#### Please solve RD Sharma class 12 chapter Differential Equation exercise 21.3 question 9 maths textbook solution

$A x^{2}+B y^{2}=1$  is a solution of differential equation

Hint:

Differentiate and substitute values of equation to obtain differential equation.

Given:

$A x^{2}+B y^{2}=1$

Solution:

Differentiating on both sides with respect to $x$

\begin{aligned} &\frac{d\left(A x^{2}+B y^{2}\right)}{d x}=\frac{d}{d x}(1) \\\\ &\frac{d\left(A x^{2}\right)}{d x}+\frac{d\left(B y^{2}\right)}{d x}=\frac{d}{d x}(1) \end{aligned}

$2 A x+2 B y \frac{d y}{d x}=0$                                ..........(i)

Differentiate equation (i) with respect to $x$

\begin{aligned} &\frac{d}{d x}\left(2 A x+2 B y \frac{d y}{d x}\right)=\frac{d}{d x}(0) \\\\ &2 A \frac{d}{d x}(x)+2 B \frac{d}{d x}\left(y \frac{d y}{d x}\right)=0 \end{aligned}

\begin{aligned} &2 A+2 B\left[\frac{d y}{d x} \frac{d y}{d x}+y \frac{d}{d x}\left(\frac{d y}{d x}\right)\right]=0 \\\\ &2 B\left[\left(\frac{d y}{d x}\right)^{2}+y \frac{d^{2} y}{d x^{2}}\right]=-2 A \end{aligned}

$\left[\left(\frac{d y}{d x}\right)^{2}+y \frac{d^{2} y}{d x^{2}}\right]=\frac{-2 A}{2 B}$                        .........(ii)

Using equation (i), we can find the values of$\frac{-2 A}{2 B}$

\begin{aligned} &2 A x+2 B y \frac{d y}{d x}=0 \\\\ &2 B y \frac{d y}{d x}=-2 A x \\\\ &\frac{y}{x} \frac{d y}{d x}=\frac{-2 A}{2 B} \end{aligned}                                    ..........(iii)

Now put equation (ii) in (iii) ,

\begin{aligned} &{\left[\left(\frac{d y}{d x}\right)^{2}+y \frac{d^{2} y}{d x^{2}}\right]=\frac{y d y}{x d x}} \end{aligned}

$x\left[\left(\frac{d y}{d x}\right)^{2}+y \frac{d^{2} y}{d x^{2}}\right]=y \frac{d y}{d x}$

Hence proved.

Thus, $A x^{2}+B y^{2}=1$ is a solution of differential equation.