Form the differential equation having $y = (\sin^{-1} x)^2 + A \cos^{-1} x + B$, where $A$ and $B$ are arbitrary constants, as its general solution.

$(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - 2 = 0$

The correct answer is Option (4) → $(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - 2 = 0$ ##

Given that, $y = (\sin^{-1} x)^2 + A \cos^{-1} x + B$

On differentiating w.r.t. $x$, we get

$\frac{dy}{dx} = \frac{2 \sin^{-1} x}{\sqrt{1 - x^2}} + \frac{(-A)}{\sqrt{1 - x^2}}$

$\Rightarrow \sqrt{1 - x^2} \frac{dy}{dx} = 2 \sin^{-1} x - A$

Again, differentiating w.r.t. $x$, we get

$\sqrt{1 - x^2} \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{-2x}{2\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}}$

$\Rightarrow (1 - x^2) \frac{d^2y}{dx^2} - \frac{x}{\sqrt{1 - x^2}} \cdot \sqrt{1 - x^2} \frac{dy}{dx} = 2$

$\Rightarrow (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} = 2$

$\Rightarrow (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - 2 = 0$

which is the required differential equation.