If $x = a\cos α + b\sin α$ and $y = a\sin α - b\cos α$, then $\left(x\frac{dy}{dx}-y^2\frac{d^2y}{dx^2}\right)$ is equal to:

$y$

The correct answer is Option (4) → $y$

Given $x=a\cos\alpha+b\sin\alpha,\quad y=a\sin\alpha-b\cos\alpha$.

$\frac{dx}{d\alpha}=-a\sin\alpha+b\cos\alpha=-y,\qquad \frac{dy}{d\alpha}=a\cos\alpha+b\sin\alpha=x$

$\frac{dy}{dx}=\frac{dy/d\alpha}{dx/d\alpha}=\frac{x}{-y}=-\frac{x}{y}$

$\frac{d^2y}{dx^2}=\frac{d}{d\alpha}\!\left(\frac{dy}{dx}\right)\Big/\frac{dx}{d\alpha} =\frac{d}{d\alpha}\!\left(\frac{x}{-y}\right)\Big/(-y) =\frac{(-y)^2+(x)^2}{(-y)^3}=-\frac{x^2+y^2}{y^3}$

$\displaystyle x\frac{dy}{dx}-y^2\frac{d^2y}{dx^2} = x\left(-\frac{x}{y}\right)-y^2\left(-\frac{x^2+y^2}{y^3}\right) = -\frac{x^2}{y}+\frac{x^2+y^2}{y}=\frac{y^2}{y}=y$

Answer: ${y}$