This document provides an overview of ordinary differential equations with constant coefficients. It defines key terms like order, degree, homogeneous and non-homogeneous equations. It describes the general forms of linear differential equations and how to find the complementary function and particular integral to determine the general solution. Specifically, it outlines four cases for determining the complementary function based on whether the roots of the auxiliary equation are real/complex and distinct/repeated. It also includes two examples of solving second and fourth order linear differential equations.

1. UNIT-III
ORDINARY DIFFERENTIAL
EQUATIONS
COURSE CODE-MAIR 11

2. Contents
Linear Higher Order Ordinary Differential
Equation with Constant Coefficients
Solution of Homogeneous and Non-
Homogeneous Equations
Method of Undetermined Coefficients
Method of variation of parameters
Equations Reducible to Linear Equations with
Constant Coefficients

3. Differential Equation
 A Differential Equation is an equation containing the
derivative of one or more dependent variables with respect to
one or more independent variables.
 Example: 3
2
2
2 0
d y dy
dx
dx
 
 
 
 
2 2
2 2
u u u
t
x y
  
 

 
3 2
2
3 2
0
d y d y
x
dx dx
 
 
 
 
 
 
2 2
z z
z x y
x y
 
  
 

4. Types of Differential Equation
 Ordinary Differential Equation: A differential equation
involving derivatives with respect to a single independent
variables is called Ordinary Differential Equation or (ODE) for
short.
 Partial Differential Equation:
A differential equation involving partial derivatives with
respect to more than one independent variables is called Partial
Differential Equation or (PDE) for short.
 Example: (ODE)
(PDE)
3
2
2
2 0
d y dy
dx
dx
 
 
 
 
2 2
2 2
u u u
t
x y
  
 

 

5. Order and Degree of a Differential equation
 Order of a differential equation: Order of a differential
equation is defined as the order of the highest order derivative
involved in the given differential equation.
 Degree of a differential equation: Degree of a differential
equation is defined as the degree of the highest order
derivative involved in the given differential equation after
made it free from radicals and fractions as far as.
 Examples : Order-2, Degree-1
Order-3, Degree-1
2
2
0
d y
y
dx
 
3 2
2
3 2
0
d y d y
x
dx dx
 
 
 
 
 

6. Linear and Non-linear Differential Equations
 Linear differential Equation:
A differential equation is called linear if every dependent
variable and every derivative involved occurs in the first
degree only and no products of dependent variables and
derivatives occur.
 Non-Linear differential Equation:
A differential equation which is not linear is called a non-
linear differential equation.
Example:
3 5
2
2
( )
d y dy
x y x Non Linear
dx
dx
   
   
   
   
 
2
2
0 ( )
d y
y Linear
dx
 

7. Solution of a differential equation
 Any relation between the dependent and independent variables, when
substituted in the differential equation, reduces it an identity is called a
solution or integral of the differential equation.
 A solution of a differential equation does not involve the derivatives of the
dependent variable with respect to the independent variable.
 Example: is a solution of because by putting
in the given differential equation reduces to the identity. Observe that
is a solution of the given differential equation for any real constant c which
is called an arbitrary constant.
2x
y ce
 2
dy
y
dx
 2x
y ce

2x
y ce


8. General Solution and Particular Solution
 Let
be an nth order ordinary differential equation.
 General Solution: A solution of (1) containing n arbitrary
constants is called a general solution.
 Particular Solution: A solution of (1) obtained from a general
solution of (1) by giving particular values to one or more of
the n independent arbitrary constants is called a particular
solution of (1).
   
, , , , ,....... 0 1
F x y y y y
   

9. Existence and Uniqueness theorem
 Statement:
 
1 2
1 2
1 2
.................. 1
n n n
n
n n n
d y d y d y
a a a y X
dx dx dx
 
 
    
1
0
0 1 1
2 ( ) continuous
on an interval I which contains a point x .
, ,......., tan
( ), ( ),........ (
(
)
. ,
n
n
where and X x are functions
Let c c c be n cons ts Then thereexists a unique solution
on I of the nth order equ
a x a x a x
ation


 
1
0 0 0 1 0 1
1) .
(x ) , (x ) ,..........., (x )
n
n
satisfying theinitial conditions
c c c
   


  
Consider a nth order linear differential equation

10. Linear Differential Equations with Constant Coefficients
Linear Equations: Linear differential equations are those in which
dependent variable and its differential coefficients occur only in
the first degree and are not multiplied together.
The general linear differential equation of nth order is of the form
Linear Differential Equations with Constant Coefficients
where are constants and X is any function of x.
 
1 2
1 2
1 2
.................. 1
n n n
n
n n n
d y d y d y
a a a y X
dx dx dx
 
 
    
 
1 2
1 2
1 2
.................. 2
n n n
n
n n n
d y d y d y
a a a y X
dx dx dx
 
 
    
1 2
, ,........ n
a a a
1 2
, ,....... .
. n
where and X are functions of x o
a a nly
a

11. Homogeneous and non-homogeneous linear
differential equations with constant coefficients
 If X=0 in equation (2), then diff. equation is called
homogeneous otherwise it is called non-homogeneous linear
diff. eq. with constant coefficient.
 For example:
2
2
2
2
5 6 0
6 9 sin 2
d y dy
y Homogeneous
dx
dx
d y dy
y x non Homogeneous
dx
dx
       
        

12. Operator D:
The operators are denoted by
The index of D indicates the number of times the operation of
differentiation must be carried out. For example, shows that we
must differentiate three times.
Eq. (2) in symbolic form can be written as
f(D) now acts as operator and operates on y to yield X
Auxiliary Equation (A.E.):
Suppose linear diff. eq. (L.D.E.)
Auxiliary equation (A.E.) is
2 3
2 3
, , ,.......
d d d
dx dx dx
2 3
, , ,........
D D D
 
1 2
1 2 ............... 3
n n n
n
D y a D y a D y a y X
 
    
    1 2
1 2
. . ...............
n n n
n
i e f D y X where f D D a D a D a
 
     
 
f D y X

  0
f D 
1 2
1 2 ............... 0
n n n
n
D a D a D a
 
    
3 4
D x
4
x

13. General Solution of L.D.E.
1 2
1 1 2 2
1 2
( ) 0 (1)
, ,..........,
(1) ..........
(1), , ,.....,
n
n n
n
Consider thedifferential equation f D y
If y y y arenlinearlyindependent solutions of given
differential equation then c y c y c y is also a
solutionof wherec c c being ar

  
1 1 2 2
tan .
.......... ( )
(1).
( ) 0 (2)
( ) (3)
( ) (4)
n n
bitrary cons ts
Thus c y c y c y u say isthe general solutionof
equation
f D u
Let vbeany particular solutionof f D y X
and hence f D v X
   
 



14. 1 1 2 2
1 1 2 2
, ( )( ) ( ) ( ) 0 ( sin (2) (4)).
( ) . . ..........
(3
( .
).
.......... .)
,
n n
n n
Now f D u v f D u f D v X u g and
This showsthat u v i e c y c y c y v isthe general
solutionof
The part c y c y c y is know Complimentary function
na C F
and v notin
s
    
    
  
 
 
 
tan . . .
0.
2.
. . . .
:1. . .
i
volving any arbitrarycons t iscalled Particular integral P I
General solution of is given by
y C F P I
Note Complimentary functionis the solutio
f D y X
n of the D E
Partic a
D y
l r
f
u
 


 
nt egral appears d f
ue D
t X in y
o X

General Solution of L.D.E.(Continue)

15. Determination of Complimentary function(C.F.)
 Consider the linear diff. eq. (L.D.E.)
Write A.E.
Solve A.E.
 Suppose are ‘n’ roots of the A.E.
Case I. If all the roots of the Auxiliary equation(A.E.) are real and
distinct then complete solution is
Case II. If two roots of the A.E. are equal (i.e. )
then complete solution is
  0
f D y 
  0
f D 
1 2
1 2
. . ............... 0
n n n
n
i e D a D a D a
 
    
1 2 3
, , ,............., n
m m m m
1 2
1 2 .......... n
m x
m x m x
n
y c e c e c e
   
1 2
m m

  3
1
1 2 3 .......... n
m x m x
m x
n
y c c x e c e c e
    

16. If three roots of the A.E. are equal (i.e. )
then complete solution is
Case III. If one pair of roots of A.E. are complex i.e.
then complete solution is
Case IV. If two pairs of roots of A.E. are complex and equal i.e.
then complete solution is
1 2 3
m m m
 
  1 4
2
1 2 3 4 .......... n
m x
m x m x
n
y c c x c x e c e c e
     
1 2
,
m i m
  
  
  3
1 2 3
cos sin .......... n
m x m x
x
n
y e c x c x c e c e

 
    
    5
1 2 3 4 5
cos sin .......... n
m x m x
x
n
y e c c x x c c x x c e c e

 
 
      
 

17.  
  
2
2
2
2
3 4
1 2
:1. 7 12 0.
: .
7 12 0
. . 7 12 0
3 4 0
3,4
, x x
d y dy
Example Solve y
dx
dx
Solution Given eq in symbolic form is
D D y
A E is D D
D D
D
Hence the complete solutionis y c e c e
  
  
  
  

 

18.  
 
 
   
  
   
   
4
4
4 2 2
2 2
2
2 2
2 2
1 2 3 4
:2. 4 0
: . . 4 0
4 4 4 0
2 2 0
2 2 2 2 0
2 2 0 2 2 0
1 1
,
cos sin cos sin
t x
Example Solve D y
Solution A E is D
D D D
D D
D D D D
Either D D or D D
D i and D i
Hence the complete solution is
x e c x c x e c x c x

 
 
   
  
    
     
    
   

19. Problems
 Solve the following differential equations:
 
   
 
 
 
4 2
1 2 3 4
4 3 2
3 3 2
1 2 3 4
2
1. 8 16 0.
: cos2 sin 2
2. 4 5 36 36 0.
:
3. 4 0, (0) 2 (0) 0.
: 2cos2
x x x
D D y
Ans y c c x x c c x x
D D D D y
Ans y c e c e c c x e
Solve D y giventhat y and y
Ans y x
 
  
   
    
   

   


20. Inverse Operator:
If then is the inverse operator of .
The particular integral of the equation is
1.
2.
Since we require only a particular integral, we shall never add a
constant of integration after integration is performed in
connection with any method of finding P.I. Hence P.I. never
contain any arbitrary constant.
 
 
1
f D X X
f D
 
 

 
 
   
1
f D
 
f D
 
f D y X

1
X X dx
D
 
1 ax ax
X e X e dx
D a


 
 
1
X
f D

21. Particular Integral in some special cases
 Consider the equation
 Symbolic form of the given eq. is
 Particular Integral (P.I.) =
Case I. When where ‘a’ is constant.
1 2
1 2
1 2
..................
n n n
n
n n n
d y d y d y
a a a y X
dx dx dx
 
 
    
 
1 2
1 2 ...............
n n n
n
D a D a D a y X
 
    
  1 2
1 2
1 1
...............
n n n
n
X X
f D D a D a D a
 

   
ax
X e

   
 
1 1
. . 0
ax ax
P I e e provided f a
f D f a
  

22. If it is a case of failure and then we proceed as
If , then
and so on.
  0
f a 
   
 
1 1
. . 0
ax ax
P I e x e provided f a
f D f a

  

  0
f a
 
   
 
2
1 1
. . 0
ax ax
P I e x e provided f a
f D f a

  

 
   
3 2
3 2
2
1: . . 6 11 6
1
: . .
6 11 6
, 0 1
1
. .
3 12 11 2
x
x
x x
Example Find P I of D D D y e
Solution P I e
D D D
Clearly f a where a Case of failure
x
P I x e e
D D
   

  
 
 
 

23.  
 
 
 
 
   
   
 
2
2
2
2
1 2
2
2 2
2 2
1 2
2 : 4 4 2sinh 2
2
: . . 4 4 0
2 0
2, 2
. .
1
. .
2
1 1
2 2
1 1
1 2 1 2
1
9
,
. . . .
x x
x x
x
x x
x x
x x
x x
e e
Example D D y x e e
Solution A E is D D
D
D
C F c c x e
P I e e
D
e e
D D
e e
e e
Hence complete solution is
y C F P I
c c x







 

     
 
 
  
 
  
 
 

 
 
 
  
 
 
  2 1
9
x x x
e e e
 
 

24. Case II.
Replace in
If , then above rule fails and we proceed further.
If ,then
   
sin cos
When X ax b or ax b
  
 
1
. .
P I X
f D

2 2
D by a
  
f D
 
 
 
   
2
2 2
1 1
sin sin , 0
ax b ax b provided f a
f D f a
    

 
2
0
f a
 
 
 
 
   
2
2 2
1 1
sin sin , 0
ax b x ax b provided f a
f D f a

    
 
 
2
0
f a
  
 
 
 
   
2 2
2 2
1 1
sin sin , 0
ax b x ax b provided f a
f D f a

    
 

25.  
 
2
2
2
1 2
2
2 2
2 2
1: 9 cos 2 sin 2
: . . 9 0
3
. . cos3 sin 3
1
. . cos 2 sin 2
9
1 1
cos 2 sin 2
9 9
cos 2 sin 2 cos 2 sin 2
5 5
2 9 2 9
1
cos 2 sin 2
5
,
d y
Example Solve y x x
dx
Solution A E is D
D i
C F c x c x
P I x x
D
x x
D D
x x x x
x x
Hence complete solution is
y
  
 
 
  
 

 
 
   
   
 
 
1 2
. . . .
1
cos3 sin 3 cos 2 sin 2
5
C F P I
c x c x x x
 
   

26.   
  
 
4
4
4
4 4
2 2 2 2
2 2 2 2
2 2
2 2
2 3
2: . . sin .
1
: . . sin
1
sin
1 1
sin
1 1 1
sin sin
2
2 2
1
cos cos
2
2 4
d y
Example Find P I of m y mx
dx
Solution P I mx
D m
mx
D m D m
mx
D m m m
x
mx mx
D
m m
D m
x x
mx mx
m
m m
 



 

  
 
     
 

 
    
 
 

27. Case III. where m is a positive integer
 Bring out the lowest degree term from f(D) so that the
remaining factor in the denominator is of the form
or , n being a positive integer.
 Take in the numerator so that it takes
the form
 Expand by Binomial expansion.
m
When X x

 
 
1
1
. . m m
P I x f D x
f D

   
 
 
1
n
D


 
 
 
1
n
D


 
 
   
1 1
n n
D or D
 
 
   
   
   
1 1
n n
D or D
 
 
 
   
   
   
1 1
n n
D or D
 
 
 
   
   

28.  
 
 
 
1 2 3 4
1 2 3 4
2 2 3 4
2 2 3 4
1 1 ......................
1 1 ......................
1 1 2 3 4 5 ...............
1 1 2 3 4 5 ...............
x x x x x
x x x x x
x x x x x
x x x x x




      
      
      
      
The following Binomial expansions should be remembered.
Some Useful Formulae:

29.  
  
 
   
 
3 4
3
2
2
1 2 3
4 4
3
3
1
3 4 3 4
1: 8 2 1.
: 8 0
2 2 4 0
2, 1 3
. . cos 3 sin 3
1 1
. . 2 1 2 1
1
8 8 1
8
1 1 1 1
1 2 1 1 ...... 2
8 8 8 8
x x
Example D y x x
Solution Theauxiliaryequationis D
D D D
D i
C F c e e c x c x
P I x x x x
D D
D x x D x


   
 
   
  
   
     
 
 
 
 
   
       
   
   
 
     
 
4 3 4 4
4
2 4
1 2 3
1
1 1 1
2 1 2 1 2 1 3
8 8 8
1
1
8
, . . . .
1
cos 3 sin 3 1
8
x x
x
x x D x x x x x
x x
Hence thecompletesolutionis y C F P I
c e e c x c x x x


   
         
   
 
 
  
 
 
 
     
 

30. Problems
 Solve the following differential equations:
 
 
 
   
2
1 2
4 2
1 2 3 4
1. 2 1 1.
: 1
2. 8 16 16 256.
: cos2 sin 2 16
x
D D y x
Ans y c c x e x
D D y x
Ans y c c x x c c x x x
   
   
   
     

31. Case IV. , where V is any function of x.
ax
When X e V

 
   
1 1
. . ax ax
P I e V e V
f D f D a
 

 
   
2
2
2
2
1: . . 2 4 cos
1
: . . cos
2 4
1
cos
1 2 1 4
1
cos cos
2
3
x
x
x
x
x
Example Find P I of D D y e x
Solution P I e x
D D
e x
D D
e
e x x
D
  

 

   
 


32.  
 
 
 
 
2 2
2
2
1 2
2 2
2 2
2 2
2 2
3 3 4
2 4
2: 2 1 .
: 2 1 0
1 0
1,1
. .
1 1
. .
2 1 1
1 1
1 1
1 1 1 1
3 3 3 4 12
,
x
x
x x
x x
x
x x x x
Example D D y x e
Solution Theauxiliary equationis D D
D
D
C F c c x e
P I x e x e
D D D
e x e x
D
D
x x e x
e x dx e e dx e x
D D D
Hence thecomplete solu
  
  
 

  
 
  
 
 
    
 
  4
1 2
. . . .
1
12
x x
tionis y C F P I
c c x e e x
 
  

33. Case V. When X is any other function of x
Resolve into partial fractions and operate each partial
fraction on X
 
1
. .
P I X
f D

 
1
f D
1 ax ax
X e Xe dx
D a


 
 
2
2
2
2 2
2 2
1 2
: sec
: .
sec
. . 0
. . cos sin
d y
Example Solve a y ax
dx
Solution Symbolic form of given eq is
D a y ax
A E is D a D ia
C F c ax c ax
 
 
    
 

34.   
   
 
 
 
2 2
1 1
. . sec sec
1 1 1
sec
2
1
sec sec
cos sin
cos
1 tan log cos
1
sec log cos
1
. .
2
iax iax
iax
iax iax
iax
iax
P I ax ax
D ia D ia
D a
ax
ia D ia D ia
ax e ax e dx
D ia
ax i ax
e dx
ax
i
e i ax dx e x ax
a
i
ax e x ax
D ia a
i
P I e x
ia


 
 

 
 
 
 
 




 
   
 
 
 
 
 
  
 



2
2
1 2 2
log cos log cos
1
log cos
2 2
1
sin log cos cos
. . . .
1
cos sin sin log cos cos
iax
iax iax iax iax
i
ax e x ax
a a
x e e e e
ax
a i a
x
ax ax ax
a a
y C F P I
x
c ax c ax ax ax ax
a a

 
 
   
 
   
 
   
 
   
 
 
   
   
   
 
 
   

35. Case VI. When , V is any function of x
P.I.=
X xV

 
 
   
1 1 1
d
xV x V V
f D f D dD f D
 
  
 
 
 
 
   
2
2
2
2
1 2
2 2 2
2
2
: 4 sin
: 4 sin
. . 4 0 2
. . cos 2 sin 2
1 1 2
. . sin sin sin
4 4 4
1 2 sin 2
sin sin cos
1 4 3 9
1 4
,
d y
Example Solve y x x
dx
Solution Symbolic form is D y x x
A E is D D i
C F c x c x
D
P I x x x x x
D D D
D x x
x x x x
Hence thecomplete solution
 
 
    
 
  
  
   
   
1 2
. . . .
sin 2
cos 2 sin 2 cos
3 9
is y C F P I
x x
c x c x x
 
   

36. The formula that is used in Case-VI should not be used when
V=sinax or cosax and vanishes by putting for
 In such situations we shall apply the following alternative
method. This alternative method can also be used even when
 
2
f D 2
a
 2
D
 
2
0
f a
 
   
 
 
   
 
 
sin cos
1 1
. . cos cos sin
1
. .
1 1
. . sin Im cos sin
1
. .
. . m m
m m
m ax
m m
m ax
A X x ax or x ax
P I x ax Real part of x ax i ax
f D f D
R P of x e
f D
P I x ax aginary part of x ax i ax
f D f
lternative Method for finding
D
I P of x e
h
D
P w n
f
I e 
  

  


37.  
 
 
2 2
2
1 2
2 2 2
2 2
2 2 2 2
2 2
2 2
2
2
1: 1 sin 2 .
: 1 0
. . cos sin
1 1
. . sin 2 . .
1 1
1 1
. . . .
4 3
2 1
1
. . . .
3 1 4 / 3
ix
ix ix
ix
Example D y x x
Solution Theauxiliaryequationis D
D i
C F c x c x
P I x x I P of x e
D D
I P of e x I P of e x
D iD
D i
e e
I P of x I P of
iD D
 
 
 
  
 
 
 
 
 
 
  
 
 
1
2
2
2 2 2
2
4
1
3 3 3
4 4
. . 1 .......
3 3 3 3 3
ix
ix
iD D
x
e iD D iD D
I P of x

 
 
 
 
 
 
  
 
 
 
   
     
 
   
   
  
   
 

38.  
2 2 2
2
2 2
2
2
2
4 16
. . . . 1 .......
3 3 3 9
4 13
. . 1 .......
3 3 9
cos2 sin2 8 26
. .
3 3 9
1 26 8
sin2 cos2
3 9 3
, .
ix
ix
e iD D D
P I I P of x
e iD D
I P of x
x i x ix
I P of x
x x x x
Hence the complete solutionis y C F
 
    
 
  
 
 
   
 
  
 
  
  
 
  
 
 
   
 
 
 
 

2
1 2
. . .
1 26 8
cos sin sin2 cos2
3 9 3
P I
c x c x x x x x

 
 
    
 
 
 
 

39. Method of Undetermined Coefficients:
 This method is used to find the P.I. of L.D.E.
 Assume a trial solution containing unknown constants which
are determined by substitution in given eq.
 The trial solution to be assumed in each case, depends on the
form of X.
Function Trial Solution
1.
2.
3.
4.
5.
6.
 
f D y X

x
X e
 x
Ae
cos sin
X x or x
 
 sin cos
A x B x
 

n
X x
 1
0 1 .........
n n
n
A x A x A

  
cos sin
x x
X e x or e x
 
 
  
sin cos
x
e A x B x

 

x n
X e x

  
1
0 1 .........
x n n
n
e A x A x A
 
  
sin cos
n n
X x ax or x ax
  
 
1
0 1
1
0 1
......... sin
......... s
n n
n
n n
n
A x A x A ax
A x A x A co ax


  
   

40. Note: This method holds as long as no term in the trial solution
appears in the C.F. If any of term of the trial appears in the C.F.,
we multiply this trial solution by lowest positive integral power
of x so that no term in the trial solution and C.F. are same.
 
 
2
2
1 2
1 2
1 2
1 2 1 2
: 1 sin
: . . 1 0
. . cos sin
cos sin .
. . .
cos sin
cos sin sin co
Example Solve D y x
Solution A E is D
D i
C F c x c x
Let trial solutionis a x a x
Sincetheseterms appear in C F somultiply abovetrial sol by x
y x a x a x
Dy a x a x x a x a
 
 
 
 

  
     
 
   
1 2 2 1
s
cos sin
x
a a x x a a x x
   

41.     
   
     
2
1 2 2 2 1 1
2 1 1 2
2 1 1 2 1 2
2 1
1 2
1
sin cos cos sin
2 cos 2 sin
.,
2 cos 2 sin cos sin sin
,
2 cos 2 sin sin
2 1, 2 0
D y a a x x a x a a x x a x
a a x x a a x x
Substituting thesevaluesin giveneq we get
a a x x a a x x x a x a x x
On solving we get
a x a x x
a a
a
       
   
     
 
   

  2
1 2
1
, 0
2
,
. .
cos sin cos
2
a
Hence completesolutionis
y C F y
x
c x c x x

  
  

42. Problems
• Solve the following differential equations by using
the method of undetermined coefficient.
 
 
 
 
3 2 2
2 2
1 2 3
2
1 2
1. 3 2 4 8.
: 33 3 2
12
2. 2 1 .
1
: 2
4
x x
x
x x
D D D y x x
x
Ans y c c e c e x x
D D y x e
Ans y c c x e x e
 

    
     
   
    

43. Method of Variation of Parameter:
 This method is used to find the particular integral of second
order non-homogeneous L.D.E.(including variable coeff. as
well as constant coeff. ) whose C.F. is known.
 This method is applicable to equation of the form
where p, q and X are the functions of x .
where and are the two L.I. solutions of eq. (1).
and is called Wronskian of
 
2
2
1
d y dy
p qy X
dx
dx
  
2 1
1 2
. .
y X y X
P I y dx y dx
W W
  
 
1
y 2
y
1 2
1 2
y y
W
y y

 
1 2
, .
y y

44.  
2
2
2
2 2
2 2
1 2
1 2
1 2
1 2
2 1
1 2
: cos
: cos
. . 0
. . cos sin
cos , sin , cos
cos sin
0
sin cos
. .
c
d y
Example a y ecax
dx
Solution Symbolic form is D a y ecax
A E is D a D ai
C F c ax c ax
Let y ax y ax X ecax
y y ax ax
W a
y y a ax a ax
y X y X
P I y dx y dx
W W
 
 
    
 
  
   
  
  
 
 
2
1 2 2
sin cos cos cos
os sin
sin
cos logsin
. . .
sin
cos sin cos logsin
ax ecax ax ecax
ax dx ax dx
a a
x ax
ax ax
a a
y C F P I
x ax
c ax c ax ax ax
a a

  
 
   
 

45. Problems
• Solve the following differential equations by using
the method of variation of parameter.
 
   
2
1 2
2
1 2
1. 4 4cos 2 .
: cos2 sin 2 1 cos2 log tan
2. 2 2 tan .
: cos sin cos log sin tan
x
x x
y y ec x
Ans y c x c x x x
D D y e x
Ans y e c x c x e x x x
 
   
  
   

46. Equations reducible to linear equations with constant
coefficients
Cauchy Homogeneous Linear Differential Equation:
 Cauchy’s Homogeneous Linear Differential Equation:
where are constants and X is any function of x
is called Cauchy’s Homogeneous Linear Differential Equation.
 Here, index of x and the order of the derivative is same in each
term of such equations.
 Such equations can be reduced to L.D.E. with constant coeff.
by introducing a new variable t such that
1 2
1 2
1 2
1 2
..................
n n n
n n n
n
n n n
d y d y d y
x a x a x a y X
dx dx dx
 
 
 
    
1 2
, ,........ n
a a a
log
t
x e t x
  

47.  
  
1
1
2
2 2
2
2 2
2
2
1 1
2
3
3
1 1 1
3
1
1 1 1
1
1
, 1 2 .
d d
Let D and D
dt dx
dy dy dt dy dy dy
x D y
dx dt dx dt x dx dt
d y d dy dy d dy dt
dx x dt dt x dt dt dx
dx x
d y dy
dt
x dt
d y
x D D y
dx
d y
Similary x D D D y and soon
dx
 
    
   
   
   
   
 
 
 
 
 
 
  

48.    
    
 
  
3 2
3 2
3 2
3 3 2 2
2 3
2 3
1 1 1 1 1 1
2 3
1
1 1 1
: 3 log
: 3 1 log 1
log
, 1 , 1 2
. 1
1 2
t
d y d y dy
Example Solve x x x y x x
dx
dx dx
Solution Symbolic form is x D x D xD y x x
Put x e or t x
dy d y d y
sothat x D y x D D y x D D D y
dx dx dx
d
where D
dt
Eq canbewrittenas
D D D
    
    
 
     

   
 
  
1 1 1
3
1
3
1
2
1 1 1
1
3 1 1
, 1
. . 1 0
1 1 0
1 3
1,
2
t
t
D D D y e t
On Solving D y e t
A E is D
D D D
i
D
     
 
 
  
 
   

  

49.    
 
2
1 2 3
1 1 2
1 2 3
1
3
1
3 3
1 1
3
1
1 1 2
1 2
3 3
. . cos sin
2 2
3 3
cos log sin log
2 2
1 1
. . 1
1 1
1 1 1
1 ....... log
2 2 2
. . . .
3
cos log
2
t t
t t
t t
C F c e e c t c t
c x x c x c x
P I e t e D t
D D
e D t e t x x
y C F P I
c x x c x




 
  
 
 
 
   
  
 
   
   
 
   
 
    
 
       
 
 
  


3
3 1
sin log log
2 2
c x x x
 
 
  
 
  
  
 
  
 

50. Equations reducible to linear equations with constant
coefficients
 Legendre’s linear Equation:
where are constants and X is any function of x
is called Legendre’s linear Differential Equation.
 Here, index of and the order of the derivative is same
in each term of such equations.
 Such equations can be reduced to L.D.E. with constant coeff.
by introducing a new variable t such that
     
1 2
1 2
1 2
1 2
............
n n n
n n n
n
n n n
d y d y d y
ax b a ax b a ax b a y X
dx dx dx
 
 
 
       
1 2
, ,........ n
a a a
 
ax b

 
log
t
ax b e ax b t
    

51.  
 
 
   
    
1
1
2 2
2 2
2 2
2 2
2
2 2
1 1
2
3
3 3
1 1 1
3
1
, 1 2
d d
Let D and D
dt dx
dy dy dt dy a dy dy
ax b a a D y
dx dt dx dt ax b dx dt
d y a dy a dy a d dy dt
ax b dt dt ax b dt dt dx
dx ax b
a d y dy
dt
dt
ax b
d y
ax b a D D y
dx
d y
Similarly ax b a D D D
dx
 
     

   
   
   
 
   

 
 
 
 
  
  
    .
y and soon

52.        
       
 
         
 
     
4 3 2
3 2
4 3 2
3 2
3 2 2
3 2
2
2
1 1 1
2
1
: 1 2 1 1 1
1
1
: 1 2 1 1 1
1
1 ,
1 2 1 1 1 1 1
1 log 1
1 , 1 1
t
Example x D x D x D x y
x
Solution x D x D x D x y
x
Dividing both sidesby x we get
x D x D x D y x
Put x e or t x
dy d y
sothat x D y x D D
dx dx

 
       
  
 
       
  

 
       
 
   
    
    
 
    
3
3
1 1 1
3
1
2
1 1 1 1 1 1
,
1 1 2
. 1
1 2 2 1 1 t
y
d y
x D D D y
dx
d
where D
dt
Eq canbe writtenas
D D D D D D y e
   

      
 
 

53.  
   
 
 
    
   
   
 
 
      
3 2
1 1 1
2
1 1
1
1 2 3
1
1 2 3
2
2
1 1
2
2
2
2
1 2
1 2 3
. . 1 0
1 1 0
1,1, 1
. .
log 1 1 1
1
. .
1 1
1
2 1 2 1
1 1
1
9 9
. . . .
1
log 1 1 1 1
9
t t
t
t
t
A E is D D D
D D
D
C F c c t e c e
c c x x c x
P I e
D D
e
e x
y C F P I
c c x x c x x






 
   
   
  
  
     

 

   
    
 
        

54. Problems
• Solve the following differential equations:
     
       
 
2
1 2
2 2 2
2
2
1 2
1. 1 1 4cos log 1 .
: cos log 1 sin log 1 2log 1 sin log 1
2. 3 5 sin log .
: coslog sin log log coslog
2
x y x y y x
Ans y c x c x x x
x D y x Dy y x x
x
Ans y x c x c x x x
 
     
      
  
  