Chapter: 2
Formation of differential equation:
If an equation having n numbers of independent variables then the equation
need to differentiate n times. Then the independent variables required
eliminating. This is way of formation of differential equation.
Example: Formation of differential equation of Φ (x, y, a) =0
Solution: Given equation is Φ (x, y, a) =0…………………….(1)
Differentiating (1) with respect to x, we get
Φ(x, y, a,
𝑑𝑦
𝑑𝑥
) =0 …………………………….(2)
Eliminating “a” from equation (1) and (2) we get,
Φ(x, y,
𝑑𝑦
𝑑𝑥
) =0; which is the required differential equation.
Example: Find the differential equation from a straight line y=mx
Solution: Given equation
y=mx………………..(1)
By differentiating both sides of (1) with respect to x;
𝑑𝑦
𝑑𝑥
=m…………………………….(2)
From (2), putting the value of m in equation (1)
y=𝑥
𝑑𝑦
𝑑𝑥
; this is the required differential equation.
Example: Form the differential equation of all circles passing through
origin and having their Centre’s on the x-axis.
Solution: Let the equation of a circle whose Centre is (a, 0) and radius =a
where a is an arbitrary constant is
(x − a)2 + (y − 0)2 = 𝑎2
Or, 𝑥 2 -2ax+𝑎2 +𝑦 2 =𝑎2
Or, 𝑥 2 + 𝑦 2 = 2ax ………………………..(1)
By differentiating (1) with respect to x,
𝑑𝑦
2x+2y =2a………………….(2)
𝑑𝑥
Multiplying both sides by x of equation (2),
𝑑𝑦
2𝑥 2 + 2xy
=2ax
𝑑𝑥
𝑑𝑦
Or, 2𝑥 2 + 2xy
𝑑𝑥
𝑑𝑦
𝑥 2 - 𝑦 2 + 2xy
𝑑𝑥
=𝑥 2 + 𝑦 2 ……………………………. [from equation (1)]
=0;
This is the required differential equation.
Example: Form a differential equation from relation y= Acos 𝑥+Bsin 𝑥
Solution: Given Equation,
y= Acos 𝑥+Bsin 𝑥…………………..(1)
By differentiating equation (1) with respect to x we get
𝑑𝑦
𝑑𝑥
= -Asin 𝑥+ Bcos 𝑥………………….(2)
Again differentiating (2) with respect to x we get,
𝑑2𝑦
d𝑥 2
= - [Acos 𝑥+ Bsin 𝑥]
Or,
𝑑2𝑦
Or,
𝑑2𝑦
d𝑥 2
d𝑥 2
= -y……….[ From equation (1)]
+y=0;
This is the required differential equation.
Example: Show that the differential equation of A𝒙𝟐 +B𝒙𝟐 =1 is
𝑑2𝑦
𝑑𝑦
𝑑𝑦
d𝑥
𝑑𝑥
𝑑𝑥
x[y
+ ( )2 ] = y
2
Solution: Given equation,
A𝑥 2 +B𝑥 2 =1………………………(1)
By differentiating equation (1) with respect to x we get,
𝑑𝑦
2Ax+2By
Or, By
Or,
𝑑𝑦
𝑑𝑥
𝑦 𝑑𝑦
𝑥 𝑑𝑥
𝑑𝑥
=0
=-Ax
𝐴
= - …………………………….(2)
𝐵
Again differentiating equation (2) with respect to x we get,
𝑦 𝑑 𝑑𝑦
𝑑
𝑦
𝑑𝑦
𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑥 𝑥
𝑑𝑥
( )+
Or,
𝑦 𝑑2𝑦
Or,
𝑦 𝑑2𝑦
Or,
𝑦 𝑑2𝑦
𝑥
𝑥
+ [
d𝑥 2
𝑥 d𝑥 2
+
d𝑥 2
Or, 𝑥𝑦
𝑑2𝑦
d𝑥 2
𝑑2𝑦
Or, x[y
d𝑥 2
( )
=0
𝑑𝑦
)−y.1
𝑑𝑥
𝑥2
x(
1
𝑥2
𝑑𝑦
[x
𝑑𝑥
].
𝑑𝑦
𝑑𝑥
=0
− 𝑦] =0
1 𝑑𝑦
𝑦 𝑑𝑦
𝑥 𝑑𝑥
𝑥 2 𝑑𝑥
+ ( )2 =
𝑑𝑦
𝑑𝑦
𝑑𝑥
𝑑𝑥
+x( )2 =y
𝑑𝑦
𝑑𝑦
𝑑𝑥
𝑑𝑥
+ ( )2 ] = y
(showed)
Example: Prove the differential equations of all parabolas whose axes
are parallel to y-axis is
𝒅𝟑 𝐲
𝐝𝒙𝟑
=0.
Solution: The equation of all parabolas, whose axes are parallel to y-axis is
(x − h)2 = 4a(y-k)……………………….(1)
When a, h, k are arbitrary constants.
Differentiating both sides of (1) with respect to x (1) we get,
𝒅𝒚
2(x-h) = 4a
𝒅𝒙
𝒅𝒚
Or, (x-h) = 2a
𝒅𝒙
Again differentiating both sides with respect to x we get,
𝑑2𝑦
1=2a
Or,
d𝑥 2
𝑑2𝑦
d𝑥 2
=
1
2𝑎
Again differentiating with respect to x we get,
𝑑3y
d𝑥 3
=0
(proved)
Example: Form a differential equation for a cardio ide r=a (1+𝐜𝐨𝐬 𝜽)
Solution: Given equation,
r=a (1+cos 𝜃)……………….(1)
Differentiating both sides with respect to 𝜃 we get,
𝑑𝑟
𝑑𝜃
= -asin 𝜃
Or, a= -
1
𝑑𝑟
sin 𝜃 𝑑𝜃
Now putting the value of “a” in equation (1) we get,
r=-
1
sin 𝜃
(1 + cos 𝜃)
rsin 𝜃= -(1 + cos 𝜃)
𝑑𝑟
𝑑𝜃
𝑑𝑟
𝑑𝜃
rsin 𝜃+(1 + cos 𝜃)
𝑑𝑟
𝑑𝜃
=0
This is the required differential equation.