Question

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

Answer 1

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

Differentiating w.r.t to $x$, we have

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

$\Rightarrow y_1=\frac{1}{x}[-3\sin (\log x)+4\cos (\log x\left)\right]$

Again differentiating w.r.t to $x$, we have
$\frac{d^{2}y}{d{x}^2}=\frac{x[\frac{-3\cos \left(\log x\right)}{x}-\frac{4\sin \left(\log x\right)}{x}]-[-3\sin (\log x)+4\cos (\log x\left)\right]}{x^2}$

$=\frac{-3\cos \left(\log x\right)-4\sin \left(\log x\right)+3\sin \left(\log x\right)+4\cos \left(\log x\right)}{x^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{-\sin \left(\log x\right)-7\cos \left(\log x\right)}{x^2}$

$\Rightarrow y_2=\frac{-\sin \left(\log x\right)-7\cos \left(\log x\right)}{x^2}$

Now, L.H.S = $x^2{y}_2+x{y}_1+y$
$=x^2\left(\frac{-\sin \left(\log x\right)-7\cos \left(\log x\right)}{x^2}\right)+x\times \frac{1}{x}[-3\sin (\log x)+4\cos (\log x)+3\cos (\log x)+4\sin (\log x\left)\right]$

$=-\sin \left(\log x\right)-7\cos \left(\log x\right)-3\sin \left(\log x\right)+4\cos \left(\log x\right)+3\cos \left(\log x\right)+4\sin \left(\log x\right)=0$ = RHS

Hence proved.