Write the correct alternative in the following:
If $y=\sin \left(m \sin ^{-1} x\right)$, then $\left(1-x^{2}\right) y_{2}-x y_{1}$ is equal to
A. $m^{2} y$
B. my
C. $-m^{2} y$
D. none of these
Given:
$y=\sin \left(m \sin ^{-1} x\right)$
$\frac{d y}{d x}=m \cos \left(m \sin ^{-1} x\right) \frac{1}{\sqrt{\left(1-x^{2}\right)}}$
$x \frac{d y}{d x}=\cos \left(m \sin ^{-1} x\right) \frac{m x}{\sqrt{\left(1-x^{2}\right)}}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$
$=m\left\{\frac{-m \sin \left(m \sin ^{-1} x\right) \sqrt{1-x^{2}} \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{2 \sqrt{\left(1-x^{2}\right)}}(-2 x) \cos \left(m \sin ^{-1} x\right)}{\left(\sqrt{\left.\left(1-x^{2}\right)\right)^{2}}\right.}\right\}$
$=\frac{m}{\left(1-x^{2}\right)}\left\{-m \sin \left(m \sin ^{-1} x\right)+\frac{x}{\sqrt{\left(1-x^{2}\right.}} \cos \left(m \sin ^{-1} x\right)\right\}$
$\left(1-x^{2}\right) y_{2}=m\left\{-m \sin \left(m \sin ^{-1} x\right)+\frac{x}{\sqrt{\left(1-x^{2}\right.}} \cos \left(m \sin ^{-1} x\right)\right\}$
$=-m^{2} \sin \left(m \sin ^{-1} x\right)+\frac{m x}{\sqrt{\left(1-x^{2}\right.}} \cos \left(m \sin ^{-1} x\right)$
$\left(1-x^{2}\right) y_{2}-x y_{1}$
$=-m^{2} \sin \left(m \sin ^{-1} x\right)+\frac{m x}{\sqrt{\left(1-x^{2}\right.}} \cos \left(m \sin ^{-1} x\right)-\cos \left(m \sin ^{-1} x\right) \frac{m x}{\sqrt{\left(1-x^{2}\right)}}$
$=-m^{2} \sin \left(m \sin ^{-1} x\right)$
$=-m^{2} y$