Answer
$xy'=-sin x -y$ and $y(\pi/2)=0$
Work Step by Step
Apply the quotient rule of differentiation:
Thus, $y'=\dfrac{-x \sin x -\cos x}{x^2}=(\dfrac{-\sin x}{x}) -(\dfrac{1}{x})\dfrac{\cos x}{x}$
Since, we have $y=\dfrac{\cos x}{x}$
Therefore, $y'=(\dfrac{-\sin x}{x}) -\dfrac{y}{x}$
or, $xy'=-sin x -y$
Apply the initial conditions:
$y(\dfrac{\pi}{2})=\dfrac{\cos \dfrac{\pi}{2}}{\dfrac{\pi}{2}}=0$
