Problem Statement
If $y = (\sin^{-1} x)^2$, prove that:
$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - 2 = 0$
Verified Solution & Marking Scheme
Compute First Derivative
Given $y = (\sin^{-1} x)^2$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 (\sin^{-1} x) \cdot \frac{d}{dx}(\sin^{-1} x) = \frac{2 \sin^{-1} x}{\sqrt{1 - x^2}}$
Clear the Radical by Cross-Multiplying and Squaring
$\sqrt{1 - x^2} \frac{dy}{dx} = 2 \sin^{-1} x$
Squaring both sides:
$(1 - x^2) \left( \frac{dy}{dx} \right)^2 = 4 (\sin^{-1} x)^2 = 4y$
Differentiate Implicitly with Respect to x
Differentiating both sides using product rule:
$\frac{d}{dx} \left[ (1 - x^2) \left( \frac{dy}{dx} \right)^2 \right] = \frac{d}{dx}[4y]$
$(1 - x^2) \cdot 2 \left( \frac{dy}{dx} \right) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 (-2x) = 4 \frac{dy}{dx}$
$2 \frac{dy}{dx} \left[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} \right] = 4 \frac{dy}{dx}$
Dividing both sides by $2\frac{dy}{dx}$ (since $\frac{dy}{dx} \neq 0$):
$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 2$
$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - 2 = 0$.