Question

If the transformation $x=\cos \theta$. reduces the differential equation $(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+y=0$ into $\frac{d^{2}y}{d{\theta }^2}+Ay=0$, then the value of A is

Answer 1

Since $\frac{dx}{d\theta }=-\sin \theta$, so
$\frac{dy}{dx}=\frac{dy}{d\theta }\frac{d\theta }{dx}=-cosec\theta \frac{dy}{d\theta }$
Also,
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{d\theta }\left(\frac{dy}{dx}\right)\frac{d\theta }{dx}=\frac{d}{d\theta }\left(cosec\theta \frac{dy}{d\theta }\right)cosec\theta$
$=(-cosec\theta \cot \theta \frac{dy}{d\theta }+cosec\theta \frac{d^{2}y}{d{\theta }^2})cosec\theta$
$=-cose{c}^2\theta \cot \theta \frac{dy}{d\theta }+cose{c}^2\theta \frac{d^{2}y}{d{\theta }^2}$
Putting these in the given differential equation, we get
$(1-x^{})\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+y=0$
$\Rightarrow$ $(1-{\cos }^2\theta )(cose{c}^2\theta \cot \theta \frac{dy}{d\theta }+cose{c}^2\theta \frac{d^{2}y}{d{\theta }^2})-\cos \theta (-cosec\theta \frac{dy}{d\theta })+y=0$
$-\cot \theta \frac{dy}{d\theta }+\frac{d^{2}y}{d{\theta }^2}+\cot \theta \frac{dy}{d\theta }+y=0\Rightarrow \frac{d^{2}y}{d{\theta }^2}+y=0$
Hence $A=1$