The function \(y(x)\) represented by \(x=\sin \mathrm{t}\), \(y=a e^{\mathrm{t} \sqrt{2}}+b \mathrm{e}^{\mathrm{t} \sqrt{2}}, \mathrm{t} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) satisfies the equation \(\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}=\mathrm{k} y\), then the value of \(k\) is

\(x =\sin t \)
\( \therefore \frac{d x}{d t} =\cos t \)
\( y =a e^{t \sqrt{2}}+b e^{t \sqrt{2}} \)
\( =e^{t \sqrt{2}}(a+b)\)
\(\therefore \frac{\mathrm{d} y}{\mathrm{dt}}=(\mathrm{a}+\mathrm{b}) \cdot \mathrm{e}^{\mathrm{t} \sqrt{2}} \cdot \sqrt{2} \)
\( \therefore \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{\sqrt{2}(\mathrm{a}+\mathrm{b}) \mathrm{e}^{\mathrm{t} \sqrt{2}}}{\cos \mathrm{t}}\)
\(\therefore \frac{d^2 y}{d x^2}=\frac{d}{d t}\left(\frac{\sqrt{2}(a+b) e^{t \sqrt{2}}}{\cos t}\right) \cdot \frac{d t}{d x} \)
\( =\sqrt{2}(a+b)\left[\frac{\cos t\left(e^{t \sqrt{2}} \cdot \sqrt{2}\right)-e^{t \sqrt{2}}(-\sin t)}{(\cos t)^2}\right] \cdot \frac{1}{\cos t}\)
\(=\sqrt{2}(\mathrm{a}+\mathrm{b})\left[\frac{\mathrm{e}^{\mathrm{t} \sqrt{2}}(\sqrt{2} \cos \mathrm{t}+\sin t)}{\cos ^3 t}\right] \)
\(\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}\)
\(=\left(1-\sin ^2 t\right) \sqrt{2}(a+b)\left[\frac{e^{t \sqrt{2}}(\sqrt{2} \cos t+\sin t)}{\cos ^3 t}\right]\) \(-\sin t\left[\frac{\sqrt{2}(a+b) \mathrm{e}^{\mathrm{t} \sqrt{2}}}{\cos t}\right]\)
\(=\left(\cos ^2 t\right) \sqrt{2}(a+b)\left[\frac{e^{t \sqrt{2}}(\sqrt{2} \cos t+\sin t)}{\cos ^3 t}\right] \) \( -\sin t\left[\frac{\sqrt{2}(a+b) e^{i \sqrt{2}}}{\cos t}\right]\)
\(=\frac{\sqrt{2}(a+b) e^{t \sqrt{2}}(\sqrt{2} \cos t+\sin t-\sin t)}{\cos t} \)
\( =\frac{\sqrt{2}(a+b) e^{t \sqrt{2}}(\sqrt{2} \cos t)}{\cos t} \)
\( =2\left(a e^{t \sqrt{2}}+b e^{t \sqrt{2}}\right) \)
\( = 2y \)
\( \therefore=2\)