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(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2$ and $Q = f(x)$.Case $1$: For

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$\frac{e^2 - 1}{2e^3}$

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(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2$ and $Q = f(x)$.Case $1$: For $x \in [0, 1]$,$f(x) = 1$.$\frac{dy}{dx} + 2y = 1$. The integrating factor is $IF = e^{\int 2 dx} = e^{2x}$.The solution is $y \cdot e^{2x} = \int 1 \cdot e^{2x} dx + C_1 = \frac{1}{2}e^{2x} + C_1$.So,$y(x) = \frac{1}{2} + C_1 e^{-2x}$.Given $y(0) = 0$,we have $0 = \frac{1}{2} + C_1 \Rightarrow C_1 = -\frac{1}{2}$.Thus,$y(x) = \frac{1}{2} - \frac{1}{2}e^{-2x}$ for $x \in [0, 1]$.At $x = 1$,$y(1) = \frac{1}{2} - \frac{1}{2}e^{-2} = \frac{e^2 - 1}{2e^2}$.Case $2$: For $x > 1$,$f(x) = 0$.$\frac{dy}{dx} + 2y = 0 \Rightarrow \frac{dy}{y} = -2 dx$.Integrating both sides,$\ln|y| = -2x + C_2 \Rightarrow y = C_3 e^{-2x}$.Using the continuity of $y(x)$ at $x = 1$,$y(1) = C_3 e^{-2} = \frac{e^2 - 1}{2e^2}$.$C_3 = \frac{e^2 - 1}{2e^2} \cdot e^2 = \frac{e^2 - 1}{2}$.So,$y(x) = \left(\frac{e^2 - 1}{2}\right) e^{-2x}$ for $x > 1$.For $x = \frac{3}{2}$,$y\left(\frac{3}{2}\right) = \left(\frac{e^2 - 1}{2}\right) e^{-2(\frac{3}{2})} = \left(\frac{e^2 - 1}{2}\right) e^{-3} = \frac{e^2 - 1}{2e^3}$.

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 2y = f(x)$,where $f(x) = \begin{cases} 1, & x \in [0, 1] \\ 0, & \text{otherwise} \end{cases}$. If $y(0) = 0$,then $y\left(\frac{3}{2}\right)$ is

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