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
\[\frac{dy}{dx} = \frac{d}{dx}\left[ \log \left( x + \sqrt{x^2 + a^2} \right)^2 \right]\]
\[ = \frac{d}{dx}\left[ 2 \log \left( x + \sqrt{x^2 + a^2} \right) \right]\]
\[ = 2\frac{1 + \frac{1}{2}\frac{2x}{\sqrt{x^2 + a^2}}}{x + \sqrt{x^2 + a^2}}\]
\[ = 2\frac{\frac{\sqrt{x^2 + a^2} + x}{\sqrt{x^2 + a^2}}}{x + \sqrt{x^2 + a^2}}\]
\[ = \frac{2}{\sqrt{x^2 + a^2}} ............(2)\]
Differentiating both sides of (2) with respect to x, we get
\[\frac{d^2 y}{d x^2} = 2\left( - \frac{1}{2} \right)\frac{2x}{\left( x^2 + a^2 \right)\sqrt{x^2 + a^2}}\]
\[ \Rightarrow \left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} = - \frac{2x}{\sqrt{x^2 + a^2}}\]
\[ \Rightarrow \left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} = - x\frac{dy}{dx} ...........\left[\text{Using (2)} \right]\]
\[ \Rightarrow \left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]

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