Let y=y(x) solve x,dy=(y+x^3 x),dx, y()=0.

Question

Let $y=y(x)$ solve $x\,dy=(y+x^3\cos x)\,dx$, $y(\pi)=0$. Then $y\left(\dfrac{\pi}{2}\right)$ is:

Accepted Answer

$$d\left(\frac{y}{x}\right) = x\cos x\,dx \implies \frac{y}{x} = x\sin x+\cos x+C.$$

Using $(\pi,0)$: $C=1$. At $x=\pi/2$: $\dfrac{y}{\pi/2} = \dfrac{\pi}{2}+0+1 \implies y = \dfrac{\pi^2}{4}+\dfrac{\pi}{2}$.

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