## University Calculus: Early Transcendentals (3rd Edition)

$xy'=-sin x -y$ and $y(\pi/2)=0$
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: $y(\dfrac{\pi}{2})=\dfrac{\cos \dfrac{\pi}{2}}{\dfrac{\pi}{2}}=0$