The function $y(x)$ represented by $x=\sin \mathrm{t}$, $y=a e^{\mathrm{t} \sqrt{2}}+b…

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
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Solution

$\begin{aligned} x & =\sin t \\ \therefore \quad \frac{d x}{d t} & =\cos t \\ y & =a e^{t \sqrt{2}}+b e^{t \sqrt{2}} \\ & =e^{t \sqrt{2}}(a+b)\end{aligned}$ $\begin{array}{ll}\therefore & \frac{\mathrm{d} y}{\mathrm{dt}}=(\mathrm{a}+\mathrm{b}) \cdot \mathrm{e}^{\mathrm{t} \sqrt{2}} \cdot \sqrt{2} \\ \therefore & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{\sqrt{2}(\mathrm{a}+\mathrm{b}) \mathrm{e}^{\mathrm{t} \sqrt{2}}}{\cos \mathrm{t}}\end{array}$ $\begin{aligned} \therefore \quad & \frac{d^2 y}{d x^2}=\frac{d}{d t}\left(\frac{\sqrt{2}(a+b) e^{t \sqrt{2}}}{\cos t}\right) \cdot \frac{d t}{d x} \\ & =\sqrt{2}(a+b)\left[\frac{\cos t\left(e^{t \sqrt{2}} \cdot \sqrt{2}\right)-e^{t \sqrt{2}}(-\sin t)}{(\cos t)^2}\right] \cdot \frac{1}{\cos t}\end{aligned}$ $\begin{aligned} & =\sqrt{2}(\mathrm{a}+\mathrm{b})\left[\frac{\mathrm{e}^{\mathrm{t} \sqrt{2}}(\sqrt{2} \cos \mathrm{t}+\sin t)}{\cos ^3 t}\right] \\ & \left(1-x^2\right) y^{\prime \prime}-x y^{\prime}\end{aligned}$ $=\left(1-\sin ^2 t\right) \sqrt{2}(a+b)\left[\frac{e^{t \sqrt{2}}(\sqrt{2} \cos t+\sin t)}{\cos ^3 t}\right]$ $-\sin t\left[\frac{\sqrt{2}(a+b) \mathrm{e}^{\mathrm{t} \sqrt{2}}}{\cos t}\right]$ $\begin{array}{r}=\left(\cos ^2 t\right) \sqrt{2}(a+b)\left[\frac{e^{t \sqrt{2}}(\sqrt{2} \cos t+\sin t)}{\cos ^3 t}\right] \\ -\sin t\left[\frac{\sqrt{2}(a+b) e^{i \sqrt{2}}}{\cos t}\right]\end{array}$ $\begin{aligned} & =\frac{\sqrt{2}(a+b) e^{t \sqrt{2}}(\sqrt{2} \cos t+\sin t-\sin t)}{\cos t} \\ & =\frac{\sqrt{2}(a+b) e^{t \sqrt{2}}(\sqrt{2} \cos t)}{\cos t} \\ & =2\left(a e^{t \sqrt{2}}+b e^{t \sqrt{2}}\right) \\ & = 2y \\ & \therefore=2\end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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