If \(y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2\), then \(\left(1-x^2\right) y_2-x y_1=\)

\(y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2 \)
\( \therefore \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 \sin ^{-1} x}{\sqrt{1-x^2}}-\frac{2 \cos ^{-1} x}{\sqrt{1-x^2}} \)
\( \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2\left(\sin ^{-1} x-\cos ^{-1} x\right)}{\sqrt{1-x^2}} \)
\( \Rightarrow \sqrt{1-x^2} \frac{\mathrm{d} y}{\mathrm{~d} x}=2\left(\sin ^{-1} x-\cos ^{-1} x\right)\)
Differentiating both sides w.r.t. \(x\), we get
\(\sqrt{1-x^2} \cdot \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\frac{\mathrm{d} y}{\mathrm{~d} x} \cdot \frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x) \)
\( =2\left(\frac{1}{\sqrt{1-x^2}}-\frac{(-1)}{\sqrt{1-x^2}}\right)=\frac{4}{\sqrt{1-x^2}} \)
\( \therefore \left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-x \frac{\mathrm{d} y}{\mathrm{~d} x}=4\)