If y=y(x) solves dydx+2y x= x, y(3)=0, the maximum value of y(x) over R is:

Question

If $y=y(x)$ solves $\dfrac{dy}{dx}+2y	an x=\sin x$, $y\left(\dfrac{\pi}{3}ight)=0$, the maximum value of $y(x)$ over $\mathbb{R}$ is:

Accepted Answer

(Same setup as Question 128.) $y=-2\cos^2x+\cos x = \dfrac18-2\left(\cos x-\dfrac14\right)^2$; maximum value $\dfrac18$.

Solutions · 1