Question

If $\cos x=\frac{1-t^2}{1+t^2}$ and $\sin y=\frac{2t}{1+t^2}$ where $t\in (-1,0)$, then the value of $4\frac{d^{2}y}{d{x}^2}-\frac{3}{2}\frac{dy}{dx}-\frac{1}{2}\frac{y}{x}$ at $(x,y)=(1,-1)$ is

Answer 1

$\cos x=\frac{1-t^2}{1+t^2}$
$\Rightarrow x={\cos }^{-1}\left(\frac{1-t^2}{1+t^2}\right)$
Let $t=\tan \theta$, where $\theta \in (-\frac{\pi }{4},0)$
Then $x={\cos }^{-1}\cos 2\theta$
$\Rightarrow x={\cos }^{-1}\cos (-2\theta )(\because -2\theta \in [0,\pi \left]\right)$
$\Rightarrow x=-2\theta$
$\Rightarrow x=-2{\tan }^{-1}t$
$\frac{dx}{dt}=\frac{-2}{1+t^2}$

$\sin y=\frac{2t}{1+t^2}$
$\Rightarrow y={\sin }^{-1}\left(\frac{2t}{1+t^2}\right)$
$\Rightarrow y={\sin }^{-1}\sin 2\theta$
$\Rightarrow y=2\theta$
$\Rightarrow y=2{\tan }^{-1}t$
$\frac{dy}{dt}=\frac{2}{1+t^2}$

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$
$\Rightarrow \frac{dy}{dx}=-1$
$\frac{d^{2}y}{d{x}^2}=0$
$\frac{y}{x}=\frac{2\theta }{-2\theta }=-1$

$\therefore 4\frac{d^{2}y}{d{x}^2}-\frac{3}{2}\frac{dy}{dx}-\frac{1}{2}\frac{y}{x}=2$