The document discusses methods for finding the general solution to linear differential equations of second order with constant coefficients. It presents four types of complementary functions depending on whether the roots of the auxiliary equation are real and distinct, real and equal, complex, or surd roots. It also describes four types of particular integrals depending on whether the given function is an exponential, sine, cosine, or contains an exponential term. The document provides examples of solving differential equations of each type and includes multiple choice questions to test understanding of the concepts and methods presented.

1. 1
MATHEMATICS
E-Content
On
Liner Differential Equation
Vinay M. Raut
Associate Professor
Shri Shivaji Science College, Amravati

2. 2
Liner differential Equation
Notation:
1) d dy
D Dy
dx dx
  
2
2
2 2
2 2
d y
d
D D y
dx dx
  
So on
2) P(D) = polynomial operator
Fundamental theorem of Algebra
The nth degree equation has at least one root and not more than ‘n’
distinct roots.
Linear D.E. of Secondorder with constantcoefficient
It is given by
2
0 1 2
2
d y dy
P P P y Q
dx dx
   __________(1)
Where P0, P1, P2 are constant and Q is function of x –
GeneralSolution
The solution of equation (1) is given by G.S = C.f + P.I.
Where c.f  complementary function
P.I.  Particular Integral
G.S.  General Solution
Method of find C.F.
Consider the reduced equation
2
0 1 2
2
0
d y dy
y
dx dx
P P P
  
 2
0 1 2
( ) 0
D D y
P P P
   ______________(2)
Or ( ) 0
f D y 

3. 3
To find the C.f. For this we write or replace D by m in equation (2)
then find roots. Nature of the roots gives the respective solution. If roots
are m1 & m2 say.
A) If the roots are real and different then solution is
1 1
1 2
mx mx
Y Ce C e
  where are 1
C , 2
C constant
Ex. Solve the
2
2
. . 2 0
d y dy
D
dx dx
  
Sol. The given equation (D2 – 2D)y=0
A.E. is m2 – 2m = 0
 m (m – 2) = 0
 m = 0 or m = 2
The roots are real & different
 C.f = y = C1 e0x +C2.e2x = C1 + C2 e2x
B) If the roots are real and equal
Then Solution is y = = (C1 + C2x) emx
Where m = m1 = m2 , C1 and C2 are constant
C) If the roots are complex then solution is
Supposeroots are in the form i
 
 then solution can be written as
1 2
[ cos sin ]
x
y e C x C x

 
 
Where C1 & C2 are constant
Ex. Solve the
2
2
. . 0
d y dy
D y
dx dx
   
A.E. is m2 + m + 1 = 0
1 1 4 1 3
2 2 2
m i
  
   
 Solution is
1
2
1
3 3
[ cos sin ]
2 2
x
y e c x B x

 

4. 4
D) If the roots are surd roots
Then solution is cos h sin
x
y e A bx B bx

 
 
 
(If the roots are in the form a b
 where b is irrational)
Ex. Solve the
2
2
. . 2 4 0
d y dy
D y
dx dx
   
Solution : The given D.E. is (D2+2D – 4)y = 0
A.E. is m2 + 2m – 4 = 0

2 4 16
1 5
2
m
 
   
 Solution is 1 cosh 5 sinh 5
x
y e C x B x

 
 
 
---------------------- ********* ---------------------------------
To find particular Integral (P.I.)
Formula for
1
.
( )
P I
f D
 Q
There are four types to find P.I.
Type I. If ax
Q e

Then
1 1
. . .
( ) ( )
ax ax
P I e e
f D f a
 
Where ( ) 0
f a  ie. replace D by a.
Ex. Solve the
2
4
2
. . 5 6 x
d y dy
D y e
dx dx
   
Solution: The given D.E. (D2 – 5D + 6) y = 0
 A.E. is m2 – 5m + 6 = 0
 (m – 2) (m – 3) = 0
 m = 2, 3
 c.f = c1 e2x + c2 e3x
To find P.I.
Now 4
2
1
.
5 6
x
P I e
D D

 

5. 5
4
4
2
1
(4 ) 5(4) 6 2
x
x e
e
 
 
GS = c.f + P. I
------------------- ****************** -------------------
Type II
If Q = sin ax or cos ax then
2 2
1 1
sin cos sin
( ) ( )
ax or ax ax
P D P a


i.e. replaced D2 by – a2
This rule fails if P (-a2) = 0
Ex. Find P.I. of
2
2
2 cos3
d y dy
y x
dx dx
  
Solution : 2
1 2
. x x
C f C e C e
 
 
2
1
. . cos3
2
P I x
D D

 
Replace D2 by -3
1
cos3
9 2
x
D

  
1
cos3
11
x
D


( 11){cos3 } ( 11)
D x D
     
1
cos3
( 11)( 11)
x
D D
 
 
  

( 11)
D
   2
1
cos3
121
x
D
 
 
 

( 11)
D
  
1
cos3
9 121
x
 
 
  

(again apply rule)
1
{ cos3 11cos3 }
130
D x x
 
3sin3 11cos3
130
x x
 
G.S. = C.f + P.I
-------------- **************** ------------------

6. 6
Type III
If Q = xm where ‘m’is positive integer
Rule : 1
. . .
( )
P I y Q
f D
 
Take out lowest degree term or suitable term from denominator.
After this put in Nr and apply binomial expansion and solve it.
Note : Binomial expansion for
1 2 3 4
(1 ) 1 ____
x x x x x

      
1 2 3 4
(1 ) 1 ____
x x x x x

      
Ex. Find the P.I. of
2
3
2
. . 2.
d y dy
D x
dx dx
  
Solution : 2
1 2
. x
C f C C e
 
To find P.I. : 3
2
1
. .
2
P I x
D D


 
1
3 3
1 1
1 .
2
1
2
D
x x
D D
D

  
 

 
 
2 3
3
1
1
2 4 8
D D D
x
D
 
   
 
 
3 2 3 3 3
3
1
2 4 8
Dx D x D x
x
D
 
   
 
 
2
3
1 3 6 6
2 4 8
x x
x
D
 
   
 
 
4 3 2
3 6 6
4 2 3 4 2 8
x x x
x
 
   
 
 
1
.
as Q Q dx
D
 

 
 

4 3 2
1
( 2 3 3 )
8
x x x x
   
 G.S = C.f + P.I.
-------------------- ****************-----------------

7. 7
Type IV
If Q = eax V
To evaluate 1
.
( )
ax
e
f D
V Take out eax as first at the same time replace D by
(D+a) in f (D)
This Rule is useful when
i)
1
( ) ( )
ax
ax e
e
f D f a
 When ( ) 0
f a  in this casetake V = 1
ii) 2
1
( )
P D
sin an or (cos ax) when P (-a2) = 0 in this situation
take 2
1
sin
( )
ax
P D
as I.P. of iax
e
and 2
1
cos
( )
ax
P D
as R. P. of iax
e
Because cos sin
iax
e ax i ax
 
Ex. Solve
2
2
2
3 2 x
d y dy
y e
dx dx
  
Solution : 2
1 2
. x x
C f C e C e
 
Now P.I. 2 2
2 2
1 1
.1
3 2 ( 2) 3( 2) 2
x x
e e
D D D D
 
     
( 2)
a 
2
2
1
. .1
x
e
D D


2 1
1
. (1 ) .1
x
e D
D

  2 1
. (1 ).1
x
e D
D
  (by binomial expansion)
2 1
. .1
x
e
D

2
.
x
e x

 G.S. = C.f + P.I.
---------------------- ******************* ------------------

8. 8
MCQ
1) The form of LDE of second order with constant coefficient is given
by _________
2) General solution of L.D.E. is given by _____________
3) The Auxiliary equation of L.D.E. is given by _________
4) If the roots are district and equal them c.f. is __________
5) If the roots are repeated then C.f. is ________
6) If the roots are complex them C.f. is ________
7) It the roots are surd them C.f. is _________
8) If Q = eax them P.I. of
1
( )
y
f D
 Q is ________
9) If sin
Q ax
 then P.I. of 2
1
( )
y
P D
 Q is _________
10) If Q = eax V then P.I. of
1
.
( )
ax
y e V
f D
 is __________