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 $