|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
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.
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
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:
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,
This is a required differential equation of Hyperbola.