Question

Verify that $y=c{e}^{{\tan }^{-1}x}$ 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
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=c\frac{\left(1+x^2\right)e^{{\tan }^{-1}x}{\displaystyle \frac{1}{1+x^2}}-e^{{\tan }^{-1}x}\left(2x\right)}{{\left(1+x^2\right)}^2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=c\frac{e^{{\tan }^{-1}x}-2x{e}^{{\tan }^{-1}x}}{{\left(1+x^2\right)}^2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=c\frac{\left(1-2x\right)e^{{\tan }^{-1}x}}{{\left(1+x^2\right)}^2} \\ \Rightarrow \left(1+x^2\right)\frac{d^{2}y}{d{x}^2}=c\left(1-2x\right)\frac{e^{{\tan }^{-1}x}}{\left(1+x^2\right)} \\ \Rightarrow \left(1+x^2\right)\frac{d^{2}y}{d{x}^2}=\left(1-2x\right)\frac{dy}{dx}\left[\text{Using}\left(2\right)\right] \\ \Rightarrow \left(1+x^2\right)\frac{d^{2}y}{d{x}^2}+\left(2x-1\right)\frac{dy}{dx}=0 \end{gathered}$$
Hence, the given function is the solution to the given differential equation.