Since the roots of the auxiliary equation are real and distinct,
the general solution is
y = c1 + c2 ex + c3 e2x

So the option ‘a’ is correct.

Solution 3 The differential equation is

y (iv) + 2y 00 + y = x2 cos x
The particular integral is

1 1
P. I. = x2 cos x = real part of x2 eix
(D2 + 1) 2 (D + 1)2
2

1 eix
= R.P. of eix x 2
= R.P. of x2
[(D + i)2 + 1]2 (D2 + 2Di)2
−2
eix

ix 1 2 D
= R.P. of e x = R.P. of 1+ x2
(2Di)2 (1 + D/2i)2 −4D2 2i
−2
eix eix 3i2 D2
  
iD 2
= R.P. of 1− x = R.P. of 1 + iD + + ... x2
−4D2 2 −4D2 4
eix eix
   3 
2 3 x 2 3
= R.P. of x + 2ix − = R.P. of + ix − x
−4D2 2 −4D 3 2
ix
 4 3 2

e x ix 3x
= R.P. of + −
−4 12 3 4
 4 2
 3

−1 x 3x x
= − cos x − sin x .
4 12 4 3

2
 Hence the particular integral is
1
(x4 − 9x2 ) cos x − 4x3 sin x .


y=−
48

So the option ‘c’ is correct.

Solution 4 The differential equation is

x2 y 00 − 2x(x + 1)y 0 + 2(1 + x)y = x3

converting it into standard form
2 2
y 00 − (x + 1)y 0 + 2 (1 + x)y = x
x x
Let the complete solution be

y = Ax + Bxe2x
The two equations connecting A and B are

A0 x + xe2x B 0 = 0

A0 + (1 + 2x)e2x B 0 = x
Solving these two equations, we get
1 1
A = − x + a, B = − e−2x + b
2 4
Hence the complete solution is

x2
y = (c1 + c2 e2x )x −
2

So the option ‘d’ is correct.

Solution 5 The differential equation is

y 00 − (2 tan x)y 0 + y = 0 (1)

Comparing eq (1) with y 00 + P y 0 + Qy = R

P = −2 tan x, Q = 1, R=0
To remove the first derivative from (1), we choose
1 1
R R
u = e− 2 P dx
= e− 2 2 tan xdx
= eln sec x = sec x

3
Let the required general solution be

y = uv
then v is given by the normal form

d2 v
+ Iv = S (2)
dx2
where
P 2 1 dP
I = Q− −
4 2 dx
= 1 − tan x + sec2 x = 2.
2

and
R
S= =0
u
Then equation (2) becomes
d2 v
+ 2v = 0
dx2
its auxiliary equation is
m2 + 2 = 0
√
m = ± 2i
√ √
Therefore v = c1 cos 2x + c2 sin 2x
Hence the general solution of the given equation is
√ √
y = sec x(c1 cos 2x + c2 sin 2x)
So the option ‘b’ is correct.

Solution 6 The given differential equation is

2
y 00 − 4xy 0 + (4x2 − 1)y = −3ex sin 2x. (1)
Comparing eq (1) with y 00 + P y 0 + Qy = R
2
P = −4x, Q = 4x2 − 1, R = −3ex sin 2x.
To remove the first derivative from (1), we choose
1 1 2
R R
u = e− 2 P dx
= e− 2 (−4x)dx
= ex .
Let the required general solution be

y = uv

4
then v is given by the normal form
d2 v
+ Iv = S (2)
dx2
where
P 2 1 dP
I = Q− −
4 2 dx
1 1
= 4x2 − 1 − (16x2 ) − (−4) = 1.
4 2
and
R
S= = −3 sin 2x
u
Then equation (2) becomes
d2 v
+ v = −3 sin 2x
dx2
its auxiliary equation is
m2 + 1 = 0
m = ±i
Its C.F. = c1 cos x + c2 sin x
and
1
P.I.= 2 (−3 sin 2x) = sin 2x.
D +1
So, v = c1 cos x + c2 sin x + sin 2x.
Hence, the required general solution
2
y = ex (c1 cos x + c2 sin x + sin 2x).

So the option ‘c’ is correct.

Solution 7 The given differential equation is

x2 y 00 − 2xy 0 + (x2 + 2)y = 0. (1)

Comparing eq (1) with y 00 + P y 0 + Qy = R
2 2
P =− , Q = 1 + 2, R = 0.
x x
To remove the first derivative from (1), we choose
1 1
R R
u = e− 2 P dx
=e x
dx
= x.
Let the required general solution be

y = uv

5
then v is given by the normal form
d2 v
+ Iv = S (2)
dx2
where
P 2 1 dP
I = Q− −
4 2 dx
2 1 1
= 1+ 2 − 2 − 2
x x x
= 1.

and
0
S= =0
u
Then equation (2) becomes
d2 v
+v =0
dx2
its auxiliary equation is
m2 + 1 = 0
m = ±i
v = c1 cos x + c2 sin x

Hence, the required general solution

y = x(c1 cos x + c2 sin x).

So the option ‘b’ is correct.

Solution 8 The given differential equation is

cos xy 00 + y 0 sin x − 2y cos3 x = 2 cos5 x (1)

Dividing by cos x, we have

y 00 + tan xy 0 − 2 cos2 xy = 2 cos4 x.

Comparing eq (1) with y 00 + P y 0 + Qy = R

P = tan x, Q = −2 cos2 x, R = 2 cos4 x. (2)

Choose z such that  2
dz
= 2 cos2 x
dx
dz √
= 2 cos x
dx
6
Then equation (1) transform to

d2 y dy
2
+ P1 + Q1 y = R1
dz dz
d2 z dz √ √
2
+ P − 2 sin x + tan x 2 cos x
where dx
P1 =  2 = dx = 0,
dz 2 cos2 x
dx
Q R 2 z2
Q1 =  2 = −1 and R1 =  2 = cos x = 1 − .
dz dz 2
dx dx
Hence equation (2) becomes

d2 y z2
− y = 1 − ,
dz 2 2
Its auxiliary equation is
m2 − 1 = 0
m = ±1
C.F. = c1 ez + c2 e−z
and
 
1 1 2 d
P.I. = 1− z where D1 ≡
D12 − 1 2 dz
1 1 1 1
= 2
e0.z + 2
z 2 = −1 + (1 − D12 )−1 z 2
D1 − 1 2 (1 − D1 ) 2
z2
= 1−
2

Hence the required solution is

z2
y = c1 ez + c2 e−z +
2
√ √
= c1 e 2 sin x
+ c2 e − 2 sin x
+ sin2 x.
So the option ‘c’ is correct.

Solution 9 The differential equation is

y 000 + 3y 00 + 3y 0 + 1 = 0
Its auxiliary equation is

m3 + 3m2 + 3m + 1 = 0

7
 m = −1, −1, −1
Since the roots of an auxiliary equation are real and equal, we have
the general solution is
y = (c1 + xc2 + x2 c3 )e−x .
So the option ‘d’ is correct.

Solution 10 The differential equation is

y (iv) − y = x sin x
Its auxiliary equation is

m4 − 1 = 0
m = ±1, ±i
C.F. = c1 ex + c2 e−x + c3 cos x + c4 sin x
and the particular solution is

1 1
P.I. = x sin x = imaginary part of xeix
D4 − 1 D4 − 1
1 eix
= I.P. of eix x = I.P. of x
(D + i)4 − 1 D4 + 4iD3 − 6D2 − 4iD
−1
eix D3

3D 2
= I.P. of 1+ −D − x
−4iD 2i 4i
−1
ieix iD3
 
3Di 2
= I.P. of 1+ − −D + x
4D 2 4
ieix iD3
   
3Di 2
= I.P. of 1+ +D − + ... x
4D 2 4
ieix
 
3i
= I.P. of x+
4D 2
1 2
= (x cos x − 3x sin x).
8
Hence the required solution is
1
y = c1 ex + c2 e−x + c3 cos x + c4 sin x + (x2 cos x − 3x sin x).
8
So the option ‘d’ is correct.

Solution 11 The differential equation is

y (iv) + 10y 00 + 9y = 96 sin 2x cos x.

8
Its auxiliary equation is

m4 + 10m2 + 9 = 0
m = ±i, ±3i.
Hence, C.F.= c1 cos x + c2 sin x + c3 cos 3x + c4 sin 3x.
Now,
1 1
P.I. = 96 sin 2x cos x = 48(sin 3x + sin x)
(D2 2
+ 1)(D + 9) (D + 1)(D2 + 9)
2

1 1
= 48 2 sin 3x + 48 sin x
(D + 1)(D2 + 9) (D2 + 1)(D2 + 9)
1 1
= 48 2 2
sin 3x + 48 2 sin x
(−3 + 1)(D + 9) (D + 1)(−12 + 9)
1 1
= −6 2 sin 3x + 6 2 sin x
(D + 9) (D + 1)
 
1 x
= x cos 3x − 3x cos x, since sin ax = − cos ax
D 2 + a2 2a

Hence, the general solution is

y = c1 cos x + c2 sin x + c3 cos 3x + c2 sin 3x + (x cos 3x − 3x cos x).

So option ‘a’ is correct.