Chapter-01
Md. Mohiuddin
Formation of Differential Equation
Differential Equation: An equation involving derivatives of one or more dependent variables with respect
to one or more independent variables is called a differential equation.
Example: For examples of differential equation we list the following:
2
1. d y2 + dy + y =
0
dx
dx
4
2
2. d u4 + 5 d u2 + 3u =
sin t
dt
dt
3. ∂ u2 + ∂ u2 + ∂ u2 =
0
2
2
2
∂x
∂y
∂z
4. ∂u + ∂v =
v
∂s ∂t
Classification: Differential equations are classified on the basis of type as follows:
1. Ordinary Differential Equation (ODE),
2. Partial Differential Equation (PDE).
Ordinary Differential Equation (ODE): A differential equation involving derivatives of one or more
dependent variables with respect to only one independent variable is called an ordinary differential equation.
d2 y
dy
Example: 1.
+2 + y =
0
2
dx
dx
d3y
d2y
2. 3 + 5 2 + 2 y =
sin x
dx
dx
Partial Differential Equation (PDE): A differential equation involving derivatives of one or more
dependent variables with respect to more than one independent variable is called a partial differential
equation.
Example: 1. ∂u + ∂v =
v
∂s ∂t
2
2
2. ∂ u2 + ∂ u2 =
0
∂x
∂y
Order of a differential equation: The order of the highest ordered derivative involved in a differential
equation is called the order of the differential equation.
dy
Example: 1.
0 is a first order differential equation.
+y=
dx
d 2 y dy
2.
+ +y=
0 is a second order differential equation.
dx 2 dx
Degree of a differential equation: The power of the highest ordered derivative involved in a differential
equation is called the degree of the differential equation, after the equation is freed from radicals and
fractions in its derivatives.
2
Example: 1. d y2 + dy + y =
0 is a differential equation of first degree.
dx
dx
2
2.  d y2  +  dy  + y =
0 is a differential equation of second degree.
2
 dx 
4
 dx 
2
3.
d3y
d2y
 dy  is a differential equation of first degree.
+4 2 + y =


3
dx
dx
 dx 
Linear ordinary differential equation: An ordinary differential equation of order n is called a linear
ordinary differential equation of order n if it does not contain,
1. the transcendental functions of dependent variable,
2. the product of dependent variable and
3. the product of the derivatives of dependent variable.
1
Chapter-01
It can be expressed as
a0 ( x )
Md. Mohiuddin
dny
d n −1 y
dy
a
x
b ( x)
+
+ ... ... ... + an −1 ( x )
+ an ( x ) y =
(
)
1
n
n −1
dx
dx
dx
where, a0 is not identically zero.
d2y
dy
+ 5 + 6y =
0
2
dx
dx
2
d3y
2 d y
2.
x
xe x
+
+ xy =
3
2
dx
dx
Nonlinear ordinary differential equation: A nonlinear ordinary differential equation is an ordinary
differential equation that is not linear.
d2 y
dy
Example: 1.
0
+ 5 + 6 y2 =
2
dx
dx
2
d3 y
y d y
2.
e
xe x
+
+ xy =
3
2
dx
dx
2
2
d y  dy 
3.
+   + 6y =
0
dx 2  dx 
Example: 1.
Application of differential equation: Some applications of differential equation are given below:
1. The problem of determining the motion of a projectile, rocket, satellite or planet.
2. The problem of the conduction of heat in a rod or wire in a slab.
3. The problem of determining the vibrations of a wire or a membrane.
4. The study of the rate of decomposition of a radioactive substance or the rate of growth of a
population.
Problem-01: Form the differential equation of the complete integral =
y ax + bx 2 .
Solution: Given that,
=
y ax + bx 2 … … ... (1)
Differentiating (1) with respect to x we get,
dy
= a + 2bx … … … (2)
dx
dy
⇒a=
− 2bx
dx
Again differentiating (2) with respect to x we get,
d2y
= 2b
dx 2
1 d2y
⇒b= 2
2 dx
Putting these values of ‘ a ’ and ‘ b ’ in the equation (1) we have,
d y
1 d 2 y  x2 d 2 y
y=
x  − 2 x.
+
2 dx 2  2 dx 2
 dx
d y 2 d 2 y x2 d 2 y
−x
+
dx
2 dx 2
dx 2
dy x 2 d 2 y
⇒ y= x −
dx 2 dx 2
2
dy
2 d y
⇒x
− 2x + 2 y =
0
2
dx
dx
which is a differential equation of second order and first degree.
⇒ y= x
2
Chapter-01
Md. Mohiuddin
2
Problem-02: Form the differential equation of the complete integral c ( y + c ) =
x3 .
Solution: Given that,
2
c ( y + c) =
x3 … … … (1)
Differentiating (1) with respect to x we get,
dy
2c ( y + c ) =
3 x 2 … … … (2)
dx
Dividing (1) by (2) we get,
y+c x
=
dy 3
2
dx
dy
⇒ 3( y + c) =
2x
dx
1  dy

=
⇒c
 2x − 3y 
3  dx

Putting this value of ‘ c ’ in the equation (2) we get,
2  dy
1  dy

  dy
3x 2
=
 2 x − 3 y   y +  2 x − 3 y 
3  dx
3  dx

  dx
2 x dy
 4 x dy

 dy
⇒
− 2yy +
− y
=
3x 2
3 dx
 3 dx

 dx
2
 4 x dy
 2 x  dy 
⇒
− 2y   =
3x 2
 3 dx
 3  dx 
3
2
8 x 2  dy  4 xy  dy 
⇒
3x 2
  −
  =
9  dx 
3  dx 
3
2
 dy 
 dy 
⇒ 8 x   − 12 y   =
27 x
 dx 
 dx 
which is a differential equation of first order and third degree.
Problem-03: Find the differential equation from the relation
=
y A cos x + B sin x
where A and B are arbitrary constants.
Solution: Given that,
=
y A cos x + B sin x … … … (1)
Differentiating (1) with respect to x we get,
dy
=
− A sin x + B cos x … … … (2)
dx
Again, differentiating (2) with respect to x we get,
d2y
=
− A cos x − B sin x
dx 2
d2 y
⇒ 2 =
− ( A cos x + B sin x )
dx
d2 y
⇒ 2 =
−y
dx
which is a differential equation of second order and first degree.
3
Chapter-01
Md. Mohiuddin
2x
, y Ae + Be−2 x for different values of
Problem-04: Find the differential equation of the family of curves =
A and B .
Solution: Given that,
=
y Ae 2 x + Be −2 x … … … (1)
Differentiating (1) with respect to x we get,
dy
= 2 Ae 2 x − 2 Be −2 x … … … (2)
dx
Again, differentiating (2) with respect to x we get,
d2y
=
4 Ae 2 x + 4 Be −2 x
2
dx
d2 y
⇒
= 4 ( Ae 2 x + Be −2 x )
2
dx
d2y
4y
⇒ 2 =
dx
which is a differential equation of second order and first degree.
Problem-05: Find the differential equation from the relation
=
y e x ( A cos x + B sin x )
where A and B are arbitrary constants.
Solution: Given that,
=
y e x ( A cos x + B sin x ) … … … (1)
Differentiating (1) with respect to x we get,
dy
=
e x ( − A sin x + B cos x ) + e x ( A cos x + B sin x )
dx
dy
⇒
= e x ( − A sin x + B cos x ) + y
dx
dy
⇒
− y = e x ( − A sin x + B cos x ) … … … (2)
dx
Again, differentiating (2) with respect to x we get,
d 2 y dy
−
= e x ( − A cos x − B sin x ) + e x ( − A sin x + B cos x )
dx 2 dx
d 2 y dy
⇒ 2 −
=
−e x ( A cos x + B sin x ) + e x ( − A sin x + B cos x )
dx
dx
2
d y dy
 dy

⇒ 2 −
=− y +  − y 
dx
dx
 dx

2
d y
dy
⇒ 2 − 2 + 2y =
0
dx
dx
which is a differential equation of second order and first degree.
Problem-06: Find the differential equation of the complete integral Ax 2 + By 2 =
1 for different values of A and
B.
Solution: Given that,
Ax 2 + By 2 =
1 … … … (1)
Differentiating (1) with respect to x we get,
dy
2 Ax + 2 By
=
0
dx
dy
⇒ Ax + By
=
0 … … … (2)
dx
4
Chapter-01
Again, differentiating (2) with respect to x we get,
 dy  2
d2y
A + B   + y 2  =
0
dx 
 dx 
Multiply above equation by x we get,
 dy  2
d2y
Ax + Bx   + y 2  =
0 … … … (3)
dx 
 dx 
Subtracting equation (2) from equation (3) we get,
 dy 2
d2y 
dy 
Ax + Bx   + y 2  −  Ax + By  =
0
dx  
dx 
 dx 
 dy  2
d2y
dy
⇒ Bx   + y 2  − By
=
0
dx 
dx
 dx 
 dy  2
d2y
dy
⇒ x   + y 2  − y
=
0
dx 
dx
 dx 
which is a differential equation of second order and first degree.
Exercise: Form the differential equations for the following equations:
1. =
xy ae x + be − x
2. y =ae3 x + be −2 x + sin 5 x
3. y a cos ( mx + b )
=
4. ( x − h ) = 4a ( y − k )
2
5. y = axe x
6.
=
y ae −4 x + be −6 x
5
Md. Mohiuddin