The document discusses various types of differential equations including ordinary differential equations (ODEs) and partial differential equations (PDEs). It defines key terms like order, degree, and describes several methods for solving common types of differential equations, such as separating variables, exact differentials, linear equations, Bernoulli's equation, and Clairaut's equation. It also includes sample problems and solutions for each method and concludes with multiple choice questions.

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

2. 2
Differential Equation (D.E.)
Differential Equation : Equation containing the variables and their
derivatives are called differential equation.
There are two types of D.E.
1) Ordinary D.E. (ODE)
2) Partial D.E. (PDE)
1) Ordinary D.E. (ODE) :
A. D.E. which involes only are independent variable is called
ordinary D.E.
For Ex. (1) sin log
dy x x
dx = +
(2)
2
2
2
2
d y
dx
 
+
 
 
3
2
2
2 0
d y
y
dx
 
+ =
 
 
2) Partial differential equation
A.D.E. Which involes moer than one independent variable is called PDE.
For Ex. (1)
2
2
.
x t
k
θ θ
∂ ∂
=
∂ ∂
(2)
2 2 2 2
2 2 2 2 2
1
0
v v v v
x y z c x
∂ ∂ ∂ ∂
+ + − =
∂ ∂ ∂ ∂
3) Order of D.E. :
The order of the D.E. is the order of the highest derivative involved
in the equation.
4) Degree of D.E. :
The degree of the highest order derivative is the degree of given
D.E.
For Ex. (1) sin log : 1, deg 1
dy
x x order ree
dx
= +

3. 3
(2)
3
5
4
3
( ) : 3, deg 4
dy
y
dx
d y
order ree
dx
+
=
(3)
2
2
: 2, deg 1
x t
k order ree
θ θ
∂ ∂
=
∂ ∂
Formation of a D.E.
It is obtained by eliminating the arbilary constant and function
from the given relation.
Example : Eliminate the constant from following relation
cos sin
y A mx B mx
= +
Where A and B are arbitrary constant
Now cos sin
y A mx B mx
= + _______________(1)
Differentiate w.r. to x
We get /
sin cos
y Am mx Bm mx
= − +
Again differential w.r. to x
// 2( cos sin )
y m A mx B mx
= − +
// 2y
y m
=− (by equation 1)
// 2y
y m
+ = ∂
⇒
Which is required D.E.
Example for formation of D.E. by limiting the arbitrary function
Ex. 2 2
( )
Z xy f x y
= + +
Solution : Differentiate given equation w.r. to x we gel
/
2 .
z
y x f
x
∂
= +
∂
i.e.
/
2 .
P y x f
= + ________________(1)
dz
P
x
 
=
 
∂
 
Simillarly differentiate given equation w.r. to y

4. 4
We /
2
z
x yf
y
∂
= +
∂
i.e. /
2
q x yf
= + _____________(2)
dz
q
dy
 
 
 
=
multiply equation (1) by y and equation (2) by x and subtract we get.
2 2
x
YP xq y −
− = is the required D.E.
First order ordinary differential equation :
In general an ODE of the first order and first degree is expressed in
the form / /
( ) 0, y , [ ]
dy
f x y y x a b
dx ε
= =
Method for solving first order O.D.E.
Method First :
Working Rule : Seprate the coefficient of dx and dy so that the
coefficient of dx is function x- only and coefficient of dy is function of
y-only. Further on integration we get the required solution.
For ex. (1) Solve 2
3
dy x y y
e x e
dx
− −
= +
Solution 2
( 3 )
x
e x
dy y
e
dx
+
−
=
2
( 3 )
y x
dy e e x dx
−
⇒ = +
2
( 3 )
dy
y x
e e x dx
=
⇒ +
Thus the variable are separable
∴ On integration we get
3
y x
e e x C
= + + Where C = integration constant is the required solution
of given D.E.

5. 5
Method Second : Exact differential equation
Theorem :
A necessary condition for ( ) ( ) 0
M xy dx N xy dy
+ = to be exact is
My Nx
= where M and N are continuously differential function.
Proof :
Let ( ) ( ) 0
M x y dx N x y dy
+ = to be exact differential equation
then by definition df Mdx Ndy
= + ____________ (1)
(df = total differential function of ( )
f xy )
But . .
df fx dx fy dy
= + ______________(2)
Equation (1) and (2) ⇒
M fx
= and fy N
=
differentiate partially
⇒ My Fxy
= fyx Nx
=
But mixed partial derivative are equation
∴ [ ]
My Nx Fyx Fxy
= =
∵
Working Method to solve
1) Find the values of M and N then verify My Nx
=
2) Then solution is written as
( form N not containing )
Mdx term x dy C
+ =
∫ ∫
Ex. Test the following D.E. for exactness and solve them if they are
exact. 2 2 2 2
( 4 2 ) ( 4 2 ) 0
x xy y dx y xy x dy
− − + − − =
Solution : This is of the form
0
Mdx Ndy
+ =
Here 2 2
4 2
M x xy y
= − − 2 2
4 2
N y xy x
= − −
⇒ diff. partially w.r. to y diff. partially w.r. to x
∴ 4 4
My x y
= − − 4 4
Nx y x
= − −

6. 6
∴ We have My Nx
=
⇒ Give D.E. is exact
∴ Required solution is
. ( form N not containing )
M dx term x dy c
+ =
∫ ∫
y-constant (x)
⇒ ( )
tan
2 2 2
4 2
y cons t
x xy y dx y dy C
−
− − + =
∫
∫
⇒
3
3 2 2
2 2
3 3
y
x x y xy C
− − + =
Or
3 3 6 ( )
x y xy x y c
+ − + =
is required solution
Method Third :
Lineal Equation
A linear equation of the first order is of the form
/
Y PY Q
+ =
Where P, Q are function of x
Working Rule
1) Find out integrating factor
e
Pdx
∫
2) Then Solution may be written as
.( ) .( ).
y If Q If dx C
= +
∫
⇒ y.e
Pdx
=
∫ . .
Pdx
Q e dx C
∫ +
∫
Ex. Solve | 1
2
1
y y
x
e
+ =
+
Solution : The given D.E. is linear because it is of the form

7. 7
dy
dx
py Q
+ = Where 2
1
1,
1 x
P Q
e
= =
+
Now integrating factor =
Pdx
e∫
Pdx x
e
e∫ =
=
∴ Required solution is
2
1
. .
1
. x x
x
e dx c
e
y e = +
+
∫
⇒ 1
tan
. x
t c
y e −
= + 1
tan
. x x
e c
y e −
= +
⇒
2
1
x
x
e
dx
e
+
∫
Put dn dt
x x
e t e =
= ⇒
1
2
1
tan
1
dt
t
−
∴ =
+
∫
Method Forth
Bernoullis equation
A. D.E. of the form
1 n
y Py y Q
+ = Where P and Q are the function of x.
Working Method
Given D.E. is 1 n
y Py y Q
+ = divided by yn
⇒ 1 1
.
n n
y y py Q
− −
+ =
Put 1 n
y v
−
=
Then given D. E. turns to linear D.E. equation which can be earily solved.
Ex. Solve
2
3.
dy x
xy y e
dx
−
− =
Solution : Given D. E.
2
3.
dy x
xy y e
dx
−
− =
Dividing the D.E. by 3
y

8. 8
⇒ 2 3 2
.
dy x
xy y e
dx
− − −
− = ____________________(1)
Taking 2
y V
− =
⇒
3
2 .
dv
dy
y dx dx
−
− =
⇒
1
3.
2
dy dv
y dx dx
−
− =
∴ Equation (1) ⇒
2
1
2
x
dv
xv e
dx
−
+ =
⇒ 2
2 2 x
dv xv e
dx
−
+ = ____________________(2)
This is linear (1) equation in v
∴ Solution of equation (2) is
⇒
2 2 2
. 2. .
x x x
v e e e dx c
−
= +
∫
Put the value of v
⇒ 2 2(2 )
ex y x c
= +
Method fifth
Clairaut’s Equation
The form of the equation is ( )
y px f p
= +
It’s solution is obtained just replacing P by C.
Ex. Solve log( )
P Px y
= −
Solution : Given D. Equation is
p
Px y e
− =
Or p
y Px e
= +
This is of clairants form
∴ It’s solution is y = cx+eP
2
2
.
xdx x
I f e e
∫
= =

9. 9
Objective Question
1) The D.E. // /
4 4 0
Y y y
− + = has roots ______________.
2) The D.E. Which involes only one independent variable is called an
____________.
3) The D.E. which involes more than one independent variable is called
_________.
4) The order of D.E. is the order of ________ derivative in the D.E.
5) A necessary condition for 0
Mdx Ndy
+ = to be exact is __________.
6) The form of the equation ( )
y px f p
= + is called _________.
7) The order and degree of D. E. is
2
2
2
sin 0
d y dy
x y x
dx dx
+ − − = _________.
8) A linear equation of the first order is of the form ________.
9) The D.E. of the form | n
y py y Q
+ = is called as ________.
10) Integrating factor is ___________.
11) Ordinary D.E. involves only ________ independent variable.
12) PDE which involves ________ independent variables.
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