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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P=2$ and $Q=\sin(2x)$.The Integratin

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$e^{-\pi/4}$

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P=2$ and $Q=\sin(2x)$.The Integrating Factor ($I$.$F$.) is given by $e^{\int P dx} = e^{\int 2 dx} = e^{2x}$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$.$y \cdot e^{2x} = \int e^{2x} \sin(2x) dx + C$.Using the formula $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2} (a \sin(bx) - b \cos(bx))$,we get:$y \cdot e^{2x} = \frac{e^{2x}}{2^2+2^2} (2 \sin(2x) - 2 \cos(2x)) + C = \frac{e^{2x}}{8} (2 \sin(2x) - 2 \cos(2x)) + C = \frac{e^{2x}}{4} (\sin(2x) - \cos(2x)) + C$.Given $y(0) = \frac{3}{4}$,substitute $x=0$ and $y=\frac{3}{4}$:$\frac{3}{4} \cdot e^0 = \frac{e^0}{4} (\sin(0) - \cos(0)) + C \Rightarrow \frac{3}{4} = \frac{1}{4} (0 - 1) + C \Rightarrow \frac{3}{4} = -\frac{1}{4} + C \Rightarrow C = 1$.So,$y \cdot e^{2x} = \frac{e^{2x}}{4} (\sin(2x) - \cos(2x)) + 1$,which simplifies to $y = \frac{1}{4} (\sin(2x) - \cos(2x)) + e^{-2x}$.Now,evaluate at $x = \frac{\pi}{8}$:$y\left(\frac{\pi}{8}\right) = \frac{1}{4} (\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4})) + e^{-2(\frac{\pi}{8})} = \frac{1}{4} (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) + e^{-\pi/4} = 0 + e^{-\pi/4} = e^{-\pi/4}$.

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

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