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__

- 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

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*

**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)

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)

d^{2}y/dx^{2} = -a sin(x+b) --(3)

and since we put the value in 3rd in the equation, we get d^{2}y/dx^{2} = -y

or d^{2}y/dx^{2} +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/a^{2} -2yy'/b^{2} =0

x/a^{2} – yy’/b^{2} = 0 ----(2)

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

Now substituting value of 1/a^{2} 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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