Given,

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

To find the value of $\left(1-x^2\right)y_2-x{y}_1$

As, $y=\frac{\left({\sin }^{-1}x\right)}{2}$

Differentiating it with respect to $x$, we get

$\therefore \frac{dy}{dx}=\frac{1}{2}\times 2{\sin }^{-1}x·\frac{d}{dx}\left({\sin }^{-1}x\right)$

$\therefore \frac{dy}{dx}=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}} ...\left(i\right)$

Differentiating again with respect to $x$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{\left(\sqrt{1-x^2}\right)\times \frac{1}{\sqrt{1-x^2}}-\left({\sin }^{-1}x\right)\times {\displaystyle \frac{1\left(-2x\right)}{2\sqrt{1-x^2}}}}{(1-x^2)} \left[\text{As}, \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]$

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

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

$\therefore \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x \frac{dy}{dx}=1$.