Correct option is (2) \(\mathrm{e}^{-\pi / 4}\)
\(\frac{d y}{d x}+2 y=\sin 2 x, y(0)=\frac{3}{4}\)
I.F \(=\mathrm{e}^{\int 2 \mathrm{dx}}=\mathrm{e}^{2 \mathrm{x}}\)
\(y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x\)
\(y \cdot e^{2 x}=\frac{e^{2 x}(2 \sin 2 x-2 \cos 2 x)}{4+4}+C\)
\(\mathrm{x}=0, \mathrm{y}=\frac{3}{4} \Rightarrow \frac{3}{4} \cdot 1=\frac{1(0-2)}{8}+\mathrm{C}\)
\(\frac{3}{4}=-\frac{1}{4}+\mathrm{C}\)
\(1=\mathrm{C}\)
\(\mathrm{y}=\frac{2 \sin 2 \mathrm{x}-2 \cos 2 \mathrm{x}}{8}+1 . \mathrm{e}^{-2 \mathrm{x}}\)
\(\mathrm{x}=\frac{\pi}{8}, \mathrm{y}=\frac{1}{8}\left(2 \sin \frac{\pi}{4}-2 \cos \frac{\pi}{4}\right)+\mathrm{e}^{-2\left(\frac{\pi}{8}\right)}\)
\(y=0+e^{-\frac{\pi}{4}}\)
