If $y = \left(\sin^{-1} x\right)^2$, prove that $\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} - 2 = 0$
Given: $\;$ $y = \left(\sin^{-1} x\right)^2$ $\;\;\; \cdots \; (1)$
Differentiating equation $(1)$ w.r.t $x$ we have,
$\dfrac{dy}{dx} = 2 \sin^{-1} x \times \dfrac{1}{\sqrt{1 - x^2}}$
i.e. $\;$ $\left(\dfrac{dy}{dx}\right)^2 = \dfrac{4 \left(\sin^{-1}x\right)^2}{1 - x^2}$
i.e. $\;$ $\left(1 - x^2\right) \left(\dfrac{dy}{dx}\right)^2 = 4 \left(\sin^{-1}x\right)^2$
i.e. $\;$ $\left(1 - x^2\right) \left(\dfrac{dy}{dx}\right)^2 = 4 y$ $\;\;\; \cdots \; (2)$ $\;\;\;$ [in view of equation $(1)$]
Differentiating equation $(2)$ w.r.t $x$ we have,
$\left(1 - x^2\right) \times 2 \dfrac{dy}{dx} \times \dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2 \left(-2 x\right) = 4 \times \dfrac{dy}{dx}$ $\;\;\; \cdots \; (3)$
Since $\;$ $\dfrac{dy}{dx} \neq 0$, $\;$ dividing equation $(3)$ by $\dfrac{dy}{dx}$, we get,
$\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} = 2$
i.e. $\;$ $\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} - 2 = 0$
Hence proved.