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Q.
If the differential equation $y d x-x d y+y^2 \cos x d x=0$ satisfies the initial condition $y\left(\frac{\pi}{2}\right)=1$, then $y (\pi)$ equals
Differential Equations
Solution:
$ \frac{y d x-x d y}{y^2}+\cos x d x=0 $
$d\left(\frac{x}{y}\right)+\cos x d x=0 $
$\frac{x}{y}+\sin x=c$
$x+y \sin x=c y$
$\frac{\pi}{2}+1=c $
$\Rightarrow x+y \sin x=\left(\frac{\pi}{2}+1\right) y $
$ \pi=\left(\frac{\pi}{2}+1\right) y \Rightarrow y=\frac{2 \pi}{\pi+2}$