Let y=y(x) solve xdydx+y x=4x, x(0,).

Question

Let $y=y(x)$ solve $\sin x\dfrac{dy}{dx}+y\cos x=4x$, $x\in(0,\pi)$. If $y\left(\dfrac{\pi}{2}\right)=0$, then $y\left(\dfrac{\pi}{6}\right)$ equals:

Accepted Answer

$d(y\sin x)=4x\,dx \implies y\sin x=2x^2+C$. Using $(\pi/2,0)$: $C=-\dfrac{\pi^2}{2}$.

At $x=\pi/6$: $y\cdot\dfrac12 = \dfrac{\pi^2}{18}-\dfrac{\pi^2}{2} \implies y=-\dfrac{8\pi^2}{9}$.

Solutions · 1