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Q.
If $x \, sin \left(\frac{y}{x}\right) dy=\left[y sin\left(\frac{y}{x}\right)-x\right]dx$ and $y(1) =\frac{\pi}{2}$, then the value of cos $\left(\frac{y}{x}\right)$ is equal to
Given : x sin $\left(\frac{y}{x}\right) dy=\left[y sin\left(\frac{y}{x}\right)-x\right]dx$
$\Rightarrow \frac{dy}{dx}=\frac{y sin \left(\frac{y}{x}\right)-x}{y sin \left(\frac{y}{x}\right) }$
Put$\frac{y}{x}=z \Rightarrow \frac{dy}{dx}=z.1+x. \frac{dz}{dx}$
$\therefore x. \frac{dz}{dx}+z=\frac{zx sin z x}{x sin z}=z cosec z$
$\Rightarrow x \frac{dz}{dx}=-cosec z \Rightarrow -sin z dz =\frac{dx}{x}$
$\Rightarrow cos z = log x + c$
$\Rightarrow cos\left(\frac{y}{x}\right)=log x +c$
But y$\left(1\right) =\frac{\pi }{2} \Rightarrow x= 1, y=\frac{\pi }{2}$
$\therefore cos \frac{\pi }{2}log\left(1\right)+c \Rightarrow c=0$
$\therefore cos\left(\frac{y}{x}\right)=log x$