Question:

Step 1: Identify the type of differential equation and find the integrating factor. The given differential equation is $\dfrac{dy}{dx}+2y=\sin(2x)$. This is a first-order linear differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = 2$ and $Q(x) = \sin(2x)$. The integrating factor (IF) is given by $e^{\int P(x) dx}$. $$ \text{IF} = e^{\int 2 dx} = e^{2x} $$ Step 2: Multiply the differential equation by the integrating factor and integrate. Multiply both sides of the differential equation by the integrating factor $e^{2x}$: $$ e^{2x}\dfrac{dy}{dx} + 2e^{2x}y = e^{2x}\sin(2x) $$ The left side of this equation is the derivative of the product $y \cdot e^{2x}$: $$ \dfrac{d}{dx}(y e^{2x}) = e^{2x}\sin(2x) $$ Now, integrate both sides with respect to $x$: $$ y e^{2x} = \int e^{2x}\sin(2x) dx $$ Step 3: Evaluate the integral $\int e^{2x}\sin(2x) dx$. We use the standard formula for integrals of the form $\int e^{ax}\sin(bx) dx = \dfrac{e^{ax}}{a^2+b^2}(a\sin(bx) - b\cos(bx)) + C$. Here, $a=2$ and $b=2$. $$ \int e^{2x}\sin(2x) dx = \dfrac{e^{2x}}{2^2+2^2}(2\sin(2x) - 2\cos(2x)) + C $$ $$ = \dfrac{e^{2x}}{8}(2\sin(2x) - 2\cos(2x)) + C $$ $$ = \dfrac{e^{2x}}{4}(\sin(2x) - \cos(2x)) + C $$ Step 4: Substitute the integral back into the solution and find the general solution for $y(x)$. Substitute the result of the integral back into the equation from Step 2: $$ y e^{2x} = \dfrac{e^{2x}}{4}(\sin(2x) - \cos(2x)) + C $$ Divide by $e^{2x}$ to solve for $y(x)$: $$ y(x) = \dfrac{1}{4}(\sin(2x) - \cos(2x)) + C e^{-2x} $$ Step 5: Use the initial condition $y(0) = \dfrac{3}{4}$ to find the value of the constant $C$. Substitute $x=0$ and $y=\dfrac{3}{4}$ into the general solution: $$ \dfrac{3}{4} = \dfrac{1}{4}(\sin(2 \cdot 0) - \cos(2 \cdot 0)) + C e^{-2 \cdot 0} $$ $$ \dfrac{3}{4} = \dfrac{1}{4}(\sin(0) - \cos(0)) + C e^{0} $$ $$ \dfrac{3}{4} = \dfrac{1}{4}(0 - 1) + C \cdot 1 $$ $$ \dfrac{3}{4} = -\dfrac{1}{4} + C $$ Solve for $C$: $$ C = \dfrac{3}{4} + \dfrac{1}{4} = \dfrac{4}{4} = 1 $$ Step 6: Write the particular solution $y(x)$ and evaluate $y\left(\dfrac{\pi}{8}\right)$. Substitute $C=1$ back into the general solution to get the particular solution: $$ y(x) = \dfrac{1}{4}(\sin(2x) - \cos(2x)) + e^{-2x} $$ Now, evaluate $y\left(\dfrac{\pi}{8}\right)$: $$ y\left(\dfrac{\pi}{8}\right) = \dfrac{1}{4}\left(\sin\left(2 \cdot \dfrac{\pi}{8}\right) - \cos\left(2 \cdot \dfrac{\pi}{8}\right)\right) + e^{-2 \cdot \dfrac{\pi}{8}} $$ $$ y\left(\dfrac{\pi}{8}\right) = \dfrac{1}{4}\left(\sin\left(\dfrac{\pi}{4}\right) - \cos\left(\dfrac{\pi}{4}\right)\right) + e^{-\dfrac{\pi}{4}} $$ We know that $\sin\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$ and $\cos\left(\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$. $$ y\left(\dfrac{\pi}{8}\right) = \dfrac{1}{4}\left(\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\right) + e^{-\dfrac{\pi}{4}} $$ $$ y\left(\dfrac{\pi}{8}\right) = \dfrac{1}{4}(0) + e^{-\dfrac{\pi}{4}} $$ $$ y\left(\dfrac{\pi}{8}\right) = e^{-\dfrac{\pi}{4}} $$ The final answer is $\boxed{e^{-\pi/4}}$, which corresponds to Option 3.