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Mathematics
Session
Differential Equations - 1
Session Objectives

Differential Equation

Order and Degree

Solution of a Differential Equation, General
and Particular Solution

Initial Value Problems

Formation of Differential Equations

Class Exercise
Differential Equation
An equation containing an independent
variable x,dependent variable y and the
differential coefficients of the dependent
variable y with respect to independent
variable x, i.e.
dy d2 y
,
,…
dx dx2
Examples
1
2 
dy
= 3xy
dx
d2 y
2
dx
3
+ 4y = 0
2
3
3
d y
dy
+
+ 4y = sinx
 +
3
2

dx
dx
 dx 
 4
x2 dx + y2 dy = 0
d y
Order of the Differential Equation
The order of a differential equation is the order
of the highest order derivative occurring in the
differential equation.
2
3
 d2 y 
d
y
dy
dy


Example : 3
=

 =

 dx2 
dx
dx


dx2


2
The order of the highest order derivative
d2 y
2
dx
is 2.
Therefore, order is 2
Degree of the Differential Equation
The degree of a differential equation is the degree of the highest
order derivative, when differential coefficients are made free from
fractions and radicals.
3
2
 dy   2
3
2
2

 d2 y 
d y 
 dy  
 =0  

Example :
+ 1+ 
 = 1+ 


2
2




 dx  
 dx  
dx
 dx 
2
The degree of the highest order derivative
Therefore, degree is 2.
d2 y
2
dx
is 2.
Example - 1
Determine the order and degree of the differential
2
dy
 dy 
equation: y = x
+ a 1+ 
.

dx
 dx 
2
dy
 dy 
Solution : We have y = x
+ a 1+ 

dx
 dx 
2
dy
 dy 
 y-x
= a 1+ 

dx
 dx 
2
dy 

2
 y - x
=
a
dx 

2

 dy  
1+ 
 
dx

 



Solution Cont.
2
2
dy
 dy 
2
2  dy 
 y2 - 2xy
+ x2 
=
a
+
a

 dx 
dx
 dx 


dy
The order of the highest derivative
is 1
dx
and its degree is 2.
Example - 2
Determine the order and degree of the differential
3
2
 dy   2

equation:
 c    
4

 dx  
dx
d4 y
3
2
 dy   2

d y
= c +   
Solution: We have
4

 dx  
dx
4
2
2

d y
 dy  

 = c +   
 dx4 

 dx  


4
d4 y
Here, the order of the highest order
is 4
4
dx
d4 y
and, the degree of the highest order
is 2
4
dx
3
Linear and Non-Linear
Differential Equation
A differential equation in which the dependent variable y and
dy d2 y
,
, … occur only in the
its differential coefficients i.e.
2
dx dx
first degree and are not multiplied together is called a linear
differential equation. Otherwise, it is a non-linear differential
equation.
Example - 3
i 
d2 y
dx2
-3
dy
+ 7y = 4x
dx
is a linear differential equation of order 2 and degree 1.
ii
 dy 
y×
-4=x

 dx 
is a non-linear differential equation because the dependent
variable y and its derivative
dy
are multiplied together.
dx
Solution of a Differential Equation
The solution of a differential equation is the relation
between the variables, not taking the differential
coefficients, satisfying the given differential equation and
containing as many arbitrary constants as its order is.
For example: y = Acosx - Bsinx
is a solution of the differential equation
d2 y
2
dx
+ 4y = 0
General Solution
If the solution of a differential equation of nth order contains n
arbitrary constants, the solution is called the general solution.
y = Acosx - Bsinx
is the general solution of the differential equation
d2 y
2
dx
+y =0
y  B sin x
is not the general solution as it contains one arbitrary constant.
Particular Solution
A solution obtained by giving particular values to the arbitrary
constants in general solution is called particular solution.
y  3 cos x  2 sin x
is a particular solution of the differential equation
d2 y
2
dx
+ y = 0.
Example - 4
Verify that y = x3 + ax2 +bx + c is a solution of the
differential equation
d3y
3
dx
= 6.
Solution: We have y = x3 + ax2 +bx + c
dy
= 3x2 + 2ax +b
dx
d2 y
2
dx
= 6x + 2a
…(ii)
…(iii)
…(i)
Differentiating i  w.r.t. x 
Differentiation ii w.r.t. x 
Solution Cont.
d3 y
3
dx

d3 y
3
dx
=6
Differentiating iii w.r.t x 
= 6 is a differential equation of i .
Initial Value Problems
The problem in which we find the solution of the
differential equation that satisfies some prescribed
initial conditions, is called initial value problem.
Example - 5
x
Show that y = e + 1 is the solution of the initial value
problem
d2 y
dx2
-
dy
= 0, y 0  = 2, y' 0  = 1
dx
Solution : We have y = ex + 1
dy
d2 y
x
x

=e ,
=
e
dx
dx2
x
y = e + 1 satisfies the differential equation
d2 y
dx2
-
dy
=0
dx
Solution Cont.
y = ex + 1 and
dy
 ex
dx
 dy 
 y  0  = e0 + 1 and 
= e0

 dx x=0
 y 0 = 2 and y ' 0  1
y = ex + 1 is the solution of the initial value problem.
Formation of Differential Equations
Assume the family of straight
lines represented by
y = mx
Y
dy
y
dy

=m 

dx
dx
x
dy
x
y
dx
is a differential equation of the first order.
y  mx

O
m = tan
X
Formation of Differential Equations
Assume the family of curves represented by
y = Acos  x +B
… (i)
where A and B are arbitrary constants.

dy
  A sin  x  B 
dx
and
d2 y
dx
2
...  ii 
 A cos  x  B 
[Differentiating (i) w.r.t. x]
[Differentiating (ii) w.r.t. x]
Formation of Differential Equations

d2 y
dx

2
d2 y
2
dx
 y
[Using (i)]
+y =0
is a differential equation of second order
Similarly, by eliminating three arbitrary constants, a differential
equation of third order is obtained.
Hence, by eliminating n arbitrary constants, a differential
equation of nth order is obtained.
Example - 6
Form the differential equation of the family of curves
y = a sin bx + c  , a and c being parameters.
Solution: We have
y = a sin bx + c 
dy
= ab cos bx + c 
dx


d2 y
2
dx
d2 y
dx2
= -ab 2 sin bx + c 
2
= -b y 
d2 y
dx2
[Differentiating w.r.t. x]
[Differentiating w.r.t. x]
+ b2 y = 0
is the required differential equation.
Example - 7
Find the differential equation of the family of all
the circles, which passes through the origin and
whose centre lies on the y-axis.
Solution: The general equation of a circle is
x2 + y2 + 2gx + 2ƒy + c = 0.
If it passes through (0, 0), we get c = 0
 x2 + y2 +2gx +2ƒy = 0
This is an equation of a circle with centre (- g, - f)
and passing through (0, 0).
Solution Cont.
Now if centre lies on y-axis, then g = 0.
 x2 + y2 +2ƒy = 0
…(i)
This represents the required family of circles.
2x + 2y
dy
dy
+ 2ƒ
=0
dx
dx
dy 

x

y


dx
 ƒ  

dy




 dx 
Differentiating i w.r.t. x 
Solution Cont.
dy 

x
+
y


dx
2
2
 x + y - 2y 
=0
dy




dx


Substituting the value of f 

2
2

dy
2 dy
- 2xy - 2y
=0
dx
dx

2
2

dy
- 2xy = 0
dx
 x +y
 x -y
Thank you