Question

Verify that \[y = ce^{tan^{- 1}} x\]  is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]

Answer 1

We have,

\[y = c e^{tan^{- 1}x } ............(1)\]

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

\[\frac{dy}{dx} = c e^{tan^{- 1}x } \frac{1}{1 + x^2}............(2)\]

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

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

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

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

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

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

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

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