Question

Verify that y = log ${\left(x+\sqrt{x^2+a^2}\right)}^2$ satisfies the differential equation $\left(a^2+x^2\right)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=0$.

Answer 1

We have,
$y=\log {\left(x+\sqrt{x^2+a^2}\right)}^2$ ...(1)
Differentiating both sides of (1) with respect to $x$, we get
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left[\log {\left(x+\sqrt{x^2+a^2}\right)}^2\right] \\ =\frac{d}{dx}\left[2\log \left(x+\sqrt{x^2+a^2}\right)\right] \\ =2\frac{1+{\displaystyle \frac{1}{2}}{\displaystyle \frac{2x}{\sqrt{x^2+a^2}}}}{x+\sqrt{x^2+a^2}} \\ =2\frac{{\displaystyle \frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}}}{x+\sqrt{x^2+a^2}} \\ =\frac{2}{\sqrt{x^2+a^2}}...\left(2\right) \end{gathered}$$
Differentiating both sides of (2) with respect to $x$, we get
$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=2\left(-\frac{1}{2}\right)\frac{2x}{\left(x^2+a^2\right)\sqrt{x^2+a^2}} \\ \Rightarrow \left(a^2+x^2\right)\frac{d^{2}y}{d{x}^2}=-\frac{2x}{\sqrt{x^2+a^2}} \\ \Rightarrow \left(a^2+x^2\right)\frac{d^{2}y}{d{x}^2}=-x\frac{dy}{dx}\left[\text{Using}\left(2\right)\right] \\ \Rightarrow \left(a^2+x^2\right)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=0 \end{gathered}$$
Hence, the given function is the solution to the given differential equation.