Question

If \[y = e^{\tan^{- 1} x}\] prove that (1 + x²)y₂ + (2x − 1)y₁ = 0 ?

Answer 1

Here,

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

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

\[\frac{d y}{d x} = e^{\tan^{- 1} x } \times \frac{1}{1 + x^2}\]

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

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

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

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

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

Hence proved