Introduction to ordinary differential equations

1. 1
INTRODUCTION
A differential equation is an equation that involves one or more
derivatives e.g.
2
2
2
cos5
d y dy
xy x
dx dx
   (1)
3
2
3
6 ln
d y dy
x x
dx dx
   (2)
0
v v
s t
 
 
 
(3)
There are two types of differential equations namely
1. Ordinary Differential equations (ODE)
These are differential equations in which there is only one
independent variable.
They are ordinary derivatives of one or more dependent variable
w.r.t. a single independent variable e.g.
2
2
2
d y dy
y
dx dx
  
y – Dependent variable and x – independent variable
Equation (1) and (2) are ordinary differential equations
2. Partial differential equation (PDE)
These are equation in which dependent variable depend on more
than one independent variable. The derivatives involved are partial
2 2 2
2 2 2
v v v
x y
x y z
  
   
  
v – Dependent variable and x, y, z – independent variables.
Equation (3) and (4) are partial differential equations.
DEFINITIONS
a) Order of a differential equation
The order of a differential equation is the order of the highest
ordered derivative which occurs in the equation e.g.
2
2
5
d y dy
y x
dx dx
   Order 2
3 2
4
3 2
d y d y
y x
dx dx
   Order 3
b) degree of a differential equation
The degree of the differential equation is the degree of the highest
ordered derivative which occurs in the equation e.g.
4
2
2
2cos
d y dy
y x
dx dx
 
  
 
 
Degree 2
Equation Order Degree
4 2
4 2
6 0
d y d y
y
dx dx
  
4 1
4 5
2
2
3 x
d y dy
y e
dx dx
   
  
   
 
 
2 4
3
3
5ln
d y
x
dx

3 1
4 5
7 3 2
7 3 2
5 8 0
d y d y d y
y
dx dx dx
   
   
   
   
7 4

2. 2
CLASSIFICATION OF ORDINARY DIFFERNTIAL
EQUATIONS
Ordinary equation are classified into two groups
a) Linear differential equations
A linear differential equation of order n in the dependent variable x
is one which can be expressed in the form
         
1
1 1 0
1
................
n n
n n
n n
d y d y dy
a x a x a x a x y r x
dx dx dx

 
   
Where   0
n
a x 
and which is such that
i. Dependent variable y and its derivatives a are in their first
degree
ii. No product of y or its derivatives are present
iii. No transcendental functions of y and its derivatives occur.
e.g.
2
2
7 3 0
d y dy
y
dx dx
  
(1)
4 3 2
2 3
4 3 2
d y d y d y
x x x x
dx dx dx
  
(2)
NB
Equation (1) is a linear equation with constant coefficient while
equation (2) is a linear equation with variable coefficient
b) Non-linear differential equations
If the O.D.E. is not linear then it is non-linear.
Identify if the equation is linear or non - linear
i.
3
2
3
5 6 9
d y dy
y
dx dx
  
ii.
2
2
2
7 4 0
d y dy
y
dx dx
 
  
 
 
iii.
2
2
8 3 0
d y dy
y y
dx dx
  
ORIGIN OF DIFFERENTIAL EQUATIONS
Differential equations occur in connection to numerous problems
that one encounters in various branches of science and engineering.
i. e. differential equations may arise from
a) Geometric problems
b) Physical problems
c) Primitive.
Examples
i. A falling stone from the top of a building if x is time and y is
displacement then
acc y

Hence we can write a model
y g
 
By integrating we get 0
y gx v
  
By integrating once more we get the displacement as
0
2
0 0
1
2
y gx v dx
gx v x y
 
  


3. 3
Where 0
v is the initial velocity and 0
y is the displacement when t =
0
ii. Determination of motion of a projectile. e.g. rocket
iii. Determination of the charge or current in an electric circuit
If charge  
q t and charge  
i t at time t are related by
dq
i
dt

0
di
E iR L
dt
  
Where E is voltage, L inductance and R resistor
The general solution is
 
Rt
L
E
i t Ke
R

 
iv. Determining of the vibration of a wire or a membrane
v. Heat condition in a rod or in a slab
vi. Population growth in human, animal and bacteria e.t.c
The mathematical formulation of such problems gives rise to
different equations
Primitive
A primitive is a relation between variables which involve n –
essential arbitrary constant e.g.
3
y x ax
 
4
y Ax Bx
 
NB: Arbitrary constant are said to be essential if they cannot be
replaced by smaller numbers of constants for example given
4
y Ax Bx Cx
  
A, B and C are not all essential constants because they can be
reduced to 2 constants
 
4
4
4
y Ax Bx Cx
Ax B C x
Ax Kx
  
  
 
Where k = B+C and therefore there are only 2 essential
constants
A primitive involving n-essential arbitrary constant gives rise to
differential equations of order n free of arbitrary constant.
We obtain equation by eliminating the n contest by differentiating n
times
Example 1
Obtain the differential equation associated with 2x x
y Ae Be C
  
Solution
2
2 x x
y Ae Be
   ........................................................ (i)
2
4 x x
y Ae Be
   ........................................................ (ii)
2
8 x x
y Ae Be
   ........................................................ (iii)
Subtracting (i) from (ii) we get
2
2 x
y y Ae
 
 
Subtracting (ii) from (iii) we get
 
2
y y y y
   
   
2
4 x
y y Ae
 
 

4. 4
3 2 0
y y y
  
   
Example 2
 
3 2 2 2 5 3 4
3 2 5 2 2 3 4
2 3 3 5 0
2 3 3 5 0
dy dy
xy x y x y x y
dx dx
dy
xy x y x y x y
dx
   
   
NB: The equation is a first order since there is only one essential
arbitrary constant we can also write the equation as
   
3 2 5 2 2 3 4
2 3 3 5 0
xy x y dx x y x y dy
   
Example 3
Show that a function defined by   2
1 2
x x
f x a e a e
  where 1
a and
2
a arbitrary constants is associated with a differential equation of the
form 2 0
y y y
 
  
Exercise:
1. Obtain a differential equation associated with the primitive
a) 3
1 2 3
x x x
y a e a e a e
 
  
b) 2
y Ax

c) 2 2
9
x y B
 
2. Show that a function defined by   3
1 2
x x
f x be b e
  where
1
b and 2
b are arbitrary constants associated with the differential
equation of the form
2
2
2 3 0
d y dy
y
dx dx
  
3. Show that a function defined by
  3 3 3
1 2 3
x x x
g x c e c xe c e
   where 1
c , 2
c and 3
c are arbitrary
constants associated with the differential equation of the form
3 2
3 2
3 9 27 0
d y d y dy
y
dx dx dx
   
4. Show that a function defined by x
x
x
e
c
e
c
e
c
x
f 4
3
2
2
2
1
)
( 

 
where 1
c , 2
c and 3
c are arbitrary constants associated with the
differential equation of the form
0
16
4
4 2
2
3
3



 y
dx
dy
dx
y
d
dx
y
d
SOLUTION OF DIFFERENTIAL EQUATIONS
A solution of a differential equation is any function satisfying the
equation .In other words the solution of a differential equation is the
primitive associated with it.
Types of Solutions
General solution
This is the primitive associated with the differential equation. The
number of arbitrary constant in the solution is equal to the order of
the differential equation. It is the primitive
Particular solution
This is obtained from the general solution by giving numerical
values of the arbitrary constant e.g.

5. 5
sin cos
y A x B x
  is a general solution of
2
2
0
d y
y
dx
 
If one is given initial condition, say  
0 3
y  and   2
2
y   then
we would get the particular solution as 2sin 3cos
y x x
 
Singular Solution
Addition solution which cannot be obtained from the general
solution (e g primitive)
Exercise
Verify that the given function is a solution of the corresponding
differential equations and determine value of c using given initial
conditions
a) 1
y y
  1
x
y Ce
 
 
0 2.5
y 
b) 2
xy y
  2
y Cx

 
2 12
y 
c) yy x
  2 2
y x C
 
 
0 1
y 
SEPARABLE EQUATIONS
A differential equation is called separable if it can be written in the
form
   
y A x B y
 
In such a case we can separate the variables and write
   
 
 
1
dy
A x B y
dx
dy A x dx
B y


Then solve by integration
Example 1
Given that 2 x
y y e
  find the general solution
Solution
2
2
1
0
x
x
dy
y e
dx
dy e dx y
y



 
On integrating we get
2
1 x
dy e dx
y


 
1 x
e C
y

    Or
1
x
y
e K



This is the general equation.
Example 2
Solve the differential equation 2
1
x y y
  
Solution
2
1
dy y
dx x



6. 6
2
1 1
1
dy dx
y x


On integrating we get
 
2
1
1
1 1
1
1
ln 1
1
C
x x
dy dx
y x
y C
x
y e Ae

 


   
  
 
Or
1
1
x
y Ae

 
This is the general equation.
Example 3
Given that    
4 4 2
3 3 0
x y dx x y dy
    find the general solution
Solution
3 3
1 1 1 1
2
C
y y x x
     
Example 4
Solve the initial value problem  
4 2
5 0 0 1
y x y y
  
Solution
5
1
x C
y
   
Example 5
Solve the initial value problems  
1 3
x
y y
y
  
Solution
2 2
8
y x
 
Exercise
1. Find the general solution of the following equation
a)   3
1 0
y x y
  
b)   
2
1 1
y x y
   
c)
1
tan
y
x y
 
d) 2 2
cos
x
y e y

 
e)  
2
sin 1 cos 0
x ydx x ydy
  
2. Solve the following initial value problems
a)  
3
2 0
y
y x e y

  
b)  
0 0 2
yy x y
   
c)  
1
1 1
x
y y
y

  
EXACT DIFFERENTIAL EQUATIONS
1. A 1st
order differential equation of the form
   
, , 0
dy
P x y Q x y
dx
  is called an exact equation if there exists a
function  
,
f x y such that

7. 7
   
, ,
x
P x y f x y

 and    
, ,
y
Q x y f x y


2. A differential equation    
, , 0
M x y dx N x y dy
  is said to be
exact if the LHS is a total differential of some function say  
,
f x y
i.e.
   
, , 0
M x y dx N x y dy
  .......................................................... (1)
is a differential of  
,
f x y C

f f
df dx dy
x y
 
 
 
......................................................................... (2)
The differential equation (1) can be written as 0
df 
    
, ,
f f
dx dy M x y dx N x y dy
x y
 
  
 
 
,
f
M x y
x

 

and  
,
f
N x y
y



TEST TO DETERMINE EXACT DIFFERENTIAL
EQUATIONS
Consider the differential equation    
, , 0
M x y dx N x y dy
 
Where M and N have continuous first order partial derivatives at all
points  
,
x y in a domain D then this equation is exact in D iff
M N
y x
 

 
PROOF
If    
, , 0
M x y dx N x y dy
  is exact in D then
   
, ,
M x y dx N x y dy
 is an exact differential in D, and by
definition there exists a function F such that
( , )
( , )
F x y
M x y
x



(i)
And
( , )
( , )
F x y
N x y
y



(ii)
for all D
xy 
)
( on differentiating (i) and (ii) .
.
. t
r
w y and x
respectively we have
2
( , ) ( , )
f x y M x y
y x y
 

  
and
2
( , ) ( , )
f x y N x y
x y x
 

  
.
Using the continuity of first order partial derivatives of M and N
then
2 2
( , ) ( , ) ( , ) ( , )
f x y f x y M x y N x y
y x x y y x
   
  
     
Which implies that 0
)
,
(
)
,
( 
 dy
y
x
N
dx
y
x
M is exact.

8. 8
Conversely
Assume that
   
, ,
M x y N x y
y dx
 


for all D
xy 
)
( we need to
show 0
)
,
(
)
,
( 
 dy
y
x
N
dx
y
x
M is exact
i.e. to prove that there exists a function f such that
( , )
( , )
f x y
M x y
x



(1)
( , )
( , )
f x y
N x y
y



(2)
For all ( , )
x y D
 suppose ( , )
f x y satisfies (1) then
( , )
( , )
F x y
M x y
x



integrating . . and
wr t y x partially we have
 
( , ) ( , )
F x y M x y dx h y
 
 (3)
Where  
h y is an arbitrary function of y alone. (Since f is a
function of x and y).
Now (3) represents all functions (solutions) of (1). Differentiating
(3) partially with respect to y we have
 
( , )
( , )
d h y
f x y
M x y dx
dy y dy
 
 
  (4)
Differentiating ( )
h y partially . .
wr t y is the same as total
differentiating .
.
. t
r
w y since ( )
h y as a function of y alone. Further
( , )
f x y must also satisfy (2) then.
 
( , )
( , ) ( , )
d h y
f x y
N x y M x y dx
dy y dy
 
  
  (5)
Hence
 
( , ) ( , )
d h y
N x y M x y dx
y dy

 
  (6)
This equation holds or ( )
h y exists if
( ( , ) ( , )
N x y M x y dx
y


  ) (7)
is independent of x.
Now differentiating (7) .
.
. t
r
w x yields
( , )
( , ) 0
N x y
M x y dx
x x y
 
  
 
 
  
 

or

9. 9
( , )
( , ) 0
N x y
M x y dx
x y x
  
 
 
 
  
 

which implies 0
)
(
)
(




 y
xy
M
x
xy
N
and thus (7) is independent of x for all ( , )
x y D

thus
  ( , ) ( , )
d h y
N x y M x y dx
dy y

 
  holds and on integrating
we have
  ( , ) ( , )
h y N x y M x y dx dy
y
 

 
 

 
 
This implies that )
(y
 exists and therefore
( , )
( , ) ( , ) ( , )
M x y
f x y M x y dx N x y dx dy
y
 

  
 

 
  (8)
Satisfies both
( )
( )
f xy
M xy
x



and
( )
( )
f xy
N xy
y



thus
    0
,
, 
 dy
y
x
N
dx
y
x
M is exact which completes the proof.
THE SOLUTION OF AN EXACT DIFFERENTIAL
EQUATION
From the definition, if the equation     0
,
, 
 dy
y
x
N
dx
y
x
M is
exact then there exists a function f such that
( , )
( , )
f x y
M x y
x



and
( , )
( , )
f x y
N x y
y



for all( , )
x y D
 . Thus the equation may be
written as
( , ) ( , )
0
F x y F x y
dx dy
x y
 
 
 
or simply ( , ) 0
df x y  . Then
the relation ( , )
f x y C
 where C is an arbitrary constant is a solution
of the exact equation
. Then a one parameter family of solutions of this equations is given
by ( , )
f x y C
 where f is a function such that
( , )
( , )
f x y
M x y
x



And
( , )
( , )
f x y
N x y
y



for all D
xy 
)
( and
C is an arbitrary.
Example 1
Show the differential equation is exact and hence solve it
   
2 2
0
x y dx y x dy
   

10. 10
Solution
 
2
M x y
  and  
2
N y x
 
1
M
y



1
N
x



Hence
M N
y x
 

 
the equation is exact.
By definition
f
M
x



and
f
N
y



 
2
2
3
3
f
M x y
x
f x y dx
x
xy h y

  

  
  

 
f
x h y
y


 

................................................................. (i)
But
2
f
N y x
y

  

.............................................................................(ii)
So comparing (i) and (ii)   2
h y y

 
 
3
3
y
h y C
 
 
3 3
,
3 3
x y
f x y xy C
    
The general solution is
3 3
3 3
x y
xy C
   
Example 2
Show the differential equation is exact and hence solve it
   
3 2 2 3
3 3 0
x xy dx x y y dy
   
 
3 2
3
M x xy
  and  
2 3
3
N x y y
 
6
M
xy
y



6
N
xy
x



Hence
M N
y x
 

 
the equation is exact.
By definition
f
M
x



and
f
N
y



 
 
 
3 2
3 2
4
2 2
3
3
3
4 2
f
M x xy
x
f x xy dx
x
x y h y

  

  
  

 
2
3
f
x y h y
y


 

....................................................................... (i)
But
2 3
3
f
N x y y
y

  

.................................................................... (ii)
So comparing (i) and (ii)   3
h y y

 

11. 11
 
4
4
y
h y C
 
 
4 4
2 2
3
,
4 2 4
x y
f x y x y C
    
The general solution is
4 4
2 2
3
4 2 4
x y
x y C
   
Example 3
Solve  
1
1 cos ln sin 0
y y dx x x x y dy
x
 
 
     
 
 
 
 
The general solution
ln cos
xy y x x y C
  
Example 4
Solve the equation    
2 2
3 4 2 2 0
x xy dx x y dy
   
The general solution
3 2 2
2
x x y y C
  
Exercise
1. show that the following differential equation are exact and solve
them
a) 0
y dx x dy
 
b) 3 2
3 0
y dx xy dy
 
c) cosh cos sinh sin
y y dx x x dy

d)  
2 0
y y
x e dx x e dy
  
2. a) State whether the differential equation is exact or not
b) Solve the initial value problem
i.  
2
0 1 0.2
x dx y dy y
  
ii.      
2
1 3 0 0
3
y dx x dy y
    
iii.    
2 2
2 1 2
xy dy x y dx y
  
iv.  
1
sinh cosh 0 0
x dx x dy y
y

  
v.    
2
0 1 0
y
x
e
y dx x dy y
x
   
INTEGRATING FACTOR
Most differential equations are not exact. However it is possible to
multiply a non-exact differential equation by a non-zero function
 
,
F x y chosen so that the resulting differential equation is exact
   
, , 0
P x y dx Q x y dy
  is a non-exact equation, we multiply
with  
,
F x y
   
, , 0
FP x y dx FQ x y dy
   is an exact equation
The function  
,
F x y is called an integrating factor

12. 12
Example 1
Show that the differential equation 0
y dx x dy
  is not exact but
has an integrating factor
2
1
x
And solve the equation
Solution
 
,
P x y y
 and  
,
Q x y x
 
1
dP
dy
 1
dQ
dx
 
dP dQ
dy dx
   the equation is not exact
    2
2
, ,
1
y dx x dy
FP x y dx FQ x y dy
x
y
dx dy
x x

 
 
2
y
M
x
 and
1
N
x
 
2
1
M
y x


 2
1
N
x x



Hence
M N
y x
 

 
the equation is exact.
By definition
f
M
x



and
f
N
y



 
 
2
2
f y
M
x x
f yx dx
y
h y
x


 

 
  

 
1
f
h y
y x


  

..............................................................................
(i)
1
f
N
y x

  

................................................................................ (ii)
So comparing (i) and (ii)   0
h y

 
 
h y C

 
,
y
f x y C
x
   
The general solution is y Kx
 
Example 2
Verify that   3
f x x
 is the integrating factor of
   
2 2
2sin cos 0
y dx xy y dy
  and then find the general solution
Solution
If   3
f x x
 is the integrating factor then
   
, , 0
FP x y dx FQ x y dy
  is an exact equation

13. 13
       
3 2 4 2
, , 2 sin cos 0
FP x y dx FQ x y dy x y dx x y y dy
   
 
3 2
2 sin
M x y
 and  
4 2
cos
N x y y

 
3 2
4 cos
M
x y y
y



 
3 2
4 cos
N
x y y
x



Hence
M N
y x
 

 
the equation is exact.
By definition
f
M
x



and
f
N
y



 
 
 
   
3 2
3 2
4 2
2 sin
2 sin
2
sin
4
f
M x y
x
f x y dx
x y h y

 

 
 

   
4 2
cos
f
x y y h y
y


 

......................................................... (i)
 
4 2
cos
f
N x y y
y

 

.................................................................... (ii)
So comparing (i) and (ii)   0
h y

 
 
h y C

   
4 2
1
, sin
2
f x y x y C
  
The general solution is  
4 2
sin
x y K
 
HOW TO FIND THE INTEGRATING FACTOR
The non exact equation    
, , 0
P x y dx Q x y dy
  is made exact by
multiplying with the integrating factor  
,
F x y to get
   
, , 0
FP x y dx FQ x y dy
 
(1)
   
, , 0
M x y dx N x y dy
  is an exact equation then
 
.
M FP x y
 and  
.
N FQ x y

For exact equations
M N
y x
 

 
FP FQ
y x
 
 
 
(2)
or
F P F Q
P F Q F
y y x x
   
  
   
(3)
Assuming integrating factor only depends on one variable say
 
F F x
 Then 0
F
y



Equation (3) becomes
P F Q
F Q F
y x x
  
 
  

14. 14
Dividing through by FQ
1 1 1
P F Q
Q y F x Q x
  
 
  
Or
1 1
F P Q
F x Q y x
 
  
 
 
  
 
Let R.H.S. be  
R x
 
 
 
1
1
1
F
R x
F x
dF R x dx
F
dF R x dx
F

 



 
 
 
ln
R x dx
F R x dx
F e


 

This is the integrating factor
NB
If  
F F y
 then 0
F
x



Equation (3) becomes
F P Q
P F F
y y x
  
 
  
Dividing through by FP
1 1 1
F P Q
F y P y P x
  
 
  
or
1 1
F Q P
F y P x y
 
  
 
 
  
 
Let R.H.S. be  
R y
 
 
 
1
1
1
F
R y
F y
dF R y dy
F
dF R y dy
F

 



 
 
 
ln
R y dy
F R y dy
F e


 

This is the integrating factor in terms of
Example 1
Find the integrating factor of 2 2
2sin( ) cos( ) 0
y dx xy y dy
 
 
2
2sin
P y
 And  
2
cos
Q xy y

 
2
4cos
P
y
y



 
2
cos
Q
y y
x




15. 15
  2 2
2
1
4cos( ) cos( )
cos( )
4 1 3
R x y y y
xy y
x x x
 
 
 
  
 
3
3ln 3
dx
x
x
F x e
e x


 
Example 2
Solve 2 0
y dx x dy
 
Solution
To check whether exact
2
M y
 and N x

2
M
y



1
N
x



M N
y x
 

 
hence not exact
2
P y
 and Q x

2
P
y



1
Q
x



   
1
2 1
1
R x
x
x
 

 
1
ln
dx
x
x
F x e
e x


 
2
2 0
xy dx x dy
  is an exact equation
2
M xy
 and 2
N x

By definition
f
M
x



and
f
N
y



 
2
2
2
f
M xy
x
f xy dx
x y h y

 

 
 

Now  
2
f
x h y
y


 

............................................................... (i)
But 2
f
N x
y

 

.................................................................... (ii)
So comparing (i) and (ii)   0
h y

 
 
h y C

  2
,
f x y x y C
  
The general solution is 2
x y C
 

16. 16
Example 3
Find the integrating factor of the differential equation
3 2 0
y dx x dy
  hence solve
The general solution
3
2
x y K

Example 4
Solve the initial value problem
   
2
2 3 4 0 0.2 1.5
xy dx x y dy y
    
The general solution
2 3 4
x y y C
 
Exercise
Solve the differential equations
a)    
2
2 2 0 0 2
x
xe y dx y dy y
   
b)  
2 2 0
y xy dx x dy
  
c)    
1 1 0
y dx x dy
   
d) 0
ay dx bx dy
 
e)  
2
3 0
y x dx xy dy
  
1st
ORDER LINGAR DIFFERENTIAL EQUATION
A first order differential equation is said to be linear if it can be
written as
   
y P x y Q x
 
(1)
If  
Q x is zero for all x in the interval in which we consider the
equations, the equations is said to be homogenous otherwise it is
said to be non-homogenous
HOMOGENEOUS EQUATION
1. A differential equation is homogeneous if it has the form
y
y f
x
 
   
 
e.g.
cos
x y
y
y x
 
   
 
y
y
x y
 

which may be written as
0
1
y y
x x
y x
y y
x
x x x
   
 
Solution of homogeneous equations
Substituting the variable
y
u
x
 always transform a homogeneous
into a separable one
If we write y ux
 then y u u x
 
 
Thus the general homogeneous equation

17. 17
y
y f
x
 
   
 
Becomes
 
u u x f u

 
or
 
du
x u f u
dx
 
This is a separable equation
 
 
1 1
du
x f u u
dx
du dx
f u u x
 


This can be integrated to give the solution
Example 1
Solve the homogeneous equation
2 2
dy
xy x y
dx
 
Solution
Diving the equation through by 2
x
2
1
y dy y
x dx x
 
  
 
Let
dy du
y
u y ux u x
x dx dx
     
On substituting in equation
2
1
du
u u x u
dx
 
  
 
 
2
2
2 2
1
1
1
ln ln
2
2ln ln
du
ux
dx
u du dx
x
u x C
y
u Cx Ax
x


 
 
  
 
 
 
Example 2
Solve the homogeneous equation 2 2 2
dy
x x xy y
dx
  
Solution
Diving the equation through by 2
x
2
1
dy y y
dx x x
 
   
 
Let
dy du
y
u y ux u x
x dx dx
     
On substituting in equation
2
1
du
u x u u
dx
   

18. 18
2
2
1 1
1
1
1
tan ln ln tan ln
du
x u
dx
du
dx
u x
u x C u Cx
 
 


   
 
1
tan ln
y
Cx
x
  
 
 
 
Example 3
Solve the homogeneous equation 2 2
dy
xy x y
dx
 
Solution
 
2 2 2
2
x x y A
 
Exercise
Solve the following homogeneous equation
a) 2 2
3
dy
x y
dx

b)  
2 2 dy
x y xy
dx
 
c)
dy
x x y
dx
 
d)  
4 4
dy
x y x
dx
 
e)  
2 2 2
dy
xy y x x y
dx
  
NON- HOMOGENEOUS EQUATION
These are the equations of the form
   
y P x y Q x
  Where   0
Q x 
They are solved using the integrating factor
Let  
F x be the integrating factor then
   
dy
F FP x y FQ x
dx
  (1)
is an exact equation. We can write from the first terms on the LHS
 
d dy dF
Fy F y
dx dx dx
  (2)
Thus (1) may also be written as
dy dF
F y FQ
dx dx
  (3)
Comparing (1) and (3)
dF dF
y FPy FP
dx dx
   
By separation of variables

19. 19
 
 
 
1
ln
P x dx
dF P x dx
F
F P x dx
F e




 

This is the integrating factor
Example 1
Solve the differential equation 2
3 x
dy
y e
dx
  given that   6
0
5
y 
Solution
2
3 x
dy
y e
dx
 
 
3 3
3
dx x
P x F e e

   
3 3 5
3
x x x
dy
e e y e
dx
 
 
3 5
x x
d
ye e
dx

3 5
5
1
5
x x
x
ye e dx
e C
 
 

The general solution is given by 2 3
1
5
x x
y e Ce
 
6 6 1
0 1
5 5 5
x y C C
      
2 3
1
5
x x
y e e
  
Example 2
Solve the initial value problem 2
dy
xy x
dx
  given that   1
0
2
y  
Solution
2
dy
xy x
dx
 
 
2
2
2
x dx x
P x x F e e

   
2 2 2
2
x x x
dy
e xe y xe
dx
 
 
2 2
x x
d
ye xe
dx

2 2
2
1
2
x x
x
ye xe dx
e C
 
 

The general solution is given by
2 2
1
2
x x
y e Ce
 
1 1 1
0 1
2 2 2
x y C C
         
2 2
1
2
x x
y e e
  

20. 20
Example 3
Solve tan sin 2
dy
y x x
dx
  given that  
0 1
y 
Solution
2
3cos 2cos
y x x
 
tan sin 2
dy
y x x
dx
 
 
tan lnsec
tan sec
x dx x
P x x F e e x

    
sec sec tan sec sin 2
dy
x y x x x x
dx
 
 
sec sec sin 2
d
y x x x
dx

sec sec sin 2
sec 2cos sin
2sin
2cos
y x x x dx
x x x dx
x dx
x C
 
 

  



The general solution is given by
2
2cos cos
y x C x
  
0 1 1 2 3
x y C C
       
2
3cos 2cos
y x x
  
Exercise
1. Solve the initial value problems
a)  
3 12 0 6
y y y
  
b)  
2 4 0 3
y xy x y
  
c)  
0 5
kx
y ky e y

  
2. Find the general solution on the differential equation
a) 4 0.8
y y
 
b) 3
3 x
y y e
 
c) 2 9
xy y x
 
d)
2
2 4 x
xy y e
 
e) 3
2 x
xy y x e
 
f) 3 5
xy y x
 
BERNOULLI EQUATIONS
An equation of the form
    n
dy
P x y Q x y
dx
  Where n is any real number
Is known as a Bernoulli equation
If n = 0 or 1 the equation is linear otherwise it is non-linear
Theorem
If 0
n  or 1 the transformation 1 n
v y 
 reduces the Bernoulli
equation to a linear equation

21. 21
Proof
    n
dy
P x y Q x y
dx
  (1)
Multiplying (1) by n
y
   
1
n n
dy
y P x y Q x
dx
 
  (2)
Let 1 n
v y 
 then
 
1
1
n
n
dv dy dy y dv
n y
dx dx dx n dx

   

(3)
Substituting (3) in (2)
   
1
n
y dv
P x v Q x
n dx
 

Or
       
1 1
dv
n P x v n Q x
dx
    (4)
Let      
1
M x n P x
  and      
1
N x n Q x
  then (4) may
be written in the form
   
dv
M x v N x
dx
  This is linear in v
Solution of Bernoulli equation
Steps
1. Multiply the equation by n
y
2. Define and substitute 1 n
v y 

3. Get y by differentiating v w.r.t x then substitute in the
Bernoulli equation to make it linear
Example 1
Solve the equation 3
dy
y xy
dx
 
Solution
This is a Bernoulli equation with 3
n  multiplying the equation by
3
y
we get
3 2
dy
y y x
dx
 
  .................................................. (A)
3
1 2
2
n dy y dv
v y y
dx dx
 
   

Substituting in the equation (A) yields
1
2
2 2
dy
v x
dx
dy
v x
dx
 

  
This is linear
 
2 2
dx x
F x e e
 

 
2 2 2
2 2
x x x
dy
e ve xe
dx
  
  

22. 22
 
 
2 2
2 2
2
2
x x
x x
d
ve xe
dx
ve xe dx
 
 
 
 

Integrating by parts
 
2 2 2
2 2
2
1
2
x x x
x x
xe dx xe e dx
xe xe C
  
 
  
  
 
 
 
2 2 2
2
2
2
1
2
1
2 1
2
1 1
2 1
2
x x x
x
x
ve xe xe C
v x Ce
x Ce
y
  
    
   
   
Example 2
Solve the Bernoulli equation   5
4 1 2
y y x y
  
Solution
This is a Bernoulli equation with 5
n  multiplying the equation by
5
y
we get
 
5 4
4 1 2
y y y x
 
   ................................................ (A)
5
1 4
4
n dy y dv
v y y
dx dx
 
   

Substituting in the equation (A) yields
1 2
2 1
dy
v x
dx
dy
v x
dx
   
  
This is linear
 
dx x
F x e e
 

 
 
 
2
2
2
2
x x x x
x x x
x x x
dy
e ve xe e
dx
d
ve xe e
dx
ve xe e dx
   
  
  
   
  
  

2 2
2 2
2
x x x
x x x
x x
xe e dx e C
xe e e C
xe e C
  
  
 
    
    
   

4
2 1
1
2 1
x
x
v x Ce
Ce x
y
    
   
Example 3
Solve the Bernoulli equation 2 2
1
6
y y x y
x

 
Solution
2 3 3
3
y x Cx
 

23. 23
Exercise
Solve the differential equations
a) 2
y xy xy
 
b) 4 2
1
y y x y
x
 
 
c) 2
1 2
y y y
x x
   
d) 3 5
2 10
xy y x y
 
e) 4 3
2 1
dy
x x y
dx
 
SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS
Linear differential equations with constant coefficients
A linear ordinary equation of order n and in the dependent variable y
and independent variable x of the form
1
1 1 0
1
( ) ( ) ..... ( ) ( ) ( )
n n
n n
n n
d y d y dy
a x a x a n a x y F x
dx dx dx

 
   
(1)
Where ( ) 0
n
a x   
0 1 2 3
, , , ................ n
a a a a a and F x are assumed to
be continuous real functions
If   0
F x  then the equation is a homogeneous equation otherwise
it is non homogeneous.
For n = 2 the equation (1) reduces to a second order linear equation.
i.e.
2
2 1 0
2
( ) ( ) ( ) ( )
d y dy
a x a x a x y F x
dx dx
  
If   0
F x  then the equation corresponds to second order is a
homogeneous equation otherwise it is a second order non
homogeneous equation.
A SECOND ORDER HOMOGENEOUS LINEAR EQUATION
A homogeneous linear differential equation is of the form
    0
y p x y q x y
 
   (1)
An important property of equation (1) is that its solution is given by
a linear combination of two solutions  
1
y x and  
2
y x . The solution
is given as a sum of constant multiples of these functions i.e.
   
1 1 2 2
y C y x C y x
 
Theorem
Let  
1
y x and  
2
y x be solutions of the equation
    0
y p x y q x y
 
   on an interval I, then for any constants
1
C and 2
C ,    
1 1 2 2
C y x C y x
 it also a solution on I
Proof
Substitute 1 1 2 2
y C y C y
  into the differential equation to get
       
1 1 2 2 1 1 2 2 1 1 2 2
" 0
C y C y p x C y C y q x C y C y y
 
     

24. 24
    
1 1 2 2 1 1 2 2 1 1 2 2
" 0
C y C y p x C y C y q x C y C y y
 
   
     
 
       
1 1 1 1 2 2 2 2 0
C y p x y q x y C y p x y q x y
   
   
     
   
But 1
y and 2
y are solutions of the equation
   
1 1 1 0
y p x y q x y
 
   and    
2 2 2 0
y p x y q x y
 
  
   
1 2
0 0 0
C C
  
And hence 1 1 2 2
y C y C y
  is also a solution.
A Second Order Homogeneous equation
Consider the equation with constant coefficients
0
y a y b y
 
   (1)
Where a and b are constants
Consider a first order differential equation
0
y ky
 
Integrating factor  
k dx kx
F x e e

 
 
0
0
kx kx
kx
kx
dy
e kye
dx
d
ye
dx
ye C
 

 
Or kx
y Ce

This gives the idea that 1
mx
y e
 (2)
1
mx
y me
  And 2
1 " mx
y m e

Substituting in (1) yields
 
2
2
0
0
mx mx mx
mx
m e ame be
m am b e
  
  
Hence (2) is a solution of (1) if m is a solution of the quadratic
equation
2
0
m am b
  
This equation is called a characteristic equation or auxiliary equation
of (1) its roots are
2
1
1
4
2
m a a b
 
   
 
and 2
2
1
4
2
m a a b
 
   
 
1
1
m x
y e
  and 2
2
m x
y e

1 2
1 2
m x m x
y C e C e
  
Depending on the signs of the discriminant we may know that the
nature of the root
This may be in three cases
CASE 1 2 real roots if 2
4 0
a b
 
CASEE double roots if 2
4 0
a b
 
CASE 3 complex roots if 2
4 0
a b
 
CASE 1: DISTINCT REAL ROOTS
If the roots are distinct real numbers 1
m and 2
m then 1
1
m x
y e
 and
2
2
m x
y e
 are the distinct solution of (1) and they are independent

25. 25
Example 1
Solve the differential equation
2
2
3 2 0
d y dy
y
dx dx
  
Solution
The auxiliary equation is
  
2
3 2 0
2 1 0
m m
m m
  
  
1 2
m
  And 2 1
m 
2
1
x
y e
  And 2
x
y e

The general solution is 2
1 2
x x
y C e C e
 
Example 2
Solve the differential equation " 2 0
y y y
 
  
Solution
The auxiliary equation is
  
2
2 0
2 1 0
m m
m m
  
  
1 2
m
   and 2 1
m 
2
1
x
y e
  and 2
x
y e

The general solution is 2
1 2
x x
y C e C e

 
Example 3
Solve the differential equation 2 8 0
y y y
 
  
The general solution
2 4
1 2
x x
y C e C e

 
CASSE II: REAL DOUBLE ROOTS
If roots are double 1 2
m m
 this implies that solution will be
1
1
m x
y e
 and 1
2
m x
y e

1 1
1 2
m x m x
y C e C e
  
NB
i. This is not a combination of two lineally independent solutions
because they are equal.
ii. This solution has only one essential arbitrary constant since we
can write
  1
1
1 2
m x
m x
y C C e
Ae
  

This is not the case for a second order differential equation which
should have two essential constants
We need to find 2
y
Let  
2 1
y V x y
 where V is to be determined.
Then
2 1 1
y V y V y
  
 
2 1 1 1 1 1 1 1
2
y V y V y V y V y V y V y V y
          
      
Substituting in (1) gives

26. 26
 
1 1 1 1 1 1
2 0
V y V y V y a V y V y by

    
     
   
1 1 1 1 1 1
2 0
V y V y ay V y ay by
   
      (A)
Since 1
y is a solution 1 1 1 0
y ay by

    (i)
For double roots 2
4 0
a b
 
2
1
1
4
2
m a a b
 
   
 
and 2
2
1
4
2
m a a b
 
   
 
1 2
2
a
m m
    2
1
a x
y e

  and 2
1
2
a x
a
y e

  
1 1
2y ay

   or 1 1
2 0
y ay
   (ii)
Substituting (i) and (ii) in (A) yields
1 0
V y
  Or 0
V
On integrating we get
1 1
V a x b
  Where 1 1,2,3.............
a  and 1 0,1,2..........
b 
Taking the simplest form of V
2 1
V x y xy
  
This forms the basis of the solution 1
1
m x
y e
 and 1
2
m x
y xe

1 1
1 2
m x m x
y C e C xe
  
Example 1
Solve the differential equation 8 16 0
y y y
 
  
Solution
The auxiliary equation is
 
2
2
8 16 0
4 0
m m
m
 
 
1 2 4
m m
  
4
1
x
y e
  and 4
2
x
y xe

The general solution is 4 4
1 2
x x
y C e C xe
 
Example 2
Solve the differential equation
   
4 4 0 0 3 0 1
y y y y y
  
    
Solution
The auxiliary equation is
 
2
2
4 4 0
2 0
m m
m
  
 
1 2 2
m m
  
2
1
x
y e
  and 2
2
x
y xe

The general solution is 2 2
1 2
x x
y C e C xe
 
Y = 3 when x = 0 1
3 C
 
2 2 2
1 2 2
2 2
x x x
y C e C e C xe
   
Y = 1 when x = 0
1 2
2 2
1 2
1 6 5
C C
C C
  
    
 
2 2
2
3 5
3 5
x x
x
y e xe
y x e
  
 

27. 27
Example 3
Solve the differential equation
   
6 9 0 0 4 0 14
y y y y y
  
     
Solution
  3
2 4 x
y x e
 
CASE III COMPLEX ROOTS
This is when 2
4 0
a b
 
Suppose the auxiliary equation has complex roots a ib
 as non
repeated roots where a, b are real numbers and 0
b 
Then the corresponding part of the general solution is
   
1 2
a ib x a ib x
y C e C e
 
 
From theory of complex numbers
cos sin
i
e i

 
  and cos sin
i
e i

 

 
   
 
   
   
1 2 1 2
1 2
1 2
1 2 1 2
cos sin cos sin
cos sin
a ib x a ib x ax ibx ax ibx
ax ibx ibx
ax
ax
C e C e C e e C e e
e C e C e
e C bx i bx C bx i bx
e C C bx i C C bx
  
 
  
 
   
 
 
   
 
 
 
1 2
cos sin
ax
e K bx K bx
 
Where 1 1 2
K C C
  and  
2 1 2
K i C C
 
The general solution corresponding to a complex root a ib
 is
 
1 2
cos sin
ax
y e C bx C bx
 
Example1
Given 2 10 0
y y y
 
   find the general solution
Solution
2
2 10 0
m m
  
2 4 40 2 6
1 3
2 2
i
m i
  
   
1
a
  and 3
b 
 
1 2
cos3 sin3
x
y e C x C x
 
Example 2
Solve the initial value problem
   
2 5 0 0 1 0 5
y y y y y
  
    
Solution
2
2 5 0
2 4 20 2 4
1 2
2 2
m m
i
m i
  
    
    
1
a
   And 2
b 
 
1 2
cos2 sin2
x
y e C x C x

 

28. 28
1
y  When 0
x  1
1 C
 
   
1 2 1 2
cos2 sin2 2 sin2 2 cos2
x x
y e C x C x e C x C x
 
      
5
y  When 0
x  1 2 2
5 2 3
C C C
     
 
cos2 3sin2
x
y e x x

 
Example 3
Solve the differential equation
2
2
6 25 0
d y dy
y
dx dx
  
Solution
 
3
1 2
cos4 sin4
x
y e C x C x
 
Exercise
1. find the general solution of
a) 25 0
y y
 
b) 2 0
y y
 
c) 8 16 0
y y y
 
  
d) 25 0
y y
 
e) 8 2 0
y y y
 
  
2. Solve the initial value problems
a)    
9 0 2 3
y y y y
 
 
    
b)    
4 4 0 0 3 0 10
y y y y y
  
    
c)    
11 24 0 0 1 0 4
y y y y y
  
    
d)    
16 0 1 4
y y y y
 
 
    
NON HOMOGENEOUS EQUATIONS
A non-homogeneous equation is an equation of the form
     
y p x y q x y r x
 
  
(1)
Where   0
r x 
A general solution of the non homogeneous on some open interval L
is a solution of the form
     
h p
y x y x y x
 
(2)
Where      
1 1 2 2
h
y x C y x C y x
  is the general solution of a
homogeneous equation
p
y is a solution obtained by considering the RHS of the given
equation
Example 1
Find the general solution of
2
3 x
y y e
 
Solution
Find the general solution of homogeneous equation
0
y y
 
The auxiliary equation is
2
1 0 1
m m
    
The general solution is 1 2
x x
h
y C e C e
 
To find p
y consider RHS

29. 29
Since 2x
e has derivative 2x
e times some constant, we try
2 2 2
2 "4
x x x
p p p
y Ae y Ae y Ae
 
   
Substituting in the equation it gives
2 2 2
4 3
3 3 1
x x x
Ae Ae e
A A
 
  
2x
p
y e
 
General solution of the non homogeneous equation is
2
1 2
x x x
h p
y y y C e C e e

    
Example 2
Solve the initial value problem
   
2
4 3 10 0 1 0 3
x
y y y e y y

  
     
Solution
Find the general solution of homogeneous equation
4 3 0
y y y
 
  
The auxiliary equation is
  
2
4 3 0
1 3 0
m m
m m
  
  
1 3
m
  and 2 1
m 
The general solution is 3
1 2
x x
h
y C e C e
 
To find p
y
Let 2 2 2
2 "4
x x x
p p p
y Ae y Ae y Ae
  
 
    
Substituting in the equation it gives
 
2 2 2 2
4 4 2 3 10
10 2
4 8 3 10
15 3
x x x x
Ae Ae Ae e
A A A A
   
   
     
2
2
3
x
p
y e
 
General solution of the non homogeneous equation is
3 2
1 2
2
3
x x x
h p
y y y C e C e e
     (3)
Particular equation satisfying the initial conditions by differentiation
3 2
1 2
4
3
3
x x x
y C e C e e
    (4)
From (3) and (4) and the given initial conditions
  1 2
4
0 3 1
3
y C C
   
  1 2
4
0 3 2
3
y C C
     
This gives 1
4
3
C  and 2 1
C  
3 2
4 2
3 3
x x x
y e e e
   
Example 3
Solve the differential equation 3 4 5 x
y y y e
 
  
Solution
4
1 2
x x x
y C e C e xe

  

30. 30
Exercise
Find the general solution of
a) 3x
y y e
 
b)    
3
6 9 2 0 0 0 1
x
y y y e y y
  
    
SOLUTION OF UNDETERMINED COEFFICIENT
General solution of a non-homogeneous linear equation
     
y p x y q x y r x
 
  
(A)
is a sum of the form
     
h p
y x y x y x
 
Undetermined coefficient method is used when the RHS ie.  
r x , is
an exponential function, trigonometric function, a polynomial or
sum or product of such functions
How to choose the undetermined coefficient
Given by table below
Term in  
r x Choice of p
y
1 K mx
e A mx
e
2 K n
x (n = 0, 1, 2
............)
1 1
1 1 0
..............
n n
n n
K x K x K x K


 
3 cos
K bx or sin
K bx cos sin
A bx B bx

4 cos
ax
Ke bx or
cos
ax
Ke bx
 
cos sin
a x
e A bx B bx

Rules
1. If  
r x in (1) is one of the function in the first column of above
table choose corresponding function in second column
2. If term of choice is in the solution of homogeneous equation then
multiply choice by x or by 2
x if this solution is a double root
e.g. 2
1 2
x x
p
y c e c e
  then   2x
r x e
  choice will
2x
p
y Axe

3. If  
r x is sum of functions listed in several lines of tables then
choice for p
y will be the sum of the function in the
corresponding lines in second column.
Consider,
2
4
2
2 3 2 x
d y dy
y e
dx dx
  
Solution must satisfy the above equation
The desired particular solution might be a constant multiple of 4x
e
Let 4x
p
y Ae

A is the undetermined coefficient
4 4 4
4 16
x x x
p p p
y Ae y Ae y Ae
 
   
Substituting in the equation it gives
 
4 4 4 4
16 2 4 3 2
2
5 2
5
x x x x
Ae Ae Ae e
A A
  
  
4
2
5
x
p
y e
 

31. 31
NB
If given a differential equation
2
1 0
2
mx
d y dy
a a y Ke
dx dx
  
Such that mx
e is in the solution of corresponding homogeneous
equation and then taking mx
p
y Ae
 will reduce reduces equation to
zero i.e. A = 0
Example
If given a differential equation
2
3
2
2 3 2 x
d y dy
y e
dx dx
  
The auxiliary equation is
  
2
2 3 0
1 3 0
m m
m m
  
  
1 3
m
  and 2 1
m  
The general solution is 3
1 2
x x
h
y C e C e
 
If we let 3x
p
y Ae
 ,
3 3 3
3 9
x x x
p p p
y Ae y Ae y Ae
 
   
Substituting in the equation it gives
 
3 3 3 3
3
9 2 3 3 2
0 2
x x x x
x
Ae Ae Ae e
A e
  

This is an impossible solution.
This is a case of a double root. A solution linearly independent of
mx
e is mx
xe therefore we assume a particular solution of the form
3 3 3 3 3 3
3 3 3 9
x x x x x x
p p p
y Axe y Ae Axe y Ae Ae Axe
 
      
Substituting in the equation it gives
 
3 3 3 3 3 3 3
3 3 9 2 3 3 2
x x x x x x x
Ae Ae Axe Ae Axe Axe e
     
3 3 3 3 3 3 3
3 3 9 2 6 3 2
x x x x x x x
Ae Ae Axe Ae Axe Axe e
     
1
4 2
2
A A
  
3
1
2
x
p
y xe
 
General solution of is
3 3
1 2
1
2
x x x
h p
y y y C e C e xe

    
Example 1
Solve the equation
2
2
2
4 8
d y
y x
dx
 
Solution
For h
y The auxiliary equation is
2
4 0 2
m m i
    
The general solution is 1 2
cos2 sin2
h
y C x C x
 

32. 32
For p
y choose 2
2 1 0
h
y K x K x k
  
2 1
2
2
2
h
h
y K x K
y K
  
 
Substituting in equation we get
 
 
2 2
2 0 1 0
2 2
0 1 2 0
2 4 8
4 4 2 4 8
K k x K x K x
k x K x K K x
   
  
Comparing coefficients
2 2
1 1
2 0 0 0
4 8 2
4 0 0
2 4 8 4 4 1
K K
K K
K K K K
   
  
      
2
2 1
p
y x
  
2
1 2
cos2 sin2 2 1
y C x C x x
   
Example 2
Solve the initial value problem
   
2 0 1 0 0
x
y y y e x y y
  
     
Solution
For h
y The auxiliary equation is
 
2
2
2 1 0 1 0
m m m
     
1
m 
The general solution is 1 2
x x
h
y C e C xe
 
For p
y in case of double root we choose 2 x
Ax e
For the polynomial 1 0
K x K

2
1 0
x
p
y Ax e K x K
   
2
1
2
2
2 2 2
x x
p
x x x x
h
y Axe Ax e K
y Ae Axe Axe Ax e
   
    
Substituting in equation we get
 
2 2 2
1 1 0
2 2 2
1 1 0
2 4 2 2
2 4 4 2 2
x x x x x x x
x x x x x x x
Ae Axe Ax e Axe Ax e K Ax e K x K e x
Ax e Ax e Ax e Axe Axe Ae K x K K e x
         
          
1 1 0
2 2
x x
Ae K x K K e x
    
Comparing coefficients
1
1 0 0 1 0
1
2 1
2
1
2 8 2 2
A A
K
K K K K K
   

     
2
1
2
2
x
p
y x e x
   
2
1 2
1
2
2
x x x
y C e C xe x e x
    
Now 2
1 2 2
1
1
2
x x x x x
y C e C e C xe xe x e
      
  1 1
0 2 1 1
y C C
     
  1 2 2 1
0 1 1 0
y C C C C
        

33. 33
2
1
2
2
x x
y e x e x
     
Example 3
Solve 3 2 x
y y y e
 
  
Solution
2
1 2
x x x
y C e C e xe
  
Example 4
Solve 2 5 16 sin2
x
y y y e x
 
   
Solution
 
1 2
4 1
cos2 sin 2 2 cos2 sin 2
17 17
x x
y e C x C x e x x

    
Exercise
Find the general solution
2
3
y y x
 
2 x
y y y e x
 
   
2
2 5 5 4 2
y y y x x
 
    
9 cos3
y y x
 
2 2 x
y y y e
 
  
METHOD OF VARIATION OF PARAMETER
The method of variation of parameter is a method for finding
particular solution nth
order non – homogeneous linear differential
equation
Consider a second order differential equation
       
2
2 1 0
2
d y dy
a x a x a x y Q x
dx dx
   (1)
Let 1
y and 2
y be two independent solutions of the corresponding
homogeneous equation
     
2
2 1 0
2
0
d y dy
a x a x a x y
dx dx
   (2)
Whose solution is 1 1 2 2
h
y C y C y
 
In variation of parameter the arbitrary constants 1
C and 2
C are
replace by functions 1
V and 2
V which are to be determined such that
   
1 1 2 2
V x y V x y
 (3)
(3) is a solution of (2)
We assume a solution of the form
   
1 1 2 2
p
y V x y V x y
  (4)
       
1 1 1 1 2 2 2 2
p
y V x y V x y V x y V x y
    
    (5)
We impose a second condition we simplify p
y by demanding that
   
1 1 2 2 0
V x y V x y
 
  (6)
   
1 1 2 2
p
y V x y V x y
  
  (7)

34. 34
       
1 1 1 1 2 2 2 2
p
y V x y V x y V x y V x y
      
    (8)
Substituting (4), (7) and (8) in (1)
         
   
       
2 1 1 1 1 2 2 2 2
1 1 1 2 2
0 1 1 2 2
a x V x y V x y V x y V x y
a V x y V x y
a x V x y V x y Q x
     
  
 
 
 
 
 
 
  
 
 
(9)
Or
       
       
       
1 2 1 1 1 0 1
2 2 '2 1 2 0 2
2 1 2 2
V x a x y a x y a x y
V x a x y a x y a x y
a x V x y V x y Q x
 
 
 
 
 
  
 
 
   
  
 
 
(10)
Since 1
y and 2
y are solutions of the equation (2), (10) becomes
   
 
 
1 2 2
2
Q x
V x y V x y
a x
   
 
Therefore the two imposed conditions require that the function  
1
V x
and  
2
V x be chosen such that the system of equation
   
1 1 2 2 0
V x y V x y
 
 
   
 
 
1 2 2
2
Q x
V x y V x y
a x
   
 
Solve the system of equations using Crammers rule
 
 
 
2
2
2
1
1 2
1 2
0 y
Q x
y
a x
V x
y y
y y

 
 
and  
 
 
2
2
2
2
1 2
1 2
0 y
Q x
y
a x
V x
y y
y y

 
 
Then obtain the function  
1
V x and  
2
V x
Example 1
Solve the differential equation
2
2
5 6 x
d y dy
y e
dx dx
  
Solution
For h
y
2
2
2
1 2
5 6 0 5 6 0
2 3
d y dy
y m m
dx dx
m m
      
 
2 3
1 2
x x
h
y C e C e
  
We assume
2 3
1 2
x x
p
y V e V e
 
The system of equations
   
1 1 2 2 0
V x y V x y
 
 
   
 
 
1 1 2 2
2
Q x
V x y V x y
a x
   
 

35. 35
From the solution of homogeneous equation
2
1
x
y e
 and 3
2
x
y e
 2
1 2 x
y e
  and 3
2 3 x
y e
 
Substituting in the system of equations
   
2 3
1 2 0
x x
V x e V x e
 
 
   
2 3
1 2
2 3
x x x
V x e V x e e
 
 
On solving the equation
 
3
3 4
1 5 5
2 3
2 3
0
3
3 2
2 3
x
x x x
x
x x
x x
x x
e
e e e
V x e
e e
e e
e e

   

 
1 3
x x
V x e dx e C
 
   

 
2
2 3
2
2 5 5
2 3
2 3
0
2
3 2
2 3
x
x x x
x
x x
x x
x x
e
e e e
V x e
e e
e e
e e

   

  2
2 4
1
2
x x
V x e dx e C
 
   

  2 2 3
3 4
2 3
3 4
2 3
3 4
1
2
1
2
3
2
x x x x
p
x x x x
x x x
y e C e e C e
e e C e C e
e C e C e
 
 
     
 
 
    
   
2 3 2 3
1 2 3 4
2 3
3
2
3
2
x x x x x
h p
x x x
y y y C e C e e C e C e
Ae Be e
      
  
Where 1 3
A C C
  and 2 4
B C C
 
Example 2
Solve the equation
2
2
tan
d y
y x
dx
 
Solution: The auxiliary equation is 0
1
2


m thus the roots are

1 the solution to homogeneous equation is
2 2
cos sin
h
y C x C x
 
Assume 1 2
( )sin( ) ( )cos ..................( )
p
y V x x V x x a
 

36. 36
Where 1( )
V x and 2 ( )
V x as above are chosen such that
1 2
( )sin( ) ( )cos 0
V x x V x x
 
 
1 2
( )cos( ) ( )sin tan
V x x V x x x
 
 
Which on solving gives
1
0 cos
tan sin
( ) cos tan sin
sin cos
cos sin
x
x x
v x x x x
x x
x x
    

2
sin 0
cos tan
( ) sin tan sin tan cos sin
1
x
x x
v x x x x ax x x
      

Thus 1 3
( ) sin cos
v x x dx x C
  

)
(
2 x
v 4
( cos sec ) sin ln(sec tan 1 )
x dx x x x C
      

3
3 4
( cos )sin (sin lnsec tan )cos
sin cos cos ln sec tan
p
y x c x x x x x
C x C x x x x
      
   
The general solution is
sin cos cos ln sec tan
h p
y y y A x B x x x x
     
Where 1 3
A C C
  and 2 4
b C C
 
EXERCISE
1. Find the general solution of each of the following using method
of variation of parameter:
a)
2
2
cot
d y
y x
dx
 
b)
2
2
2
4 5 sec
x
d y dy
y e x
dx dx

  
c)
2 3
2 3
6 9
x
d y dy e
y
dx dx x

  
d)
2
2 2
1
3 2
1 x
d y dy
y
dx dx e
  


37. 37
DIFFERENTIAL OPERATORS
Let r be an n-times differentiable function of independent variable t,
then the operation of differentiation with respect r is denoted by D
i.e.
dt
d
D  , D is known as the differential operator. The derivative
dt
dx
is denoted by Dx i.e.
dt
dx
Dx  likewise 2
2
2
dt
x
d
x
D  and the th
P
derivative of x with respect to t is denoted by p
p
p
dt
x
d
x
D  . (p=1 2
…….)
This operator notation can be extended to
cx
dt
dx
x
C
D 

 )
( And ( )
n m
n m
n m
d x d x
aD bD c x a cx
dt dt
    
Note
1. In this notation the general linear differential expression with
constant coefficients 0
a 1
a ….. n
a
x
a
dt
dx
a
dt
x
d
a
dt
x
d
a n
n
n
n
n
n



 


1
1
1
1
0 ...... is written as
x
a
D
a
D
a
D
a n
n
n
n
)
........
( 1
1
1
0 


 

2. In this case n
D 1

n
D … D does not represent quantities to be
multiplied by x but indicate operations (differentiation) to be
carried on the function x.
Definition: n
n
n
n
a
D
a
D
a
D
a 


 

1
1
1
0 ..
.......... where
1,2...............
i n
 are constants is known as a linear differential
operator of order n with constant coefficients
e.g 2
3 4 6
D D
  is a linear operator of order 2 with constant
coefficients.
NB. 
2
2
2
3 4 6 3 4 6
d x dx
D D x x
dt dt
    
Example
Solve      
2 2
4 3 7 4 0 6 0 8
D D y x y y
     
Solution
The auxiliary equation is 2
4 3 0
m m
   1
m  and 3
m 
3
1 2
x x
h
y C e C e
  
2
2 1 0
p
y K x K x K
    2 1
2
p
y K x K

   and 2
2
p
y K
 
Substituting in equation

38. 38
   
2 2
2 2 1 2 1 0
2 2
2 1 2 2 1 0
2 8 4 3 3 3 7 4
3 3 8 2 4 3 7 4
K K x K K x K x K x
K x K K x K K K x
      
      
Comparing coefficients
2 2
7
3 7
3
K K
   
1 2 1
8 7 56
3 8 0
3 3 9
K K K
     
1 2 0 0
112 76
2 4 3 4 4
9 9
K K K K
       
2
7 56 76
3 9 9
p
y x x
   
3 2
1 2
7 56 76
3 9 9
x x
y C e C e x x
     
Exercise
Solve
a)  
2 4
2 7 5 x
D D y e
  
b)    
2 2 3
2 8 4 x x
D D y e e
   
c)  
2
5 14 16 20 x
D D y x e
   
INVERSE DIFFERENTIAL OPERATOR
A particular integral of a linear differential equation    
F D y Q x

with constant coefficient is given by
 
 
1
y Q x
F D

Consider an exponential function
2 2
...............
ax ax ax r r ax
y e Dy ae D y a e D y a e
   
   
ax ax
F D e F a e
 D a
 
Hence
   
1 1
a x a x
y e e
F D F a
 
Example 1
Solve  
2 3
3 2 x
D D y e
  
Solution
The auxiliary equation is 2
3 2 0
m m
   1
m  and 2
m 
2
1 2
x x
h
y C e C e
  
3
D a
 
Hence
     
3 3
1 1
1 2 3 1 3 2
x x
p
y e e
D D
 
   
3
1
2
x
p
y e
 

39. 39
2 3
1 2
1
2
x x x
y C e C e e
   
Example 2
Solve 
2 4
2 7 5 x
D D y e
  
Solution:
2 5 4
1 2
5
6
x x x
y C e C e e

   
Example 3
Solve   
2 2
5 6 2 x x
D D y e e
   
Solution:
2 3 2
1 2
1
2
10
x x x x
y C e C e e e
    
CASE 2
For trigonometric functions
     
2 2
cos sin cos
y ax b Dy a ax b D y a ax b
       
2 2
D y a y
  2 2
D a
  
Hence
 
 
 
 
2 2
1 1
cos cos
y ax b ax b
F D F a
   

Example 1
Solve  
2
3 4 sin 2
D D y x
  
Solution
The auxiliary equation is 2
3 4 0
m m
   1
m  and 4
m  
4
1 2
x x
h
y C e C e
  
2 2
4
D a
    Hence
  
  
  
2 2
1
sin 2
1 4
1 4
sin 2
1 16
p
y x
D D
D D
x
D D

 
 

 

40. 40
  
 
   
2
2
2
3 4
sin 2
4 1 4 16
3 4 sin 2
20
sin 2 3 sin 2 4sin 2
20
4sin 2 6cos2 4sin 2
20
D D
x
D D x
D x D x x
x x x
 

   
 

 

  

 
1
4sin 2 3cos2
50
p
y x x
   
 
4
1 2
1
4sin 2 3cos2
50
x x
y C e C e x x

    
Example 2
Solve  
2
4 cos3
D y x
 
Solution
The auxiliary equation is 2
4 0
m   2
m i
 
1 2
cos2 sin2
h
y C x C x
  
2 2
9
D a
    Hence
 
2
1
sin3
4
1
sin3
9 4
1 1
sin3 sin3
5 5
p
p
y x
D
x
x y x



 
  
 
1 2
cos2 sin2
y C x C x
  
Example 3
Find particular integral of    
3
1 cos 2 1
D y x
  
 
 
2
1
cos 2 1
1
p
y x
D D
 

Solution:
   
1
cos 2 1 8sin 2 1
65
p
y x x
    
 
 
For a polynomial
When   1 1
1 1 0
............
n n
n n
Q x a x a x a x a


   
 
 
 
   
1
1
p
y Q x
F D
F D Q x




41. 41
We expand this binomial up to the term raised to power n then
operate on the function tem by term.
Example 1
Solve  
2 2
4 3 1
D y x x
   
Solution
The auxiliary equation is 2
4 0
m   2
m  
2 2
1 2
x x
h
y C e C e
  
 
   
   
1
1
p
y Q x F D Q x
F D

 
 
2
2
1
3 1
4
p
y x x
D
   

   
 
1
2 2
1
2
1 2
4 3 1
4 1 3 1
4
p
y D x x
D
x x



   
 
    
 
 
 
2
1 2
4 1 ....... 3 1
4
D
x x
  
     
 
 
 
 
 
2 2
2
2
3 1
1
3 1
4 4
1 2
3 1
4 4
D x x
x x
x x
 
 
 
    
 
 
 
    
 
 
2
1 3 3
4 4 8
p
y x x
 
    
 
 
2 2 2
1 2
1 3 3
4 4 8
x x
h
y C e C e x x

     
Example 2
Solve  
2 2
2 3 1
D D y x
   
Solution
The auxiliary equation is 2
2 3 0
m m
   1
m   and 3
m 
3
1 2
x x
h
y C e C e

  
 
   
   
1
1
y Q x F D Q x
F D

 
 
2
1
2 3
y Q x
D D

 
   
 
1
2 2
1
1 2 2
2 3 1
2 1
3 1 1
3 3
y D D x
D D x



   
 
 
   
 
 
 
 
 
2
2 2 2
1 2 1 2 1
1 ... 1
3 3 3 3 3
D D D D x
 
   
      
 
   
 
   
 

42. 42
 
2 2 2
1 2 1 4
1 1
3 3 3 9
D D D x
 
    
 
 
 
2 2
1 2 1
1 1
3 3 9
D D x
 
   
 
 
     
     
2 2 2 2
2
2
1 2 1
1 1 1
3 3 9
1 2 1
1 2 2
3 3 9
1 1 4 2
3 3 9 27
x D x D x
x x
x x
 
     
 
 
 
   
 
 
   
3 2
1 2
1 4 11
3 9 27
x x
y C e C e x x

     
Example 3
Solve  
2 2
1
D y x
 
Solution:
2
1 2 2
x x
y C e C e x

    
Example 4
Solve  
2
2 2 1
D y x
  
Solution:
2 2
1 2
1
2
x x
y C e C e x

    
Exercise
Find particular integral
a)  
2
1 x
D y e
 
b)  
2 3 2
1 3 5
x x
D D y e e
    
c)  
2
4 sin 2
D y x
 
d)  
2 2
5 6
D D y x
  
e)  
2
2 sin
D y x
 
f)  
2 2
2 2 3 2 1
D D y x x
    
EQUATION WITH VARIABLE COEFFICIENTS
In the case of differential equation with variable coefficients there is
no general method for solving them unless the equation is of special
kind e.g. Euler type of differential equation. In others the method of
approximations e.g. power series e.t.c are applied.
The Euler type of equation are in the form.
       
1
1
1 1 0
1
...................
n n
n n
n n
n n
d y d y dy
K ax b K ax b K ax b K y Q x
dx dx dx


 
       
Where 1 0
...................
n n
K K K
 are constants
This equation can be reduced to a linear differential equation with
constant coefficients by putting t
ax b e
  or  
ln
t ax b
 
Theorem

43. 43
The transformation t
ax b e
  reduces the equation the second order
differential equation
     
2
2
2 1 0
2
d y dy
K ax b K ax b K y Q x
dx dx
     ........................... (A)
to a linear equation with constant coefficients.
Proof:
Let t
ax b e
   
ln
t ax b
  
t
dt a
ae
dx ax b

 

t
dy dy dt dy
ae
dx dt dx dt

    (a)
2
2
t t
d y d dy d dy dt
ae ae
dx dx dt dt dt dx
 
   
 
   
   
2
2
t t t
d y dy
ae ae ae
dt dt
  
 
 
 
 
2
2 2
2
t d y dy
a e
dt dt
  
 
 
 
(b)
Substituting (a) and (b) in (A)
 
2
2 2 2
2 1 0
2
t t t t
d y dy dy
K e a e K e ae K y F t
dt dt dt
 
 
   
 
 
 
2
1 2 0
2
d y dy
A A K y F t
dt dt
   ............................................ (B)
Where 2
1 2
A K a
 and 2
2 1 2
A K a K a
 
Equation (B) is a second order linear differential equation with
constant coefficients.
Note
a) This result can be generalized to a th
n order differential
equation.
b) The leading coefficients  
n
n
K ax b
 in (A) is zero for
0
ax b
  thus the theorem does not include 0
ax b
 
c) In the above prove assumption is that   0
ax b
  . If
  0
ax b
  the substitution   t
ax b e
   is the correct
one. In general unless stated we assume   0
ax b
  in order
to obtain a general solution of Cauchy Euler equation.
Example 1:
Obtain the general solution of
2
2 3
2
2 2
d y dy
x x y x
dx dt
  
Solution: This equation is Euler’s equation with 2

n assuming
0

x let t
e
x  thus x
t ln

t
dy dy
e
dx dt

 
2 2
2
2
t
d y d y dy
e
dt dx dt
  
 
 
 
Substituting in the equation we have t
e
y
dt
dy
dt
y
d 3
2
2
2
3 

 (since
t
e
x  )
The auxiliary equation is 0
2
3
2


 m
m whose roots are 2

m and
1

m
2
1 2
t t
h
y c e c e
  

44. 44
To find p
y let 3 3
3
t t
p p
y Ae y Ae

   3
9 t
p
y Ae
  thus substituting
in the equation we have
2
1

A thus t
p e
y 3
2
1
 the general solution
is 2 3
1 2
1
2
t t t
y c e c e e
  
Replacing t
e by, x we have the general solution as
2 3
1 2
1
2
y c x c x x
  
Example 2:
Obtain the general solution of
     
2
2
2
1 2 1 2 2sin sin 1
d y dy
x x y x
dx dt
     
 
 
1 t
x e
  thus  
ln 1 1 1
t x a b
   
t t
dy dy dy
ae e
dx dt dt
  
2 2 2
2 2 2
2 2
t t
d y d y dy d y dy
a e e
dt dx dt dx dt
   
   
   
   
Substituting in the equation we have
2
2
2sin
d y
y t
dt
 
The auxiliary equation is 2
1 0
m   whose roots are m i
 
1 2
cos sin
h
y C t C t
  
To find p
y let
cos sin
cos sin sin cos
sin sin cos cos cos sin
2 sin cos 2 cos sin
p
p
p
p
y At t Bt t
y A t At t B t Bt t
y A t A t At t B t B t Bt t
y A t At t B t Bt t
 
    
       
     
Substituting in the equation
2 sin cos 2 cos sin 2 sin cos sin 2sin
2 sin 2 cos 2sin
A t At t B t Bt t A t At t Bt t t
At t Bt t t
       
   
2 2 1
2 0 0
A A
B B
     
 
cos
p
y t t
 
Solution is 1 2
cos sin cos
y C t C t t t
   
Replacing t by  
ln 1
x  , we have the general solution as
     
   
1 2
cos ln 1 sin ln 1 ln 1 cos ln 1
y C x C x x x
       
     
     
Find the general solution
2
2
2
5 8 4ln
d y dy
x x y x
dx dx
  
Solution
2 4
1 2
1 3
ln
2 8
y c x c x x
   

45. 45
EXERCISE
1. Find the general solution
a)
2
2
2
4 0
d y dy
x x y
dx dx
  
b)      
2
2
2
1 2 1 2 2 ln 1 2
d y dy
x x y x
dx dx
     
c)
2
2
2
3 4 3 0
d y dy
x x y
dx dx
  
2. Solve the initial value problems
a)
2
2
1
2
2 10 0 (1) 5 4
d y dy
x x y y y
dx dx

    
b)
2
2
1
2
5 3 0 (1) 1 5
d y dy
x x y y y
dx dx
     
THE LAPLACE TRANSFORM
Suppose  
f t is defined for 0
t 
The Laplace transform  
L f t
 
  of  
f t is a function defined by
   
0
st
L f t s e f t dt



 
  
For all s such this improper integral converges
We use t as the independent variable of a function defined by a
lower case letter such as f, g, or h.
The Laplace transform of such function is denoted by
 
F L f t s
  
   
G L g t s
  
  Or  
H L h t s
  
 
Example
Find the Laplace transform of
(i) a (ii) at
e (iii) cosbt (iv) sinbt (v) t
(vi) at
t e
Solution
i.   1
f t  when 0
t 
   
0
0
st
st
L f t L a e a dt
a
e
s
a
s




  
 
 
 
 
 
 


ii.   at
f t e
 when 0
t 

46. 46
 
 
0
0
0
1
1
at st at
s a t
st
L f t L e e e dt
e dt
e
s a
s a



 


 
  
 
   


 
  

 




iii.   cos
f t bt
 when 0
t 
   
0
cos cos
st
L f t L bt e bt dt


  
 
  
Using integration by parts
cos
1
sin
st
st
u bt dv e
du b bt v e
s


 
   
0 0
0
2
2 2
0
1
cos cos sin
1 1
cos sin cos
1
cos sin cos
st st st
st st st
st st st
b
e bt dt e bt e bt dt
s s
b b
e bt e bt e bt dt
s s s s
b b
e bt e bt e bt dt
s s s
 
  

  

  
     
 
    
 
 
   
 


 
 
2
2
0
2 2
2
0
0
2
2 2
0
2
2 2
1
1 cos cos sin
1
cos cos sin
1
cos
1
st st
st st
st
b
e bt dt e bt bt
s s
s b
e bt dt e bt bt
s s
s
e bt dt
s s b
s
s s b

 


 


 
   
 
 
 
  
  
   
 
 
 
  

 
 
  

 



  2 2
cos
s
L bt
s b
 

  2 2
sin
b
L bt
s b


  2
a
L at
s

 
2
1
at
L t e
s a
  
 

LINEARITY
if  
L f t
 
  and  
L g t
 
  are defined for s a
 and  and  are
any number then
      
L f g s L f L g
   
  

47. 47
Proof
   
   
   
0
0 0
st
st st
L f g e f g dt
e f t dt e g t dt
L f t L g t
   
 
 


 
 
  
 
 
   
   

 
INVERSE LAPLACE TRANSFORM
The Laplace transform of a function is unique for continuous
functions
If    
L F t F s

 
 
Then
   
1
L F t f t


 
  and we call  
f t the inverse Laplace transform.
Examples
1. Evaluate
i. 1 3
1
3
t
L e
s
  

 

 
ii. 1
2
2
sin 2
4
L t
s
  

 

 
iii. 1 3
1
3
t
L e
s
  

 

 
iv. 1 5
4 7
4 7
5
t
L e
s s
  
  
 

 
2. Evaluate 1
2
1
4
L
s
  
 

 
  
   
2
1 1
4 2 2 2 2
1 2 2
A B
s s s s s
A s B s
  
    
   
1
2 1 4
4
1
2 1 4
4
s A A
s B A
       
   
1 1 1
2
2 2
1 1 1 1 1
4 4 2 4 2
1 1
4 4
t t
L L L
s s s
e e
  

     
  
   
 
  
     
  
Find the inverse Laplace transform of
1
2
3 7
9
s
L
s
 
 
 

 
   
2
3 7
9 3 3
3 7 3 3
s A B
s s s
s A s B s

 
  
    

48. 48
2 1
3 9 7 6
6 3
16 8
3 9 7 6
6 3
s A A
s B A
       
         
1 1 1
2
3 3
3 7 1 1 8 1
4 3 3 3 3
1 8
3 3
t t
s
L L L
s s s
e e
  


     
 
   
 
  
     
 
1 1 1
2 2 2
2 5 2 15
25 25 25
2cos5 3sin5
s s
L L L
s s s
x x
  

     
 
     
  
     
 
LAPLACE TRANSFORM OF DERIVATIVES
Suppose that  
f t is continuous for all 0
t  , the Laplace transform
of the derivative  
f t
 exist and is given by
     
0
L f t s f t f
  
 
 
Proof
   
   
   
   
0
0
0
0
0
0
st
st st
st
L f t e f t dt
e f t se f t dt
f s e f t dt
f s L f t




 


 
 
 
 
 
  
 
   
    
 



     
0
L f t s L f t f

  
   
   
LAPLACE TRANSFORM OF HIGHER DERIVATIVE
Suppose  
1
...................
n
f f f f f

   are continuous on  
0, and
that n
f is piecewise continuous on  
0,k for every positive
number k
Then
 
   
   
   
   
   
1 2 3 2
0 0 0 0 '....... 0
n n n n n n n
L f s s f s f s f s f s f s f
   
    
     
 
For case n=1
     
0
L f t s F s f
  
 
 
But    
F s f t
  
 
     
0
L f t s L f t f

  
   
   

49. 49
For case n=2
       
2 0
0 0
L f t s F s s f s f
 
  
 
 
       
2
0 0
L f t s L f t s f f
 
   
   
   
For case n=3
         
3 2
0 0 0
L f t s L f t s f s f f
  
   
   
   
And so on.
We consider second order differential equations only.
Example 1
Solve the initial value problem
     
4 1 0 1
y t y t y
   
Solution
Using Laplace transform
     
4 1
L y t L y t L
  
   
   
       
0 1
L y s L y y s L y

    
   
  
  
 
 
1
1 4
1
4 1
1 1
4 1
1
4
s L y L y
s
L y s
s
s
L y s
s s
s
L y
s s
  
  

   



 
 
1
4 4
1 4
s A B
s s s s
s A s Bs

 
 
   
If s = 0
1
1 4
4
A A
    
S = 4
5
5 4
4
B A
  
 
 
1 5
4 4 4
L y
s s
  

 
1 1
4
1 1 5 1
4 4 4
1 5
4 4
t
y L L
s s
y e
 
 
 
     
  
   
  
 
4
1
1
4
t
y e
  
Example 2
Solve the initial value problem
 
2
2 0 3
t
y y e y

  
Solution
Using Laplace transform
    2
2 t
L y L y L e
  
   
       
0 3
L y s L y y s L y

    

50. 50
   
  
  
 
 
2
2
3 2
1
2 3
2
1 3 7
2 3
2 2
3 7
2
t
s L y L y L e
L y s
s
s
L y s
s s
s
L y
s

 
    
  


   
 



   
 
2 2
3 7
2
2 2
3 7 2
s A B
s
s s
s A s B

 

 
   
S=-2. 1 1
B B
  
s=0 7 2 3
A B A
   
 
 
2
3 1
2 2
L y
s s
  
 
 
1 1
2
2 2
3 1
3
2 2
3 t t
y L L
s s
y e te
 
 
 
 
    
 

  
 
 
 
  2
3 t
y t e
  
Example 3
Use Laplace transform to solve the initial value problem
 
3 12 0 6
y y y
  
Solution
3
4 2 t
y e
 
Example 4
Solve  
4 1 0 3
y y y
   
Solution:
 
4
1
1 13
4
t
y e
 
Example 5
   
0 1 0 0
y y t y y
 
   
Solution
Using Laplace transform
     
L y L y L t
  
         
2 2
0 0
L y s L y s y y s L y s
 
    

51. 51
     
 
 
 
2
2
2
2
0
1 0
1
1
s L y s L y L
s L y s
s L y s
s
L y
s
  
 
  
 
 
 
 


1
2
1
1
y L
s
  
  

 
sin
y t
 
Example 6
Solve the initial value problem
   
6 9 0 0 4 0 12
y y y y y
  
     
Solution
Using Laplace transform
       
6 9 0
L y L y L y L
 
  
       
0 4
L y s L y y s L y
    
         
2 2
0 0 4 12
L y s L y s y y s L y s
 
     
       
 
2
2
4 12 6 24 9 0
6 9 4 12 0
s L y s s L y L y L
s s L y s
     
 
    
 
 
2
6 9 12 4
s s L y s
 
   
 
 
 
2
12 4
3
s
L y
s

 

 
 
 
2 2
4 3
12 4 4
3
3 3
s
s
s
s s


 

 
 
4
3
L y
s
 

1
3
1
4
3
4 t
y L
s
y e


 
   

 

Example 7
Solve    
4 3 0 0 0 2
t
y y y e y y
  
    
Solution:
3
1 3 7
8 4 8
t t t
y e e e
 
  
Exercise
a)  
2 1 0 4
y y t y
   
b)    
4 4 0 0 3 0 4
y y y y y
  
    
c)    
0 1 0 0
y y t y y
 
   
d)  
1 0 3
y y x y
   

52. 52
e)    
2 0 2 0 2
x
y y e y y
 
   
f)    
11 24 0 0 1 0 4
y y y y y
  
    
SYSTEMS OF DIFFERENTIAL EQUATIONS
A general 1st
order system is given by
 
 
 
1 1 1 2
2 1 1 2
1 1 2
...............
...............
.
.
.
...............
n
n
n n
dx
f t x x x
dt
dx
f t x x x
dt
dx
f t x x x
dt



e.g
2
1
1 2
2
1 2
2
3 4 sin
dx
x x t
dt
dx
x x t
dt
  
  
A general type of linear system of two first order differential
equations in two unknown functions x and y is of the form.
(1)
We restrict ourselves to a system of this type with a constant
coefficient e.g.
2
2 3 2
2 3 4 t
dx dy
x y t
dt dt
dx dy
x y e
dt dt
   
   
(2)
Definition: The ordered pair of real functions )
,
( g
f is said to be a
solution of (1) if )
(t
f
x  , )
(t
g
y  simultaneously satisfy both
equations of system (1) on same real interval b
t
a 

Example 1
Solve the system:
4
2 2 4 t
t
dx dy
x y e
dt dt
dx dy
e
dt dt
   
 
Solution
Introducing the differential operator notation we have
1 2 3 4 1
1 2 3 4 2
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
dx dy
a t a t a t x a t y f t
dt dt
dx dy
b t b t b t x b t y f t
dt dt
   
   

53. 53
( 2) ( 4) t
D x D y e
   
4t
Dx Dy e
 
To solve for x multiply the first equation with D and the second
with  
4
D  to have
( 2) ( 4) t
D D x D D y De
    ............................................. (i)
      4
4 4 4 t
D D x D D y D e
     ................................ (ii)
Adding (ii) and (ii) we have
      4
2 4 4
t t
D D D D x De D e
     
 
 
Or
2 2 4 4
2 4 4
t t t
D D D D x De De e
 
     
 
4 4
2 4 4
t t t
Dx e e e
  
Or
2 t
dx
e
dt

Which on solving yields
1
2
t
x e C
 
Similarly to solve for y apply D and 2
D  to the first and second
equations respectively
( 2) ( 4) t
D D x D D y De
    ............................................. (i)
      4
2 2 2 t
D D x D D y D e
     ................................ (ii)
Subtracting the resulting equations we have
      4
4 2 2
t t
D D D D y De D e
     
 
 
Or
2 2 4 4
4 2 2
t t t
D D D D x De De e
 
     
 
4 4
2 4 2
t t t
Dy e e e
   
Or
4
2 2
t t
dy
e e
dt
  
Which on solving yields
4
1 1
2 4
t t
y e e C
   

54. 54
Example 2
Solve the system:
2
8
3
2
2
3
2
2







y
x
dt
dy
dt
dx
t
x
dt
dy
dt
dx
Solution
Introducing the differential operator notation we have
(2 3) 2
D x Dy t
  
(2 3) (2 8) 2
D x D y
   
To eliminate y multiply the first equation with (2 8)
D  and the
second wiht 2D to have
(2 8)(2 3) 2 (2 8) (2 8)
D D x D D y D t
      ................................ (i)
2 (2 3) 2 (2 8) 2 (2)
D D x D D y D
    ............................................ (ii)
Adding (ii) and (ii) we have
[(2 8)(2 3) 2 (2 3)] (2 8) 2 (2)
D D D D x D t D
      
Or
0
8
2
)
24
16
8
( 2




 t
x
D
D
finally
2 1
( 2 3)
4
D D x t
   
Or
2
2
1
2 3
4
d x dx
x t
dt dt
   
Which on solving yields
3
1 2
t t
h
x c e c e
  and
1 11
3 36
p
x t
  
Thus the general solution is 3
1 2
1 11
3 36
t t
x c e c e t

   
Similarly to solve for y apply 3
2 
D and 3
2 
D to the first and
second equations respectively
(2 3)(2 3) 2 (2 3) (2 3)
D D x D D y D t
      .....................................
........ (i)
     
2 3 (2 3) 2 3 (2 8) 2 3 (2)
D D x D D y D
       ........................
....... (ii)
Subtracting the resulting equations we have

55. 55
 
[ 2 (2 3) (2 8)(2 3)] (2 3) 2 3 (2)
D D D D y D t D
        
2 3
2 3 1
8
D y Dy y t

   
Which on solving yields
1 5
8 12
p
y t
  and 3
1 2
t t
c
y c e c e
 
Thus the general solution is 3
1 2
1 5
8 12
t t
y c e c e t

   
Therefore the general solution of the system
3
1 2
1 11
3 36
t t
x c e c e t

   
3
1 2
1 5
8 12
t t
y c e c e t

   
Example 3
Solve
6 3
2 2 6
dx dy
x y t
dt dt
dx dy
x y t
dt dt
   
   
Solution:
6 6
1 2
5
2
6
t t
x c e c e t

   
6 6
1 2
5 1
6 3
t t
y c e c e t

   
Exercise
Q1: Find the general solution for each of the following linear system
i. 2 4 t
dx dy
x y e
dt dt
   
4t
dx dy
e
dt dt
 
ii. 3 3
dx dy
x y t
dt dt
   
2 2 3 1
dx dy
x y
dt dt
   
iii. 6 3
dx dy
x y t
dt dt
   
2 2 6
dx dy
x y t
dt dt
   

56. 56

57. 57
POWER SERIES SOLUTION OF LINEAR O.D.E
In this case the solution of the linear equation is obtained in infinite
series form about a certain part.
BASIC CONCEPTS
1. A power series in power of 0
( )
x x
 is an expression of the form
1 2
0 1 0 2 0 0 0
0
( ) ( ) ........... ( ) ( )n
n n
n
a a x x a x x a x x a x x


        

2.  
f x is said to be analytic at a point 0
x x
 if the Taylor series
about 0
x x
 exists and converges to  
f x in any interval
containing 0
x i.e. The function is single valued and possesses
derivatives of all orders at that point.
A rational function is analytic everywhere except at values of x at
which denominator is equal to zero
Eg
The function 2
1
3 2
x x
 
is analytic except at 1
x  and 2
x 
Consider a second order homogeneous difference equation of the
form
     
2
2 1 0
2
0
d y dy
a x a x a x y
dx dx
  
(1)
This may be written as
 
 
 
 
2
1 0
2
2 2
0
a x a x
d y dy
y
dx a x dx a x
  
(2)
Where  
2 0
a x 
This is called, normalized form of the equation
Let  
 
 
1
1
2
a x
P x
a x
 and  
 
 
0
2
2
a x
P x
a x

Then equation (2) becomes
   
2
1 2
2
0
d y dy
P x P x y
dx dx
  

58. 58
ORDINARY POINTS
A point 0
x x
 is said to be an ordinary if both  
1
P x and  
2
P x are
analytic at 0
x x

Example
a)  
2
2 3
2
4 0
d y dy
x x y
dx dx
   
  2
1
P x x
 and   3
2 4
P x x
   
1
P x
 and  
2
P x are analytic
everywhere therefore all points are ordinary points
b)
2
2
1
0
d y dy
x y
dx dx x
  
 
1
P x x
 and  
2
1
P x
x
  
1
P x
 is analytic at all points but
 
2
P x is not analytic at 0
x  therefore all points except 0
x  are
ordinary points
SINGULAR POINTS
If either 1 or both  
1
P x and  
2
P x is not analytic at 0
x x
 . Then we
say that they point 0
x x
 is a singular point of the differential
equation
Example 1
 
2
2
1
0
1 1
d y x dy
y
dx x dx x x
  
 
 
1
1
x
P x
x


and  
 
2
1
1
P x
x x


 
1
P x is analytic at all points except when 1
x 
 
2
P x is analytic at all points except when 0
x  or 1
The points 0
x  and 1
x  are singular point
REGULAR SINGULAR POINTS
For any point 0
x x
 if both    
0 1
x x P x
 and    
2
0 2
x x P x
 are
analytic at 0
x x
 .then 0
x is said to be a regular singular points
In the example 1 above for 1
x 
     
 
0 1 1
1
x
x x P x x x
x
   

this is analytic
     
 
2 2
0 2
1 1
1
1
x
x x P x x
x x x

   

this is analytic at the point
1
x
  is a regular singular point.
IRREGULAR SINGULAR POINT
Let 0
x be a singular point if either    
0 1
x x P x
 and
   
2
0 2
x x P x
 is not analytic at 0
x x
 then 0
x is said to be an
irregular singular point
In the example 1 above for 0
x 

59. 59
     
   
2
0 1 0
1 1
x x
x x P x x
x x
   
 
this is analytic
     
   
2 2
0 2
1
0
1 1
x
x x P x x
x x x
   
 
this is analytic at the
point
0
x
  is a regular singular point.
Example 2
Find the regular and irregular singular points of the differential
equations
 
3
2 5 0
x y xy x y
 
   
Solution
Normalized form of the equation
 
3 3
5
0
2 2
x
x
y y y
x x

 
  
 
1 3 2
1
2 2
x
P x
x x
  and  
 
2 3
5
2
x
P x
x


Equation has singular points at 0
x 
   
0 1 2
1
2 2
x
x x P x
x x
   this is not analytic at 0
x 
   
2 2
0 2 3
5 5
2 2
x x
x x P x x
x x
 
   this is not analytic at the point
0
x
  is an irregular singular point.
Example 3
Find the regular and irregular singular points of the differential
equations
   
2
2
4
2 2 0
x x y x y y
x
 
    
Solution
Singular point 0
x  and 2
x 
0
x  is an irregular point
2
x  is a regular point
SOLUTION ABOUT AN ORDINARY POINT
If x si an ordinary point of the differential equation
       
1
1 1 0
1
................ 0
n n
n n
n n
d y d y dy
a x a x a x a x y
dx dx dx

 
   
The solution is given in powers of  
0
x x
 such a solution is
denoted by
     
2
0 1 0 2 0 0
0
..........................
n
n
n
y c c x x c x x c x x


       

(A)
Method of solution
Let  
0
0
n
n
n
y c x x


 


60. 60
Then find  
1
0
1
n
n
n
y nc x x



  
    
2
0
2
1
n
n
n
y n n c x x



   

Substitute in the equation (A) , find the values of
0 1 2
, , ................. n
c c c c then write out the solution
Example 1
Solve 4 0
y y
  near the ordinary point 0
x 
Solution
Let the solution be a power series of the form
0
n
n
n
y c x


  1
1
n
n
n
y nc x



   and   2
2
1 n
n
n
y n n c x



  

Substituting in the given difference equation
  2
2 0
1 4 0
n n
n n
n n
n n c x c x
 

 
  
 
We now change the indexing so that the series has a general term
2
n
x 
by putting 2
n n
  in the 2nd summation
  2 2
2
2 2
1 4 0
n n
n n
n n
n n c x c x
 
 

 
  
 
  2 2
2
2
1 4 0
n n
n n
n
n n c x c x

 


  

  2
2
2
1 4 0
n
n n
n
n n c c x




  
 
 

For the power series to vanish identically over any interval then each
coefficient over the series must be zero
  2
1 4 0 2
n n
n n c c n

   
 
 
Or
 
2
4
1
n
n
c
c
n n




(B)
Equation (B) is a recurrence solution and is used to find n
c for 2
n 
For 2
n 
 
0
2 0
4
2
2 1
c
c c

  
For 3
n 
 
0
3 1
4 2
3 2 3
c
c c
 
 
For 4
n 
 
2
4 0
4 2
4 3 3
c
c c

 
For 5
n 
 
3
5 1
4 2
5 4 15
c
c c

 
For 6
n 
 
4
6 0
4 4
6 5 45
c
c c
 
 
For 7
n 
 
5
7 1
4 4
7 6 315
c
c c
 
 
2 3 4 5 6 7
0 1 0 1 0 1 0 1
2 2 2 4 4
................
3 3 15 15 315
y c c x c x c x c x c x c x c x
        
Or
2 4 6 3 5 7
0 1
2 4 2 2 4
1 2 ........ ........
3 15 3 15 315
y c x x x c x x x x
   
       
   
   

61. 61
Example 2
Use power series method to solve the different equation near the
ordinary point 0
x 
 
2
2 0
y xy x y
 
   
Solution
Let the solution be a power series of the form
0
n
n
n
y c x


  1
1
n
n
n
y nc x



   and   2
2
1 n
n
n
y n n c x



  

Substituting in the given difference equation
   
2 1 2
2 1 0
1 2 0
n n n
n n n
n n n
n n c x x nc x x c x
  
 
  
    
  
  2 2
2 1 0 0
1 0
n n n n
n n n n
n n n n
n n c x nc x x c x c x
   

   
    
   
  2 2
2 1 0 0
1 0
n n n n
n n n n
n n n n
n n c x nc x c x c x
   
 
   
    
   
We now change the indexing so that the series has a general term
n
x by putting 2
n n
  in the 1st summation
   2
2
0 1 0 0
2 1 0
n n n n
n n n n
n n n n
n n c x nc x c x c x
   


   
     
   
We now change the indexing so that the series has a general term
n
x by putting 2
n n
  in the 3rd summation
   2 2
0 1 2 0
2 1 0
n n n n
n n n n
n n n n
n n c x nc x c x c x
   
 
   
     
    (A)
In (A) power of x are the same but ranges of n in each of the
summaries are not the same.
Now writing (A) as
  
2 3 2 1 2
2 2 2
0 1
2
2 6 2 1
2 2 2 0
n n n
n n n
n n n
n
n
n
c xc n n c x c x nc x C x
C C x C x
  
 
  


      
   
  

on combining like powers of x we have
0 2 1 3 2 2
2
2( ) (3 6 ) [( 2)( 1) ( 2) ] 0
n
n n n
n
C C C C x n n C n C C x

 

         

for this to be valid for all x in the interval of convergence coefficient
of each power of x in left must be equal to zero i.e.

62. 62
 
 
0 2 2 0
1 3 3 1
2 0
1
3 6 0
2
c c c c
c c c c
    
    
and
    
2 2
2 1 2 0
n n n
n n c n c c
 
     
Which gives the recurrence formula
 
  
2
2
2
2 1
n n
n
n c c
c
n n


  
 
 
 
 
for 2
n 
for n = 2
   
0
0
0
0
2
4
4
1
4
12
1
4
12
1
c
c
c
c
c
c








for n = 3
 
1
1
3
5 40
3
20
5
c
c
c
c 



similarly all even coefficients and odd coefficients can be expressed
interms of c0 and c1 respectively. Now substituting values of c2, c3,
c4 and c5 interms of c0 and c1 in to the assumed we have
..
..........
40
3
4
1
2
1 5
4
0
3
1
2
0
1
0 





 x
x
c
x
c
x
c
x
c
c
y
Or
  












 .....
..........
40
3
2
1
.
..........
4
1
1 5
3
1
4
2
0 x
x
x
c
x
x
c
y
which gives the solution of the differential equation.
The two series in brackets are the power series expansion of the two
linearly independent solutions of the equation and c0, c1 are the
arbitrary constants. The given expression of y is the required general
solution in powers of x.
Example 3
Use power series method to solve the different equation near the
ordinary point 0
x 

63. 63
 
2
1 6 4 0
x y xy y

   
Solution
 
2 4 6 3 5 7
0 1
5 7
1 2 3 4 ........... 3 ...............
3 3
y c x x x c x x x x
 
       
 
 
Exercise
1. Find the power series solution of the initial value problem
   
2
2
2
1 3 0 0 4 (0) 6
d y dy
x x xy y y
dx dx

     
2. Find the power series solutions in x of the following
equations.
 
   
2
2
2
2
6 4 0
2 2 3 0
d y dy
i x y
dx dx
d y dy
ii x x y
dx dx
  
   
3. Find the power series solution in powers of x – 1 of the initial
value problem
2
2
2 0 (1) 2 (1) 4
d y dy
x y y y
dx dx

    
SOLUTIONS ABOUT SINGULAR POINTS
THE METHOD OF FROBENIUS
THEOREM: If 0
x is a singular point of the equation.
2
0 1 2
2
( ) ( ) ( ) 0
d y dy
a x a x a x y
dx dx
   then this equation has at
least one non – trivial solution of the form.





0
0
0 )
(
n
n
n x
x
c
r
x
x where r is a definite real or complex
constant which can be determined and the solution is valid in the
interval.
)
0
(
0 0 


 R
R
x
x about 0
x for instance x = 0 is
regular singular point of
2
2
2
2 4 ( ) 0,
d y dy
x x x c
dx dx
    thus this

64. 64
equation has at least one non – trivial solution of the form
0
r n
n
n
x c x


 valid in 0 < |n| <R about x = 0
METHOD OF FROBENIUS OUTLINE
(1) Suppose 0
x is a regular singular point of the equation
2
0 1 2
2
( ) ( ) ( ) 0
d y dy
a x a x a x y
dx dx
   ………...(a)
then we seek a solution valid in R
x
x 

 |
|
0 0 and assume a
solution of the form






0
0
0 )
(
)
(
n
n
n
r
x
x
c
x
x
y ………………………(b)
where 0
0 
c . This solution can be written in the form
0
0
( )n r
n
n
y c x x



 
 …..…..…………….(c)
(2) Assume term-by-term differentiation is valid then
1
0
0
( )( )n r
n
n
dy
c n r x x
dx

 

  
 ………….(d)
and
2
2
0
2
0
( )( 1)( )n r
n
n
d y
c n r n r x x
dx

 

    
 …..(e)
Then substitute (c) (d) and (e) in (a)
(3) Simplify the resulting expression to form
0
........
)
(
)
( 1
0
1
0
0 



 

 k
r
k
r
x
x
b
x
x
b ……………(f)
where k is a certain integer .....)
01
( 
i
bi are functions of r
and certain functions of n
c of solution (a)
(4) For expression (f) to be valid 0 1 .................. 0
k b b
   
(5) Equating coefficients 0
b to zero, i.e. the coefficient of the
lowest power k
r  of 0
x
x  give rise to a quadratic
equation in r known as the indicial equation of (a) whose two
roots are called exponents of (1) and are the only possible
values of constant r, in the assumed solution (1). Denoting
the roots of the indicial equation by 1
r and 2
r .
(6) Equating the coefficients to zero gives sets of conditions
involving r of which all s
n
c in the series must satisfy.
(7) Equating the coefficients to zero gives sets of conditions
involving r of which all s
n
c in the series must satisfy.
(8) Substitute root 1
r for r into conditions obtained in (6) and
choose n
c to satisfy this condition so chosen n
c give rise to a
series with 1
r
r  which is a solution of desired form.

65. 65
N/B if 1
r and 2
r are real and unequal, then 1
r is the larger of the two.
(9) If 2
1 r
r  step (f) may be repeated using 2
r r
 which gives rise
to a second desired solution in the case where 2
r and 1
r are
real and unequal the second solution may not be linearly
independent to the one obtained at step (f).
Example:
Use the method of Frobenius to find the solutions of the
difference equation
2
2
2
2 ( 5) 0
d y dy
x x x y
dx dx
    in some interval R
x 

0
Solution:
0

x is a regular singular point of the given equation and since
the solution is required for the interval R
x 

0 the n assume a
solution of the form:





0
n
r
n
n x
c
y (1)
0
0 
c
1
0
( ) n r
n
n
dy
c n r x
dx

 

 
 and
2
2
2
0
( )( 1) n r
n
n
d y
c n r n r x
dx

 

   
 (2)
Substituting in the differential equations we have
2 2 1
0 0 0
2 ( )( 1) ( ) ( 5) 0
n r n r n r
n n n
n n n
x n r n r c x x n r c x x c x
  
    
  
       
  
1
0 0 0 0
2 ( )( 1) ( ) 5 0
n r n r n r n r
n n n n
n n n n
n r n r c x n r c x c x c x
   
    
   
       
   
Change the indexing in the 3rd
summation by putting
1
n n
  and simplifying yields
1
0 1
[2( )( 1) ( ) 5] 0
n r n r
n n
n n
n r n r n r c x c x
 
 

 
       
 
or ( Evaluating the summation in 1st
term by using
0
n  )

66. 66
0 1
1
(2 ( 1) 5) [2( ) 1 5] ) 0
r n r
n n
n
r r r c n n r n r n r c c x




           

Equating the coefficients to zero we have
0
5
)
1
(
2 


 r
r
r
(3)
by assumption 0
0 
c
(3) is the indicial equation of the differential equation. i.e.
0
5
3
2 2


 r
r
whose roots are
2
5
1 
r , 1
2 

r
1
r and 2
r are the exponents of the equation and are the only
possible values of r.
now, equating to zero coefficients of higher powers of r we
have
  1
2( )( 1) 5 0
n n
n r n r n r c c 
 
       
  1

n
let 2
s
r  then we have
0
)
7
2
( 1 

 
n
n c
c
n
n 1

n
or
)
7
(
1


 
n
n
c
c n
n 1

n
which is the recurrence for n
c thus
0
1
9
c
c

 0
1
2
22 198
c
c
c


 
0
2
3
39 7722
c
c
c


 
Substituting these values in the assumed solution we have
5 7 9 31
2 2 2 2
5
2
0
0
2 3
0
1 1 1
( .....)
9 198 7722
1 1 1
(1 .....
9 198 7722
n r
n
n
y c x c x x x x
c x x x x



     
    

which the solution corresponding to the larger root 1
r
Let 1
2 

 r
r putting 1


r in the recurrence relation above we
have

67. 67
)
7
2
(
1


 
n
n
c
c n
n 1

n
Using this we have
0
2
3
0
1
2
0
0
1
90
1
7
5
1
6
30
1
5
c
c
c
c
c
c
c
c
c












Using these values of 1
c 2
c 3
c …we have
.....)
90
1
30
1
5
1
( 2
1
0 


 
x
x
x
x
c
y
Which is a solution corresponding to the smaller root 1


r . These
two solutions corresponding to 1
r and 2
r are linearly independent
thus the general solution is
5
2 2 3 1 2
0 2
1 1 1 1 1 1
(1 ..... (1 .....)
9 198 7722 5 30 90
y c x x x x c x x x x

         
Where 1
c and 2
c are arbitrary constants.
EXERCISE
1. outline the method of Frobenious hence of otherwise find
solutions near 0

x of
the differential equation
2
2 2
2
2 ( 1) 0
d y dy
x x x y
dx dx
   