If y=y(x), x(0,/2), solves ^2(2x)dydx+(8^2(2x)+24x)y = 2e^-4x(22x+2x), with y(4)=e^-, then…

Question

If $y=y(x)$, $x\in(0,\pi/2)$, solves $\sin^2(2x)\dfrac{dy}{dx}+(8\sin^2(2x)+2\sin4x)y = 2e^{-4x}(2\sin2x+\cos2x)$, with $y\left(\dfrac{\pi}{4}\right)=e^{-\pi}$, then $y\left(\dfrac{\pi}{6}\right)$ is:

Accepted Answer

$$\frac{dy}{dx}+(8+4\cot2x)y = \frac{2e^{-4x}(2\sin2x+\cos2x)}{\sin^22x}.$$

I.F. $=e^{8x}\sin^22x$. Solving: $ye^{8x}\sin^2 2x = e^{4x}\sin2x+c$. Using $y(\pi/4)=e^{-\pi}$: $c=0$.

At $x=\pi/6$: $y = \dfrac{2}{\sqrt3}e^{-2\pi/3}$.

Solutions · 1