Class 12 Maths Differential Equations Formation of differential equation whose general solution is

Formation of differential equation whose general solution is given

Now let’s learn how to form the differential equation of a given function. Please note that we need to eliminated the parameter like a,b  etc. to form a differential equation

Procedure to find a differential equation

1. If the given family F of curves depends on only one parameter then we can represent the equation of the form

F (x, y, a) = 0 ..(1)

Example: Parabola Equation y2 = 4ax (as shown in fig.) this can be represented in an equation of the form  F(x, y, a) : y2 =4ax…..(1)

Now to find out differential equation we differentiate equation (1)
with respect to x, we get an equation involving  y, y, x, and a,

i.e.,   y  dy/dx = 2a ……..(2)

We get a differential equation of a family of a parabola by putting the value of parameter “a” from (2) in equation (1)

We get         y=2x dy/dx   ………..this a differential equation.

i.e  F(x,y,y’)=0

1. 2. If two parameters are given say “a and b” in the function, then we represent the equation of the form

F (x, y, a ,b) = 0

Example:  y=asin(x+b) this can be represented by an equation of the  form

F (x, y, a, b) : y =asin(x+b)……….(1)

Now since it involve two parameter (a, b) we need to differentiate twice since it is not possible to eliminate two parameters a and b from the two equations only and so, we required a third equation.

dy/dx = a cos(x+b) --(2)

d2y/dx2 = -a sin(x+b) --(3)

and since  we put the value in 3rd in the equation, we get d2y/dx2 = -y
or  d2y/dx2 +y = 0

i.e F(x,y,y’,y”)=0

Note:- Similarly we follow the same steps for 3 or more parameters if involved.

Problem:  Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Sol. The equation of the family of hyperbolas with the centre at origin and foci along the xaxis is:

..(1)

Differentiating both sides of equation (1) with respect to x, we get:

2x/a2 -2yy'/b2  =0

x/a2 – yy’/b2 = 0   ----(2)

Again Differentiating equation (2) with respect to x, we get:

Now substituting value of 1/a2 from equation in (3) in the equation (2) we get,

Solving the equation we get,

x(y’)2+xyy’’-yy’=0

xyy’’+x(y’)2-yy’=0

This is a required differential equation of Hyperbola.

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