Question

If $y = e^{a \cos^{-1}x}, \; -1 \leq x \leq 1$, then prove that \[ (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2 y = 0. \]

Hint:

When $y = e^{a \cos^{-1}x}$, always express $\frac{dy}{dx}$ in terms of $y$ itself to simplify higher-order derivatives.

Answer 1

Step 1: Let \[ y = e^{a \cos^{-1}x}, \cos^{-1}x = \theta \implies x = \cos \theta \]

Step 2: Differentiate once. \[ \frac{dy}{dx} = e^{a \cos^{-1}x} \cdot a \cdot \frac{d}{dx}(\cos^{-1}x) \] \[ \frac{dy}{dx} = y \cdot a \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) \] \[ \frac{dy}{dx} = -\frac{ay}{\sqrt{1-x^2}} \]

Step 3: Differentiate again. \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{ay}{\sqrt{1-x^2}}\right) \] Apply product rule: \[ \frac{d^2y}{dx^2} = -a \left(\frac{dy}{dx} \cdot \frac{1}{\sqrt{1-x^2}} + y \cdot \frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right)\right) \] Now, \[ \frac{d}{dx}\left(\frac{1}{\sqrt{1-x^2}}\right) = \frac{x}{(1-x^2)^{3/2}} \] So, \[ \frac{d^2y}{dx^2} = -\frac{a}{\sqrt{1-x^2}} \cdot \frac{dy}{dx} - \frac{axy}{(1-x^2)^{3/2}} \]

Step 4: Multiply through by $(1-x^2)$. \[ (1-x^2)\frac{d^2y}{dx^2} = -a \sqrt{1-x^2}\frac{dy}{dx} - \frac{axy}{\sqrt{1-x^2}} \] Now substitute $\frac{dy}{dx} = -\frac{ay}{\sqrt{1-x^2}}$: \[ (1-x^2)\frac{d^2y}{dx^2} = -a\sqrt{1-x^2}\left(-\frac{ay}{\sqrt{1-x^2}}\right) - \frac{axy}{\sqrt{1-x^2}} \] \[ = a^2 y - x \frac{dy}{dx} \]

Step 5: Rearranging. \[ (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0 \]

Final Answer: \[ \boxed{(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0} \]