Question

If y=a cos (log x) + sin(log x), then $x^2{y}_2-x{y}_1=$

Answer 1

$y=acos\left(logx\right)+sin\left(logx\right)$ _______ (1)

Differentiating w.r.t x

$\frac{dy}{dx}=-\frac{asin\left(logx\right)}{x}+\frac{cos\left(logx\right)}{x}$ $\because \frac{d}{dx}logx=\frac{1}{x}$

$x\frac{dy}{dx}=-asin\left(logx\right)+cos\left(logx\right)$

Again differentiating w.r.t x,

$x\frac{d}{dx}\left(\frac{dy}{dx}\right)+\frac{dy}{dx}\left(1\right)=-\frac{acos\left(logx\right)}{x}-\frac{sin\left(logx\right)}{x}$

$x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-\frac{acos\left(logx\right)}{x}-\frac{sin\left(logx\right)}{x}$

$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=-\left(acos\right(logx)+sin(logx\left)\right)$

$x^2{y}_1+x{y}_1=-y$ [from (i)]