Question

If $xsin\left(\frac{y}{x}\right)dy=\left[ysin\left(\frac{y}{x}\right)-x\right]dx$ and $y(1)=\frac{\pi }{2}$, then the value of cos $\left(\frac{y}{x}\right)$ is equal to

• A. $x$
• B. $\frac{1}{x}$
• C. $logx$
• D. e${}^x$

Answer 1

Answer: (C)

Given : x sin $\left(\frac{y}{x}\right)dy=\left[ysin\left(\frac{y}{x}\right)-x\right]dx$
$\Rightarrow \frac{dy}{dx}=\frac{ysin\left(\frac{y}{x}\right)-x}{ysin\left(\frac{y}{x}\right)}$
Put$\frac{y}{x}=z\Rightarrow \frac{dy}{dx}=z.1+x.\frac{dz}{dx}$
$\therefore x.\frac{dz}{dx}+z=\frac{zxsinzx}{xsinz}=zcosecz$
$\Rightarrow x\frac{dz}{dx}=-cosecz\Rightarrow -sinzdz=\frac{dx}{x}$
$\Rightarrow cosz=logx+c$
$\Rightarrow cos\left(\frac{y}{x}\right)=logx+c$
But y$\left(1\right)=\frac{\pi }{2}\Rightarrow x=1,y=\frac{\pi }{2}$
$\therefore cos\frac{\pi }{2}log\left(1\right)+c\Rightarrow c=0$
$\therefore cos\left(\frac{y}{x}\right)=logx$