#### Provide solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.3 question 18

$y=\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{2}$  is a solution of differential equation

Hint:

Differentiate and substitute the values

Given:

$y=\log \left(x+\sqrt{x^{2}+a^{2}}\right)^{2}$

Solution:

Differentiating on both sides with respect to $x$

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

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

\begin{aligned} &\frac{d y}{d x}=\frac{2\left(x+\sqrt{x^{2}+a^{2}}\right)}{\sqrt{x^{2}+a^{2}}\left(x+\sqrt{x^{2}+a^{2}}\right)} \\\\ &\frac{d y}{d x}=\frac{2}{\sqrt{x^{2}+a^{2}}} \end{aligned}\begin{aligned} &\frac{d y}{d x}=\frac{2\left(x+\sqrt{x^{2}+a^{2}}\right)}{\sqrt{x^{2}+a^{2}}\left(x+\sqrt{x^{2}+a^{2}}\right)} \\ &\frac{d y}{d x}=\frac{2}{\sqrt{x^{2}+a^{2}}} \end{aligned}                            ...............(i)

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

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

Hence the given function is the solution to given differential equation.