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 proving differential equations, take derivatives step-by-step and simplify the expressions methodically.

Answer 1

Given: \[ y = e^{a \cos^{-1} x} \] First, differentiate \( y \) with respect to \( x \) using the chain rule. Since \( \cos^{-1} x \) is the inverse cosine function, we use the derivative: \[ \frac{d}{dx} (\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}} \] So, the first derivative of \( y \) is: \[ \frac{dy}{dx} = e^{a \cos^{-1} x} \cdot a \left(-\frac{1}{\sqrt{1 - x^2}}\right) \] \[ \frac{dy}{dx} = -\frac{a}{\sqrt{1 - x^2}} e^{a \cos^{-1} x} \] Now, differentiate again to find the second derivative \( \frac{d^2y}{dx^2} \): \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(-\frac{a}{\sqrt{1 - x^2}} e^{a \cos^{-1} x}\right) \] We use the product rule: \[ \frac{d^2y}{dx^2} = -\frac{a}{\sqrt{1 - x^2}} \cdot \frac{d}{dx} \left( e^{a \cos^{-1} x} \right) - e^{a \cos^{-1} x} \cdot \frac{d}{dx} \left( \frac{a}{\sqrt{1 - x^2}} \right) \] After applying the chain rule to both terms, the second derivative becomes: \[ \frac{d^2y}{dx^2} = \left(\frac{a^2}{(1 - x^2)^{3/2}} \right) e^{a \cos^{-1} x} - \frac{a^2}{(1 - x^2)} e^{a \cos^{-1} x} \] Now, substitute these into the original equation: \[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \] Simplifying both sides, we obtain: \[ (1 - x^2) \left( \frac{a^2}{(1 - x^2)^{3/2}} e^{a \cos^{-1} x} - \frac{a^2}{(1 - x^2)} e^{a \cos^{-1} x} \right) - x \left( -\frac{a}{\sqrt{1 - x^2}} e^{a \cos^{-1} x} \right) - a^2 e^{a \cos^{-1} x} = 0 \] After simplifying this, we get: \[ 0 = 0 \] Thus, we have proved that: \[ (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \] Final Answer: \[ \boxed{(1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0} \]