Question

The differential equation whose solution is $A{x}^2+B{y}^2=1$, where $A$ and $B$ are arbitrary constants is of order and degree respectively:

• A. $2,2$
• B. $1,2$
• C. $1,1$
• D. $2,1$

Answer 1

The correct option is D $2,1$

Given the solution $A{x}^2+B{y}^2=1$......(1).

Now differentiating (1) with respect to $x$ we get,

$2Ax+2By\frac{dy}{dx}=0$

or, $\frac{y{y}^{\prime }}{x}=-\frac{A}{B}$

Again differentiating with respect to $x$ we get,

$\frac{x(y^{\prime 2}+y{y}^{\prime \prime })-y{y}^{\prime }}{x^2}=0$

or, $xy{y}^{\prime \prime }+x{y}^{\prime 2}-y{y}^{\prime }=0$

This is the differential equation of order is $2$ and degree $1$ obtained from the solution (1).