Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& \frac{dy}{dx}+2y=\sin 2x,y(0)=\frac{3}{4}\\ & \text{ I.F }=e^{\int 2dx}=e^{2x}\\ & y\cdot e^{2x}=\int e^{2x}\sin 2xdx\\ & y\cdot e^{2x}=\frac{e^{2x}(2\sin 2x-2\cos 2x)}{4+4}+C\\ & x=0,y=\frac{3}{4}\Rightarrow \frac{3}{4}\cdot 1=\frac{1(0-2)}{8}+C\\ & \frac{3}{4}=-\frac{1}{4}+C\\ & 1=C\\ & y=\frac{2\sin 2x-2\cos 2x}{8}+1\cdot e^{-2x}\\ & x=\frac{\pi }{8},y=\frac{1}{8}\left(2\sin \frac{\pi }{4}-2\cos \frac{\pi }{4}\right)+e^{-2\left(\frac{\pi }{8}\right)}\\ & y=0+e^{-\frac{\pi }{4}}\end{array}$