Question

lf $y=$ $a\cos \left(\log x\right)+b\sin \left(\log x\right)$ , then $x^{2}y{}^{\prime \prime }+xy{}^{\prime }=$

• A. $-y$
• B. $\frac{1}{y}$
• C. $y$
• D. $y^2$

Answer 1

The correct option is A $-y$

$y=a\cos \left(logx\right)+b\sin \left(logx\right)$

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

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

$x^{2}y{}^{\prime \prime }=-y-xy{}^{\prime }$

$x^{2}y{}^{\prime \prime }+xy{}^{\prime }=-y$

Hence, option'A' is correct.