If the solve the problem

Solution:

Note: $y_{2}$ represents second order derivative i.e. $\frac{d^{2} y}{d x^{2}}$ and $y_{1}=d y / d x$

Given,

$x=\sin \left(\frac{1}{a} \log y\right)$

$(\log y)=a \sin ^{-1} x$

$y=e^{\operatorname{asin}^{-1} x} \ldots \ldots$ equation 1

to prove: $\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0$

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$

So, lets first find $d y / d x$

$\because y=e^{\operatorname{asin}^{-1} x}$

Let $\mathrm{t}=\mathrm{a} \sin ^{-1} \mathrm{x}=>\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{a}}{\sqrt{\left(1-\mathrm{x}^{2}\right)}}\left[\frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{X}=\frac{1}{\sqrt{\left(1-\mathrm{x}^{2}\right)}}\right]$

And $y=e^{t}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{d y}{d x}=e^{t} \frac{a}{\sqrt{\left(1-x^{2}\right)}}=\frac{a e^{a \sin ^{-1} x}}{\sqrt{\left(1-x^{2}\right)}} \ldots \ldots . .$ equation 2

Again differentiating with respect to $x$ applying product rule:

$\frac{d^{2} y}{d x^{2}}=a e^{a \sin ^{-1} x} \frac{d}{d x}\left(\frac{1}{\sqrt{1-x^{2}}}\right)+\frac{a}{\sqrt{\left(1-x^{2}\right)}} \frac{d}{d x} e^{a \sin ^{-1} x}$

Using chain rule and equation 2 :

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{ae}^{2 \sin ^{-1} \mathrm{x}}}{2\left(1-\mathrm{x}^{2}\right) \sqrt{1-\mathrm{x}^{2}}}(-2 \mathrm{x})+\frac{\mathrm{a}^{2} \mathrm{e}^{2 \sin ^{-1}} \mathrm{x}}{\left(1-\mathrm{x}^{2}\right)}\left[\right.$ using $\left.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{x}=\frac{1}{\sqrt{\left(1-\mathrm{x}^{2}\right)}}\right]$

$\frac{d^{2} y}{d x^{2}}=\frac{x a e^{2 \sin ^{-1} x}}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}+\frac{a^{2} e^{2 \sin ^{-1} x}}{\left(1-x^{2}\right)}$

$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} e^{a \sin ^{-1} x}+\frac{x a e^{a \sin ^{-1} x}}{\sqrt{1-x^{2}}}$

Using equation 1 and equation 2 :

$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} y+x \frac{d y}{d x}$

$\therefore\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0 \ldots \ldots$ proved