Question

If y = a cos (logx) + b sin (logx) where a, b are parameters then x${}^2$y '+ xy' =

• A. y
• B. -y
• C. 2y
• D. -2y

Answer 1

The correct option is B -y

Given,

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

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

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

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

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

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

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