If $y=y(x)$ is the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=\sin (2 x), y(0)=\frac{3}{4}$, then $y\left(\frac{\pi}{8}\right)$ is equal to:

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{d y}{d x}+2 y=\sin 2 x, y(0)=\frac{3}{4} \ & 	ext { I.F }=e^{\int 2 d x}=e^{2 x} \ & y \cdot e^{2 x}=\int e^{2 x} \sin 2 x d x \ & y \cdot e^{2 x}=\frac{e^{2 x}(2 \sin 2 x-2 \cos 2 x)}{4+4}+C \ & x=0, y=\frac{3}{4} \Rightarrow \frac{3}{4} \cdot 1=\frac{1(0-2)}{8}+C \ & \frac{3}{4}=-\frac{1}{4}+C \ & 1=C \ & y=\frac{2 \sin 2 x-2 \cos 2 x}{8}+1 \cdot e^{-2 x} \ & x=\frac{\pi}{8}, y=\frac{1}{8}\left(2 \sin \frac{\pi}{4}-2 \cos \frac{\pi}{4}ight)+e^{-2\left(\frac{\pi}{8}ight)} \ & y=0+e^{-\frac{\pi}{4}}\end{aligned}$

If $y=y(x)$ is the solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=\sin (2 x),…

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