Question

Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]

Answer 1

We have,

\[y = e^{m \cos^{- 1} x}.........(1)\]

Differentiating both sides of (1) with respect to x, we get

\[\frac{dy}{dx} = m e^{m \cos^{- 1} x} \left( \frac{- 1}{\sqrt{1 - x^2}} \right)\]

\[ \Rightarrow \frac{dy}{dx} = - \frac{m e^{m \cos^{- 1} x}}{\sqrt{1 - x^2}} .........(2)\]

Differentiating both sides of (2) with respect to x, we get

\[\frac{d^2 y}{d x^2} = \frac{d}{dx}\left( - \frac{m e^{m \cos^{- 1} x}}{\sqrt{1 - x^2}} \right)\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \left( - m \right)\left[ \frac{\sqrt{1 - x^2}m e^{m \cos^{- 1} x} \left( - \frac{1}{\sqrt{1 - x^2}} \right) - e^{m \cos^{- 1} x} \frac{1}{2}\left( - \frac{2x}{\sqrt{1 - x^2}} \right)}{\left( 1 - x^2 \right)} \right]\]

\[ \Rightarrow \left( 1 - x^2 \right)\frac{d^2 y}{d x^2} = \left( - m \right)\left[ - m e^{m \cos^{- 1} x} + \frac{x e^{m \cos^{- 1} x}}{\sqrt{1 - x^2}} \right]\]

\[ \Rightarrow \left( 1 - x^2 \right)\frac{d^2 y}{d x^2} = m^2 e^{m \cos^{- 1} x} - mx\frac{e^{m \cos^{- 1} x}}{\sqrt{1 - x^2}}\]

\[ \Rightarrow \left( 1 - x^2 \right)\frac{d^2 y}{d x^2} = m^2 y + x\frac{dy}{dx} ..........\left[\text{Using }\left( 1 \right)\text{ and }\left( 2 \right) \right]\]

\[ \Rightarrow \left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]

Hence, the given function is the solution to the given differential equation.