If y=y(x) is the solution of the differential equation d y ~d x+2 y=sin (2 x), y(0)= 3 4, then y ( π 8 ) is equal to:

$$\begin{aligned} & \frac{d y}{d x}+2 y=\sin 2 x, y(0)=\frac{3}{4} \\ & \text { I.F }=e^{\int 2 d x}=e^{2 x} \\ & y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x \\ & y \cdot e^{2 x}=\frac{e^{2 x}(2 \sin 2 x-2 \cos 2 x)}{4+4}+C \\ & x=0, y=\frac{3}{4} \Rightarrow \frac{3}{4} \cdot 1=\frac{1(0-2)}{8}+C \\ & \frac{3}{4}=-\frac{1}{4}+C \\ & 1=C \\ & y=\frac{2 \sin 2 x-2 \cos 2 x}{8}+1 \cdot e^{-2 x} \\ & x=\frac{\pi}{8}, y=\frac{1}{8}\left(2 \sin \frac{\pi}{4}-2 \cos \frac{\pi}{4}\right)+e^{-2\left(\frac{\pi}{8}\right)} \\ & y=0+e^{-\frac{\pi}{4}}\end{aligned}$$​dxdy​+2y=sin2x,y(0)=43​ I.F =e∫2dx=e2xy⋅e2x=∫e2xsin2xdxy⋅e2x=4+4e2x(2sin2x−2cos2x)​+Cx=0,y=43​⇒43​⋅1=81(0−2)​+C43​=−41​+C1=Cy=82sin2x−2cos2x​+1⋅e−2xx=8π​,y=81​(2sin4π​−2cos4π​)+e−2(8π​)y=0+e−4π​​