If $y$ is twice differentiable function of $x$, then the expression $\left(1-x^2\right)·\frac{d^{2}y}{d{x}^2}-x·\frac{dy}{dx}+y$ by means of the transformation $x=\sin t$in terms of $t$ is $\frac{d^{2}y}{d{t}^2}+\lambda y.$ Find value of $\lambda .$

Given: $\left(1-x^2\right)·\frac{d^{2}y}{d{x}^2}-x·\frac{dy}{dx}+y=\frac{d^{2}y}{d{t}^2}+\lambda y$, when $x=\sin t.$

We have, 

$\Rightarrow x=\sin t$

$\Rightarrow$$t={\sin }^{-1}x$

$\Rightarrow dt=\frac{1}{\sqrt{1-x^2}}dx$

$\Rightarrow \frac{dx}{dt}=\sqrt{1-x^2}.$

Now,

$$\begin{gathered} \frac{dy}{dx}=\left(\frac{dy}{dt}\right)·\left(\frac{dt}{dx}\right) \\ \Rightarrow \frac{dy}{dx}=\left(\frac{dy}{dt}\right)·\left(\frac{1}{\sqrt{1-x^2}}\right) \\ \Rightarrow \left(\sqrt{1-x^2}\right)\frac{dy}{dx}=\frac{dy}{dt}. \end{gathered}$$

Differentiating above equation with respect to $x$, we get:

$\left(\sqrt{1-x^2}\right)·\frac{d^{2}y}{d{x}^2}+\frac{1·(-2x)}{2\sqrt{1-x^2}}·\frac{dy}{dx}=\frac{d}{dx}\left(\frac{dy}{dt}\right)$

$\Rightarrow \left(\sqrt{1-x^2}\right)·\frac{d^{2}y}{d{x}^2}-\frac{x}{\sqrt{1-x^2}}\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{dt}·\frac{dt}{dx}$

$\Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=\left(\sqrt{1-x^2}\right)·\frac{d^{2}y}{d{t}^2}·\left(\frac{1}{\sqrt{1-x^2}}\right)$

$\Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=\frac{d^{2}y}{d{t}^2}$

Adding $y$ on both sides, we get:

$\Rightarrow \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dx}{dy}+y=\frac{d^{2}y}{d{t}^2}+y$

From given equation, we get:

$$\begin{gathered} \Rightarrow \frac{d^{2}y}{d{t}^2}+\lambda y=\frac{d^{2}y}{d{t}^2}+y \\ \Rightarrow \lambda =1. \end{gathered}$$

Therefore, solution is $\left(2\right) 1.$