Question

If  \[y = e^{a \cos^{- 1}} x\] ,prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - a^2 y = 0\] ?

Answer 1

Here,

\[y = e^{a \cos^{- 1} x} \]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d y}{d x} = - e^{a \cos^{- 1} x} \times \frac{a}{\sqrt{1 - x^2}}\]

\[\text { Differentiating again w . r . t . x, we get }\]

\[\frac{d^2 y}{d x^2} = e^{a \cos^{- 1} x} \times \frac{a^2}{1 - x^2} + \frac{2xa e^{a \cos^{- 1} x}}{2 \left( 1 - x^2 \right)^\frac{3}{2}}\]

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

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

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

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

Hence proved.