If \(m \sin ^{-1} x=\log _{e} y\), then \(\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}\) is equal to

\[
\begin{aligned}
&\text { Given, } \mathrm{m} \sin ^{-1} \mathrm{x}=\log _{\mathrm{e}} \mathrm{y} \\
&\Rightarrow \quad \mathrm{y}=\mathrm{e}^{\mathrm{m} \sin ^{-1} \mathrm{x}}
\end{aligned}
\]
On differentiating w.r.t. \(x\), we get
\(y^{\prime}=e^{m \sin ^{-1} x} \times \frac{m}{\sqrt{1-x^{2}}}\)
\(\Rightarrow \sqrt{1-\mathrm{x}^{2}} \cdot \mathrm{y}^{\prime}=\mathrm{my}\)
On squaring both sides, we get
\[
\left(1-x^{2}\right) y^{\prime 2}=m^{2} y^{2}
\]
Again differentiating, we get
\[
\begin{aligned}
&\left(1-x^{2}\right) 2 y^{\prime} \times y^{\prime \prime}+y^{\prime 2}(-2 x)=m^{2} \cdot 2 y y^{\prime} \\
\therefore \quad &\left(1-x^{2}\right) y^{\prime \prime}-x y^{\prime}=m^{2} y
\end{aligned}
\]