Question

If $y=3\cos \left(\log x\right)+4\sin \left(\log x\right)$, then show that $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0$

Answer 1

Given, $y=3\cos \left(\log x\right)+4\sin \left(\log x\right)$

Differentiate both sides with respect to $x$, we get

$\frac{dy}{dx}=-3\sin \left(\log x\right)\cdot \frac{1}{x}\cdot 1+4\cos \left(\log x\right)\cdot \frac{1}{x}\cdot 1$

$\frac{dy}{dx}=\frac{1}{x}[-3\sin \left(\log x\right)+4\cos \left(\log x\right)]$

$x\frac{dy}{dx}=-3\sin \left(\log x\right)+4\cos \left(\log x\right)$

Differentiate both sides with respect to $x$, we get

$x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-3\cos \left(\log x\right)\cdot \frac{1}{x}-4\sin \left(\log x\right)\cdot \frac{1}{x}$

$x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-\frac{1}{x}\left[3\cos \left(\log x\right)+4\sin \left(\log x\right)\right]$

$x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-y/x$

$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0$
Hence, proved.