Question

If $y=a\cos (\log x)-b\sin (\log x)$ , then the value of $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$y=a\cos (\log x)-b\sin (\log x)$
On differentiating w.r.t. $x,$
we get $\frac{dy}{dx}=a\frac{[-\sin (\log x)]}{x}-\frac{b\cos (\log x)}{x}$
$=-\frac{[a\sin (\log x)+b\cos (\log x)]}{x}$
$\Rightarrow$ $x\frac{dy}{dx}=-[a\sin (\log x)+b\cos (\log x)]$
Again, on differentiating w.r.t. $x,$ we get $x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-\left[\frac{a\cos (\log x)}{x}-\frac{b\sin (\log x)}{x}\right]=-\frac{y}{x}$
$\Rightarrow$ $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0$