1. We need to solve $\frac{dy}{dx}+2y=f(x), \qquad y(0)=0,$ where $f(x)=\begin{cases}1, & x\in[0,1],\\ 0, & \text{otherwise}.\end{cases}$

Since $f(x)$ changes at $x=1$, solve the differential equation piecewise.

2. For $0\le x\le 1$, we have $\frac{dy}{dx}+2y=1.$

This is a linear differential equation. Using integrating factor: $\text{I.F.}=e^{\int 2dx}=e^{2x}.$

So, $e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{2x}$ which gives $\frac{d}{dx}\left(ye^{2x}\right)=e^{2x}.$

Integrating, $ye^{2x}=\int e^{2x}dx=\frac{1}{2}e^{2x}+C.$

Hence, $y=\frac{1}{2}+Ce^{-2x}.$

Using $y(0)=0$, $0=\frac{1}{2}+C \implies C=-\frac{1}{2}.$

Therefore, for $0\le x\le 1$, $y(x)=\frac{1}{2}\left(1-e^{-2x}\right).$

So at $x=1$, $y(1)=\frac{1}{2}\left(1-e^{-2}\right).$

3. For $x>1$, we have $f(x)=0$, so the equation becomes $\frac{dy}{dx}+2y=0.$

Its general solution is $y=Ce^{-2x}.$

Now use continuity at $x=1$: $y(1)=Ce^{-2}=\frac{1}{2}(1-e^{-2}).$

Thus, $C=\frac{1}{2}(1-e^{-2})e^2=\frac{1}{2}(e^2-1).$

Hence for $x>1$, $y(x)=\frac{e^2-1}{2}e^{-2x}.$

4. Now evaluate at $x=\frac{3}{2}$: