#### Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.3 question 8 maths

$y^{2}=4 a x$  is a solution of differential equation

Hint:

Differentiate the given solution of differential equation on both sides with respect to $x$ and once with respect to $y$

Given:

$y^{2}=4 a x$  is a solution of the equation

Solution:

Differentiating on both sides with respect to $x$

\begin{aligned} &2 y \frac{d y}{d x}=4 a \\\\ &\frac{d y}{d x}=\frac{4 a}{2 y} \\\\ &\frac{d y}{d x}=\frac{2 a}{y} \end{aligned}                            .........(i)

Now differentiating $y^{2}=4 a x$ with respect to $y$

\begin{aligned} &2 y=4 a \frac{d x}{d y} \\\\ &\frac{d x}{d y}=\frac{2 y}{4 a} \\\\ &\frac{d x}{d y}=\frac{y}{2 a} \end{aligned}                                .........(ii)

Now put both equation (i) and (ii) in given differential equation as follows

\begin{aligned} &y=x \frac{d y}{d x}+a \frac{d x}{d y} \\\\ &R H S=x \frac{d y}{d x}+a \frac{d x}{d y} \end{aligned}

\begin{aligned} &=x\left(\frac{2 a}{y}\right)+a\left(\frac{y}{2 a}\right) \\\\ &=\frac{y^{2}}{4 a}\left(\frac{2 a}{y}\right)+a\left(\frac{y}{2 a}\right) \quad\left[y^{2}=4 a x \Rightarrow \frac{y^{2}}{4 a}=x\right] \end{aligned}

\begin{aligned} &=\frac{y}{2}+\frac{y}{2} \\\\ &=\frac{2 y}{2} \\\\ &=y \\\\ &=L H S \end{aligned}

Thus, $y^{2}=4 a x$ is a solution of differential equation.

\begin{aligned} &y=x \frac{d y}{d x}+a \frac{d x}{d y} \end{aligned}