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# 1

UNIT-I
ORDINARY DIFFERENTIAL EQUATIONS
Higher order differential equations with constant coefficients Method of variation of
parameters Cauchys and Legendres linear equations Simultaneous first order
linear equations with constant coefficients.

(i)
(ii)
(iii)

## Formation of differential equation from the given physical situation, called

modeling.
Solutions of this differential equation, evaluating the arbitrary constants
from the given conditions, and
Physical interpretation of the solution.

## HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS WITH

CONSTANT COEFFICIENTS.
General form of a linear differential equation of the nth order with constant
coefficients is
dny
d n 1 y
d n2 y
+
K
+
K
+ .............. + K n y = X
1
2
dx n
dx n 1
dx n 2
Where K 1 , K 2,........................ K n are constants.

.. (1)

## The symbol D stands for the operation of differential

dy
d 2y
d3y
2
3
(i.e.,) Dy =
, similarly D y = 2 , D y = 3 , etc...
dx
dx
dx
The equation (1) above can be written in the symbolic form

## (D n + K 1 D n 1 + .......... + K n ) y = X i.e., f(D)y = X

Where f (D) = D n + K 1 D n 1 + ........... + K n

Note
1
1. X = Xdx
D
1
2.
X = e ax Xe ax dx
Da
1
3.
X = e ax Xe ax dx
D+a
(i) The general form of the differential equation of second order is

d2y
dy
+ P + Qy = R (1)
2
dx
dx

## Where P and Q are constants and R is a function of x or constant.

(ii)Differential operators:

## The symbol D stands for the operation of differential

(i.e.,) Dy =

dy
d2y
, D2 y = 2
dx
dx

1
Stands for the operation of integration
D
1
Stands for the operation of integration twice.
D2
(1) can be written in the operator form
D 2 y + PDy + Qy = R (Or) ( D 2 + PD + Q) y = R
(iv)

## Complete solution = Complementary function + Particular Integral

PROBLEMS
1. Solve (D 2 5 D + 6) y = 0
Solution: Given (D 2 5 D + 6) y = 0
The auxiliary equation is m 2 5m + 6 = 0`
i.e., m = 2,3
C.F = Ae 2 x + Be 3 x
The general solution is given by
y = Ae 2 x + Be 3 x

d2y
dy
6 + 3y = 0
2
dx
dx
2
Solution: Given (D 6 D + 3 y ) = 0
The auxiliary equation is m 2 6m + 13 = 0
6 36 52
i.e., m =
2
= 3 2i
Hence the solution is y = e 3 x ( A cos 2 x + B sin 2 x)
2. Solve

## 3. Solve (D 2 +1) = 0 given y(0) =0, y(0) = 1

3
Solution: Given (D 2 +1) = 0

A.E is m 2 + 1 = 0
M = i
Y = A cosx + B sinx
Y(x) = A cosx + B sinx
Y(0) = A =0
Y(0) =B =1
A = 0, B = 1
i.e., y = (0) cosx + sinx
y = sinx
3. Solve ( D 2 4 D + 13) y = e 2 x
Solution: Given ( D 2 4 D + 13) y = e 2 x
The auxiliary equation is m 2 4m + 13 = 0
4 16 52 4 36
m=
=
= 2 3i
2
2
C.F = e 2 x ( A cos 3x + B sin 3x )
1
P.I. = 2
e2x
D 4 D + 13
1
e2x
=
4 8 + 13
1
= e2x
9
y = C.F +P.I.
1
y = e 2 x ( A cos 3x + B sin 3 x ) + e 2 x
9
5. Find the Particular integral of y- 3y + 2y = e x e 2 x
Solution: Given y- 3y + 2y = e x e 2 x

( D 2 3D + 2) y = e x e 2 x
1
P.I 1 = 2
ex
D 3D + 2
1
=
ex
1 3 + 4
1
= ex
0
1
=x
ex
2D 3
1 x
=x
e
23
= xe x

1
e2x
D 3D + 2
1
=e2x
46+2
1
=-x
e2x
2D 3
1 2x
=-x
e
43
= - xe 2 x

P.I 2 =

## P.I. = P.I 1 + P.I 2

= xe x + (- xe 2 x )
= -x( e x + e 2 x )

d2y
dy
6. Solve
4 + 5 y = 2 cosh x
2
dx
dx
d2y
dy
Solution: Given 2 4 + 5 y = 2 cosh x
dx
dx
2
The A.E is m 4m + 5 = 0
4 16 20
m=
= 2 i
2
C.F = e 2 x ( A cos x + B sin x )
e x + ex
1
1
(
)
2
cosh
2

x
=

D 2 + 4D + 5
D 2 + 4D + 5 2
1
1
ex + 2
e x
= 2
D + 4D + 5
D + 4D + 5
x
x
e
e
=

1+ 4 + 5 1 4 + 5
e x ex
=

10
2
y = C.F + P.I

P.I =

= e 2 x ( A cos x + B sin x ) -

e x ex

10
2

1
1
sin ax(or )
cos ax
f ( D)
f ( D)
Replace D 2 by a 2

## Problems based on P.I =

7. Solve

d2y
dy
+ 3 + 2 y = sin 3x
2
dx
dx

5
d2y
dy
+ 3 + 2 y = sin 3x
2
dx
dx
2
The A.E is m +3m + 2 = 0
(m+1)(m+2) = 0
M = -1, m = -2
C.F = Ae x + Be 2 x
1
P.I = 2
sin 3 x
D + 3D + 2
1
=
sin 3 x (Replace D 2 by a 2 )
2
3 + 3D + 2
1
1 (3D + 7)
=
sin 3x =
sin 3x
3D 7
3D 7 (3D + 7)
3D + 7
sin 3x
=
(3D) 2 (7) 2
3D + 7
=
sin 3x
9 D 2 49
3D + 7
=
sin 3x
9(3 2 ) 49
3D + 7
=
sin 3x
130
1
(3D(sin 3x) + 7 sin 3x )
=
130
1
(9 cos 3x + 7 sin 3x )
=
130

Solution: Given

y = C.F + P.I
Y = Ae x + Be 2 x

1
(9 cos 3x + 7 sin 3x )
130

## 8. Find the P.I of (D 2 +1) = sin x

Solution: Given (D 2 +1) = sin x
1
P.I. = 2
sin x
D +1
1
=
sin x
1+1
1
=x
sin x
2D
x 1
=
sin x
2D
x
= sin xdx
2
x cos x
P.I =2

## 9. Find the particular integral of (D 2 +1) y = sin 2 x sin x

Solution: Given (D 2 +1) y = sin 2 x sin x
1
=- (cos 3x cos x )
2
1
1
= - cos 3x + cos x
2
2
1 1

P.I 1 = 2
cos 3x

D +1 2

1 1
=
cos 3 x
2 9 +1
1
= cos 3x
16

1 1

cos x

D +1 2

1 1
=
cos x
2 1+1
1 1
= x
cos x
2 2D
x
= cos xdx
4
x
= sin x
4
x
1
P.I =
cos 3x + sin x
16
4
Problems based on R.H.S = e ax + cos ax(or )e ax + cos ax
P.I 2 =

## 10. Solve (D 2 4 D + 4) y = e 2 x + cos 2 x

Solution: Given (D 2 4 D + 4) y = e 2 x + cos 2 x
The Auxiliary equation is m 2 4m + 4 = 0
(m 2 ) 2 = 0
m = 2 ,2
2x
C.F = (Ax +B)e
1
P.I 1 = 2
e2x
D 4D + 4
1
=
e2x
48+ 4
1
= e2x
0
1
=x
e2x
2D 4
1 2x
=x e
0

7
1
= x 2 e2x
2
1
cos 2 x
P.I 2 = D 4 D + 4
2

1
cos 2 x
2 4D + 4
1
cos 2 x
=
4D
1 1

cos 2 x
=

4 D

1 sin 2 x
sin 2 x
=
=
4
2
8
=

y = C.F + P.I
y = (Ax +B)e 2 x

sin 2 x
8

## Problems based on R.H.S = x

Note: The following are important
(1 + x) 1 = 1 x + x 2 x 3 + .......

(1 x) 1 = 1 + x + x 2 + x 3 + .................
(1 + x) 2 = 1 2 x + 3x 2 4 x 3 + ..............
(1 x) 2 = 1 + 2 x + 3x 2 + 4 x 3 + ..............

## 11. Find the Particular Integral of (D 2 +1) y = x

Solution: Given (D 2 +1) y = x
A.E is (m 2 1) = 0
m = 1
x
C.F = Ae + Be x
1
P.I = 2
x
D 1
1
=
x
1 D2

= 1 D2

= 1 + D + (D 2 ) + ......... x
= [x + 0 + 0 + 0...........]
=-x

12. Solve: D 4 2 D 3 + D 2 y = x 3

Solution: Given D 4 2 D 3 + D 2 y = x 3

The A.E is m 2m + m 2 = 0
m 2 (m 2 2m + 1) = 0
m 2 (m 1) 2 = 0
m = 0,0 , m = 1,1
C.F = (A + Bx)e 0 x +(C + Dx)e x
4

1
x3
3
2
D 2D + D
1
= 2
x3
D 1 + (D 2 2 D )
1
1
= 2 1 + (D 2 2 D ) x 3
D
2
3
1
= 2 1 D 2 2 D + D 2 2 D D 2 2 D + ...............
D
1
= 2 1 + 2 D + 3D 2 + 4 D 3 + D 4 x 3
D
1
= 2 x 3 + 6 x 2 + 18 x + 24
D

1 x 4 6 x 3 18 x 3
= +
+
+ 24 x
D 4
3
2

P.I =

[
[

]
]

) (

) (

x 5 6 x 4 18 x 3 24 x 2
+
+
+
20 12
6
2
5
4
x
x
=
+
+ 3x 3 + 12 x 2
20 2

y = C.F + P.I
y = (A + Bx)e 0 x +(C + Dx)e x +

x5 x4
+
+ 3x 3 + 12 x 2
20 2

## Problems based on R.H.S = e ax x

1
1
e ax x = e ax
x
P.I =
f ( D)
f ( D + a)
13. Obtain the particular integral of ( D 2 2 D + 5) y = e x cos 2 x
Solution: Given ( D 2 2 D + 5) y = e x cos 2 x

1
e x cos 2 x
D 2D + 5

1
= ex
cos 2 x(Re placeDbyD = 1)
2
(D + 1) 2(D + 1) + 5

P.I =

9
1

= ex 2
cos 2 x
D + 2 D + 1 2 D 2 + 5
1
cos 2 x
= ex 2
D + 4
1
= ex
cos 2 x
4+4
1
cos 2 x
= ex x
2D
xe x
=
cos 2 xdx
2
xe x sin 2 x
P.I =
4
14. Solve ( D + 2) 2 y = e 2 x sin x
Solution: Given ( D + 2) 2 y = e 2 x sin x
A.E is (m 2 +1) = 0
m = -2, -2
C.F. = (Ax + B)e 2 x
1
e 2 x sin x
P.I =
2
(D + 2 )
1
= e 2 x
sin x
2
D 2+2
1
= e 2 x 2 sin x
D
1
= e 2 x sin x
1
2 x
= -e sin x

y = C.F + P.I
y = (Ax + B)e 2 x - e 2 x sin x

## Problems based on f(x) = x n sin ax(or ) x n cos ax

To find P.I when f(x) = x n sin ax(or ) x n cos ax
1
x n sin ax(or ) x n cos ax
P.I =
f ( D)
d
1
1
1
( xV ) = x
V +
V
f ( D)
f ( D)
dD f ( D)
i.e.,

f ' (D ) 1
1
1
( xV ) = x
V
V
f ( D)
f ( D)
f ( D) f ( D)
1
1
f ' ( D)
xV = x
V
V
f ( D)
f ( D)
[ f ( D)]2

10

## 15. Solve ( D 2 + 4 D + 3) y = e x sin x + xe 3 x

Solution: The auxiliary equation is m 2 +4m + 3 = 0
m= -1,-3
x
C.F =A e + Be 3 x
1
e x sin x
P.I 1 =
2
( D + 4 D + 3)
1
=
sin x
2
(D 1) + 4(D 1) + 3
1
= e x 2
sin x
D + 2D
(1 + 2 D ) sin x
= e x
1 + 4D 2
ex
=
[2 cos x + sin x]
5
1
P.I 2 = e 3 x
x
2
D + 3) + 4( D + 3) + 3
1
x
= e 3x 2
D + 10 D + 24

D 2 + 10 D
1 +
x
24

3x
e 5D
=
1
x
24
12
e3x
5
=
x
24
12
General solution is y = C.F + P.I
e3x
=
24

y = A e x + Be 3 x

3x
ex
[2 cos x + sin x] + e x 5
24
12
5

d2y
dy
+ 2 + y = x cos x
2
dx
dx
Solution: A.E : m 2 +2m + 1 = 0
m = -1,-1
C.F = ( A + Bx)e x
1
P.I =
( x cos x)
( D + 1) 2

16 Solve

2( D + 1)
1
(cos x )
= x
2
2
(
)
(
)
D
+
1
D
+
1

11

2
1
(cos x )
= x

2
(D + 1) D + 2 D + 1

2
1

(cos x )
= x

D + 1 ( 1 + 2 D + 1)

2 sin x

= x
D + 1 2

x sin x
=
2
x sin x sin x

=
2
D +1
x sin x (D 1)sin x
=

2
D2 1
x sin x cos x sin x
=
+
2
2

+
2
2

## 17. Solve D 2 + 1 y = sin 2 x

Solution: A.E : m 2 +1 = 0
m = i
C.F =A cosx +B sinx
1
P.I = 2
sin 2 x
D +1
1 1 cos 2 x
= 2

2
D +1

1 1
1

= 2
e0x 2
cos 2 x
2 D +1
D +1

1 1

= 1 + cos 2 x
2 3

1 1
= + cos 2 x
2 6
1 1
y = A cosx +B sinx + + cos 2 x
2 6

d2y
y = x sin x + (1 + x )e x
dx 2
Solution: A.E : m 2 1 = 0
m = 1
C.F = A e x + Be x
1
(xV ) = x f ' ( D) 1 (V )
P.I 1 =
f ( D)
f ( D) f ( D)

18. Solve

12
2D
1

(sin x )
= x 2 2
D 1 (D 1)

2 D sin x

= x 2
D 1 2

x sin x
2 cos x
=
+

2
2(D 2 1)

x sin x cos x
=

2
2
1
1+ x2 ex
P.I 2 = 2
D 1
1
= ex
(1 + x 2 )
2
(D + 1) 1
1
= ex 2
(1 + x 2 )
D + 2D

(
e (2 x
=
x

y = A e x

3x 2 + 9 x )
12
e x (2 x 3 3x 2 + 9 x )
+ Be x +
12

d2y
y = xe x sin x
dx 2
Solution: A.E : m 2 1 = 0
m = 1
C.F = A e x + Be x
1
P.I =
xe x sin x
2
D 1
1
(x sin x )
= ex
(D + 1)2 1
1
(x sin x )
= ex 2
D + 2D
19. Solve

1
2D + 2
= e x x 2
sin x
sin x
2
(2 D 1)
(D + 2 D )

2
2
sin x = sin x, = 1, putD = = 1

(2 D + 2) sin x
1
= e x x
sin x

(2 D 1)2
2D 1

(1 + 2 D ) sin x (2 D + 2)sin x
= e x x
2
(4 D 2 4 D + 1)
(1 4 D )

Put D 2 = 1
(1 + 2 D )

(2 D + 2)(3 4 D )
= e x x
sin x +
sin x
2

5
9 16 D

13
2
x

## [sin x + 2 cos x] + 8D 2 D2 + 6 sin x

= ex
9 16 D

x
(sin x + 2 cos x ) + (14 2 D ) sin x
= ex
25
5

x
(sin x + 2 cos x ) + (14 sin x 2 cos x )
P.I = e x
25
5

Complete Solution is
y = A e x + Be x +
x
(sin x + 2 cos x ) + (14 sin x 2 cos x )
ex
25
5

## METHOD OF VARIATION OF PARAMETERS

This method is very useful in finding the general solution of the second order
equation.

d2y
dy
+ a1
+ a 2 y = X [Where X is a function of x] .(1)
2
dx
dx
The complementary function of (1)
C.F = c 1 f 1 + c 2 f 2
Where c 1 , c 2 are constants and f 1 andf 2 are functions of x
Then P.I = Pf 1 +Qf 2
f2 X
dx
f 1 f f1' f 2
f1 X
dx
'
f1 f 2 f1' f 2

P=
Q=

1
2

y = c1 f 1 + c 2 f 2 + P.I

1. Solve D 2 + 4 y = sec 2 x
Solution: The A.E is m 2 +4 = 0
m = 2i
C.F = C 1 cos 2 x + C 2 sin 2 x
f 1 = cos 2 x
f 2 = sin 2 x

f 1' = 2 sin 2 x
f '2 = 2 cos 2 x
f 1 f 2' f 1' f 2 = 2 cos 2 2 x + 2 sin 2 2 x
= 2 cos 2 2 x + sin 2 2 x
= 2 [1]
=2
f2 X
P=
dx
1
f 1 f 2 f1' f 2

14
sin 2 x sec 2 x
dx
2
1
1
= sin 2 x
dx
2
cos 2 x
1 2 sin 2 x
=
dx
4 cos 2 x
1
= log(cos 2 x)
4
f1 X
dx
Q=
'
f1 f 2 f1' f 2
=

cos 2 x sec 2 x
dx
2
1
1
dx
= cos 2 x
2
cos 2 x
1
= dx
2
1
= x
2
P.I = Pf 1 +Qf 2
1
1
=
log(cos 2 x) (cos2x) + x sin2x
4
2
=

## 2. Solve by the method of variation of parameters

Solution: The A.E is m 2 +1 = 0
m = i
C.F = C 1 cos x + C 2 sin x
Here f1 = cos x
f 2 = sin x
'
'
f1 = sin x
f 2 = cos x
'
'
2
f1 f 2 f 1 f 2 = cos x + sin 2 x = 1
f2 X
dx
f 1 f f1' f 2
sin x( x sin x)
=
dx
1
= x sin 2 xdx

P=

= x

1
2

(1 cos 2 x ) dx
2

1
(x x cos 2 x )dx
2
1
1
= xdx + x cos 2 xdx
2
2

d2y
+ y = x sin x
dx 2

15

=
=

Q =

1 x 2 1 sin x
cos 2 x

(1)
+ x
2 2 2 2
4

1
x2 x
+ sin 2 x + cos 2 x
4 4
8
f1 X
dx
f1 f 2' f1' f 2

(cos x) x(sin x)

1
= x sin x cos xdx
sin 2 x
= x
dx
2
=

dx

1
x sin 2 xdx
2
1 cos 2 x
sin 2 x
= x
(1)

2
2

4
x
1
= cos 2 x + sin 2 x
4
8
=

P.I = Pf 1 +Qf 2
x2 x

1
1
x

=
+ sin 2 x + cos 2 x cos x +
cos 2 x + sin 2 x sin x
4
8
8
4

## 4. Solve (D 2 4 D + 4) y = e 2 x by the method of variation of parameters.

Solution: The A.E is m 2 4m + 4 = 0
(m 2)2 = 0
m = 2,2
C.F = ( Ax + B )e 2 x
= Axe 2 x + Be 2 x
f1 = xe 2 x
f 2 = e2x
f1' = xe 2 x 2 + e 2 x , f 2' = 2e 2 x
f1 f 2' f 2 f1' = 2 x(e 2 x ) e 2 x (2 xe 2 x + e 2 x )
2

= 2x (e 2 x ) 2 x(e 2 x ) (e 2 x )
2

= (e 2 x ) [2 x 2 x 1]
2

= (e 2 x )
= e4x

P=

f X
dx
f f f f2

2
1
'
1 2
1
2x 2x

e e
dx
e 4 x

16
= dx = x
Q=

f1 X
dx
'
1 f f1 f 2
'
2

xe 2 x e 2 x
e 4 x dx
= xdx
=

x2
2

P.I = x 2 e 2 x

x 2 2x x 2 2x
e =
e
2
2

y = C.F + P.I
x 2 2x
= (Ax +B) e +
e
2
5. Use the method of variation of parameters to solve (D 2 + 1)y = sec x
Solution: Given (D 2 + 1)y = sec x
The A.E is m 2 + 1 = 0
m = i
C.F = c1 cos x + c 2 sin x
= c1 f1 + c 2 f 2
f1 = cos x, f 2 = sin x
f1' = sin x, f 2' cos x
f1 f 2' f 1' f 2 = cos 2 x + sin 2 x = 1
2x

f2 X
dx
f 1 f f1' f 2
sin x sec x
=
dx
1
sin x
=
dx
cos x
= tan xdx

P=

1
2

= log (cos x)
f1 X
Q=
dx
'
f1 f 2 f1' f 2
cos x sec x
dx
1
= dx
=

=x
P.I = Pf 1 + Qf 2
= log(cos x) cos x + x sin x
y = c1 cos x + c 2 sin x + log(cos x) cos x + x sin x

17
6. Solve (D 2 + a 2 )y = tan ax by the method of variation of parameters.
Solution: Given (D 2 + a 2 )y = tan ax
The A.E is m 2 + a 2 = 0
m = ai
C.F = c1 cos ax + c 2 sin ax
f1 = cos ax, f 2 = sin ax
f1' = a sin x, f 2' = a cos ax
f1 f 2' f 2 f1' = a cos ax cos ax sin ax( a sin ax)
= a cos 2 ax + a sin 2 ax
= a(cos 2 ax + sin 2 ax)
=a
P.I = Pf 1 + Qf 2
f2 X
dx
f 1 f f1' f 2
sin ax tan ax
=
dx
a
1 sin 2 ax
=
dx
a cos ax
1 1 cos 2 ax
=
dx
a cos ax
1
(sec ax cos ax )dx
=
a
1 1
sin ax
(
)
=
log
sec
ax
+
tan
ax

a a
a
1
= 2 [log(sec ax + tan ax ) sin ax ]
a
1
= 2 [sin ax log(sec ax + tan ax )]
a

P=

Q=
=

1
2

f1 X
dx
'
1 f f1 f 2
'
2

cos ax tan ax
dx
a

1
sin axdx
a
1
= 2 cos ax
a
P.I = Pf1 + Qf 2
1
1
= 2 cos ax[sin ax log(sec ax + tan ax )] 2 sin ax[cos ax ]
a
a
=

18
1
[cos ax log(sec ax + tan ax )]
a2
y = C.F + P.I
1
= c1 cos ax + c 2 sin ax 2 [cos ax log(sec ax + tan ax )]
a
=

d2y
7. Solve
+ y = tan x by the method of variation of parameters.
dx 2
Solution: The A.E is m 2 + 1 = 0
m = i
C.F = c1 cos x + c 2 sin x
Here f1 = cos x, f 2 = sin x
f1' = sin x, f 2' = cos x
f1 f 2' f 2 f1' = cos 2 x + sin 2 x = 1
f2 X
P=
dx
1
f 1 f 2 f1' f 2
sin x tan x
=
dx
1
sin 2 x
=
dx
cos x
1 cos 2 x
=
dx
cos x
= (sec x cos x )dx

## = log(sec x + tan x) + sin x

f1 X
Q=
dx
'
f1 f 2 f1' f 2
cos x tan x

1
= sin dx
=

dx

= cos x
P.I = Pf1 + Qf 2
= cos x log(sec x + tan x)
y = C.F + P.I
= c1 cos x + c 2 sin x cos x log(sec x + tan x)
8. Solve by method of variation of parameters y ''

4 ' 4
y + 2 y = x2 +1
x
x

4 ' 4
y + 2 y = x2 +1
x
x
2 ''
i.e., x y 4 xy ' + 4 y = x 4 + x 2
i.e., x 2 D 2 4 xD + 4 y = x 4 + x 2 .(1)
Put x = e z

## Solution: Given y ''

19

Logx = log e z
=z
So that XD = D '
x 2 D = D ' (D ' 1)

## (1) D ' (D ' 1) 4 D ' + 4 y = (e z ) + (e z )

[D

'2

5D ' + 4 y = e 4 z + e

2z

A.E is m 2 5m + 4 = 0
(m 4)(m 1) = 0
m = 1,4
C.F = c1e 4 z + c 2 e z
Here f1 = e 4 z , f 2 = e z
f1' = e 4 z , f 2' = e z
f1 f 2' f 2 f1' = e 5 z 4e 5 z = 3e 5 z
P.I = Pf1 + Qf 2
f2 X
P=
dx
1
f 1 f 2 f1' f 2

e z e4z + e2z
dx
3e 5 z
e5z + e3z
=
dx
3e 5 z
1
= 1 + e 2 z dz
3
1
e 2 z
= z +

3
2

1
1
z e 2 z
3
6
f1 Z
dz
f1 f 2' f 1' f 2
=

Q=

e 4 z (e 4 z + e 3 z )
3e 5 z dz
1 e8z + e 6 z
=
dz
3
e5z
1
= e 3 z + e z dz
3

1 e3z
=
+ ez
3 3

1
1
= e3z e z
9
3
1
1
1

1
P.I = z e 2 z e 4 z + e 3 z e z e z
6
3
3

20
1
1
1
1
= ze 4 z e 2 z e 4 z e 2 z
3
6
9
3

y = C.F + P.I
1
1
1
1
= c1e 4 z + c 2 e z + ze 4 z e 2 z e 4 z e 2 z
3
6
9
3
4
4
2
2
1
1
1
1
= c1 (e z ) 4 + c 2 e z + z (e z ) (e z ) e z (e z )
3
6
9
3
4
4
2
4
x
x
x
= c1 (e z ) + c 2 (e z ) +
log x

3
9
2

( )

## DIFFERENTIAL EQUATIONS FOR THE VARIABLE COEFFICIENTS

(CAUCHYS HOMOGENEOUS LINEAR EQUATION)
Consider homogeneous linear differential equation as:
n 1
dny
y
n 1 d
+
a
x
+ ........ + a n y = X ....................(1)
1
n
x 1
dx
dx
(Here as are constants and X be a function of X) is called Cauchys
homogeneous linear equation.
dny
d n 1 y
dy
a n x n n + a n 1 n 1 + .............. + a1 x + a 0 y = f ( x)
dx
dx
dx
is the homogeneous linear equation with variable coefficients. It is also known
as Eulers equation.
Equation (1) can be transformed into a linear equation of constant Coefficients
by the transformation.
x = e z , (or ) z = log x
Then
dy dy dz 1 dy
Dy =
=
=
dx dz dx x dz
d
d

D = dx , = dz
d
d
x
= xD =
=
dx
dz
Similarly
d 2 y d dy d 1 dy
= =

dx 2 dx dx dx x dz
1 dy 1 d dy
1 dy 1 d dy dz
= 2
+
+
= 2

x dz x dx dz
x dz x dz dz dx

an

d2y
1 dy 1 d 2 y
=

+
dx 2
x 2 dz x 2 dz 2
d 2 y d 2 y dy
x2 2 = 2
dz
dx
dz

21

x 2 D 2 = 2 = ( 1)
Similarly,
x 3 D 3 = ( 1)( 2)

x 4 D 4 = ( 1)( 2)( 3)

and soon.
xD =
x 2 D 2 = ( 1)

x 3 D 3 = ( 1)( 2)

and so on.

dy
d2y
+ x + y = 4 sin(log x)
2
dx
dx
Solution: Consider the transformation

1. Solve x 2

x = e z , (or ) z = log x
xD =
x 2 D 2 = ( 1)
(x 2 D 2 + xD + 1)y = 4 sin(log x)
( 2 + 1)y = 4 sin(z )
R.H.S = 0 : ( 2 + 1)y = 0
A.E : m 2 + 1 = 0, m = i
C.F = A cosz + B sinz
C.F = A cos (log x) +B sin (log x)
1
4(sin z )
P.I = 2
+1
z cos z
= 4
= 2 z cos z
2

## P.I = 2 log x cos(log x)

Complete solution is:
y = A cos (log x) +B sin (log x) 2 log x cos(log x)
y = ( A 2 log x ) cos(log x) 2 log x cos(log x)
d2y
dy
+ 4 x + 2 y = x log x
2
dx
dx
Solution: Given (x 2 D 2 + 4 xD + 2)y = x log x (1)
Consider: x = e z , (or ) z = log x
xD =
x 2 D 2 = ( 1)
(1) : ( ( 1) + 4 + 2) y = e z z

2. Solve x 2

22

+ 3 + 2 y = ze z
A.E : m 2 + 3m + 2 = 0
M = - 2, -1
2
1
C.F = Ae 2 z + Be z = Ae log x + Be log x
A B
C.F = 2 +
x
x
1
P.I = 2
ez z
+ 3 + 2
1
=e z
z
2
( + 1) + 3( + 1) + 2
1
= ez 2
z
+ 5 + 6
2

)( )

ez
=
6

2 + 5
1 +

6 z

ez 5
ez
5
1 z = z
6 6
6
6
log x
e
5

=
log x
6
6
x
5
= log x
6
6
Complete solution is y = C.F +P.I
A B x
5
= 2 + + log x
x 6
6
x

## 3. Solve ( ( x 2 D 2 3xD + 4) y = x 2 , giventhat y(1) = 1 and y(1) = 0

Solution: Given x 2 D 2 3xD + 4 y = x 2 (1)
Take x = e z , (or ) z = log x
xD =
x 2 D 2 = ( 1)
(1) : 2 4 + 4 y = e 2 z
A.E : m 2 4m + 4 = 0, m = 2,2
C.F = ( A + Bz )e 2 z = ( A + B log x )x 2
1
1
P.I =
e2z = e2z
(1)
2
( 2)
( + 2 2)2
1
= e 2 z 2 (1)

( )

P.I = e 2 z

z2
2

23

x 2 (log x )
2

(2)

x 2 (log x )
2

## Apply conditions: y(1) =1,y(1) = 0 in (2)

A = 1, B = -2

x 2 (log x )
Complete solution is y = (1 2 log x )x +
2

## EQUATION REDUCIBLE TO THE HOMOGENEOUS LINEAR

FORM (LEGENDRE LINEAR EQUATION)
It is of the form:

(a + bx )n d

dx n

+ A1 (a + bx )

n 1

d n 1 y
+ ........... + An y = f ( x) ..(
dx n 1

1)

## A1 , A2,......................., An are some constants

It can be reduced to linear differential equation with constant
Coefficients,
by taking: a + bx = e z (or ) z = log(a + bx)
d
d
Consider
= D, = , gives
dx
dz
dy
dy
(a + bx ) = b (a + bx )Dy = b ( y )
dx
dz
2
Similarly (a + bx ) D 2 y = b 2 ( 1) y (2)

(a + bx )3 D 3 y = b 3 ( 1)( 2) and so on

## Substitute (2) in (1) gives: the linear differential equation of

constant Coefficients.
Solve (2 x + 3) y ' '(2 x + 3) y '12 y = 6 x
Solution: This is Legendres linear equation:
(2 x + 3)2 y' '(2 x + 3) y '12 y = 6 x .(1)
Put z = log (2x + 3) , e z = 2 x + 3
(2 x + 3)D = 2
(2 x + 3)2 D 2 = 4 2 , = d
dz
2
Put in (1) : 4 6 12 y = 3e z 9
R.H.S =0
A.E : 4m 2 6m 12 = 0
3 + 57
3 57
m1 =
, m2 =
4
4
C.F = Ae m1z + Be m2 z
C.F = A(4 x + 3) m1 + B(2 x + 3) m2
2

24
3e z
3
= (2 x + 3)
2
14
4 6 12
z
9e
P.I 2 =
2
4 6 12
9
3
=
=
12
4
P.I 1 =

Solution is

(3+
y = A(2 x + 3)

57 / 4

+ B(2 x + 3)

3 57

3
(2 x + 3) 3
14
4

## SIMULTANEOUS FIRST ORDER LINEAR EQUATIONS WITH

CONSTANT COEFFICIENTS

dx
dy
+ 2 x + 3 y = 2e 2t , + 3x + 2 y = 0
dt
dt
dx
dy
Solution: Given
+ 2 x + 3 y = 2e 2t , + 3x + 2 y = 0
dt
dt
d
Using the operator D =
dt
2t
(D + 2)x + 3 y = 2e (1)
3 x + (D + 2 ) y = 0 .(2)
Solving (1) and (2) eliminate (x) :
3 (1) (2) ( D + 2) (D 2 + 4 D 5)y = 6e 2t ..............(3)
A.E : m 2 + 4m 5 = 0
m = 1,-5
C.F = Ae t + Be 5t
6e 2t
6
P.I = 2
= e 2t
7
D + 4D 5
6
y = Ae t + Be 5t e 2t
7
1
put in (1) : x = [(D + 2) y ]
3
8
x = Ae t + Be 5t + e 2t
7
solution is :
1. Solve the simultaneous equations,

8
x = Ae t + Be 5t + e 2t
7
6
y = Ae t + Be 5t e 2t
7
dx
dy
+ y = sin t , + x = cos t , giventhat t=0, x = 1, y =0
dt
dt
Solution: Dx + y = sin t ..(1)
x + Dy = cost (2)
Eliminate x : (1) (2)D y D 2 y = sin t + sin t

2. Solve

25

(D

1 y = 2 sin t...............(3)
m 1 = 0, m = 1
C.F = Ae t + Be t
sin t
sin t
P.I = 2 2
= (2)
= sin t
11
D 1
y = Ae t + Be t + sint
(2) : x = cost D(y)
d
Ae t + Be t + sin t
x = cost dt
x = cos t Ae t + Be t cos t
x = Ae t + Be t
Now using the conditions given, we can find A and B
t = 0, x = 1 1 = A + B
t = 0, y = 0 0 = A + B
1
1
B= ,A=
2
2
Solution is
1
1
x = e t + e t = cosh t
2
2
2

1
1
y = e t + e t + sin t = sin t sinh t
2
2
dx
dy
+ 2 y = sin t , 2 x = cos t
dt
dt
Solution: Dx + 2y = -sin t ..(1)
- 2x +Dy = cos t ...(2)
(1) 2 + (2) D 4 y + D 2 y = 2 sin t + D(cos t )
(D 2 + 4) = 3 sin t
m 2 + 4 = 0, m = i 2
C.F = A cos 2t + B sin 2t
3 sin t
3 sin t
P.I = 2
=
= sin t
D + 4 1+ 4
y = A cos 2t + B sin 2t - sin t
1
(2) : x = [Dy cos t ]
2
1d

2 dt

Solution is :

3. Solve

## x = A cos2t +Bsin2t cost

y = A cos 2t + B sin 2t - sin t

26
dx dy
dx dy

+ 2 y = cos 2t , +
2 x = sin 2t
dt dt
dt dt
Solution: Dx + (-D +2)y = cos 2t ..(1)
(D 2)x +Dy = sin 2t (2)
Eliminating y from (1) and (2)
(D 2 2D + 2)x = 2 sin 2t + cos 2t
R.H.S = 0 m 2 2m + 2 = 0
m = 1 i
t
C.F = e ( A cos t + B sin t )
( 2 sin 2t ) = sin 2t
P.I 1 = 2
D 2D + 2 1 + D
(1 D ) sin 2t = sin 2t D(sin 2t )
=
1+ 4
1 D2
sin 2t 2 cos 2t
=
5
1
(cos 2t )
P.I 2 = 2
D 2D + 2
(cos 2t + 2 sin 2t )
=
10
4. Solve

x = e t ( A cos t + B sin t ) +

## sin 2t 2 cos 2t (cos 2t + 2 sin 2t )

5
10

dx
+ 2 y 2 x = cos 2t + sin 2t
dt
dx
2y = cos2t +sin 2t + 2x - 2
dt
1
dx
y = cos 2t + sin 2t + 2 x 2 (3)
2
dt
Substitute x in (3)
sin 2t
y = e t ( A cos t B sin t )
2
Solution is :
(1) +(2) 2

## sin 2t 2 cos 2t (cos 2t + 2 sin 2t )

5
10
sin 2t
y = e t ( A cos t B sin t )
2

x = e t ( A cos t + B sin t ) +
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