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:

We start from the first-order linear differential equation

$$\frac{dy}{dx}+2y=f(x),\qquad\text{with}\qquad f(x)=\begin{cases}1,& 0\le x\le 1\\[4pt]0,& x\gt 1\end{cases}$$

and the initial condition $$y(0)=0.$$

For such a linear equation, the integrating-factor method is standard. The integrating factor is obtained from the coefficient of $$y$$:

$$\mu(x)=e^{\int 2\,dx}=e^{2x}.$$

Multiplying the whole differential equation by this integrating factor gives

$$e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{2x}f(x).$$

The left side is now the derivative of the product $$e^{2x}y$$, because we have the identity $$\displaystyle\frac{d}{dx}\bigl(e^{2x}y\bigr)=e^{2x}\frac{dy}{dx}+2e^{2x}y$$. Thus

$$\frac{d}{dx}\bigl(e^{2x}y\bigr)=e^{2x}f(x).$$

We integrate both sides from $$0$$ to $$x$$, using the initial condition $$y(0)=0$$ (so $$e^{2\cdot0}y(0)=0$$):

$$e^{2x}y(x)-e^{0}y(0)=\int_{0}^{x}e^{2t}f(t)\,dt.$$

Because $$y(0)=0$$, the term $$e^{0}y(0)$$ vanishes, leaving

$$e^{2x}y(x)=\int_{0}^{x}e^{2t}f(t)\,dt.$$

Solving for $$y(x)$$ gives the general expression

$$y(x)=e^{-2x}\int_{0}^{x}e^{2t}f(t)\,dt.$$

Now we evaluate the integral depending on the position of $$x$$ relative to $$1$$, since $$f(t)$$ changes there.

Case 1: $$0\le x\le 1$$. On the entire interval of integration we have $$f(t)=1$$, so

$$\int_{0}^{x}e^{2t}f(t)\,dt=\int_{0}^{x}e^{2t}\cdot1\,dt =\Bigl[\tfrac12 e^{2t}\Bigr]_{0}^{x} =\tfrac12\bigl(e^{2x}-1\bigr).$$

Hence for $$0\le x\le 1$$,

$$y(x)=e^{-2x}\cdot\tfrac12\bigl(e^{2x}-1\bigr) =\tfrac12\bigl(1-e^{-2x}\bigr).$$

Case 2: $$x\gt 1$$. Now $$f(t)=1$$ only up to $$t=1$$ and becomes $$0$$ afterwards. Therefore

$$\int_{0}^{x}e^{2t}f(t)\,dt =\int_{0}^{1}e^{2t}\cdot1\,dt+\int_{1}^{x}e^{2t}\cdot0\,dt =\int_{0}^{1}e^{2t}\,dt.$$

The second integral is zero because $$f(t)=0$$ for $$t\gt 1$$. Evaluating the first integral:

$$\int_{0}^{1}e^{2t}\,dt=\Bigl[\tfrac12 e^{2t}\Bigr]_{0}^{1} =\tfrac12\bigl(e^{2}-1\bigr).$$

Thus for every $$x\ge1$$ we have

$$y(x)=e^{-2x}\cdot\tfrac12\bigl(e^{2}-1\bigr) =\frac{e^{2}-1}{2}\,e^{-2x}.$$

We are asked to find $$y\!\left(\tfrac32\right)$$. Since $$\tfrac32=1.5\gt 1$$, we must use the second ( $$x\gt 1$$ ) formula:

$$y\!\left(\tfrac32\right)=\frac{e^{2}-1}{2}\,e^{-2\left(\tfrac32\right)} =\frac{e^{2}-1}{2}\,e^{-3} =\frac{e^{2}-1}{2e^{3}}.$$

Comparing with the given options, this matches Option A.

Hence, the correct answer is Option A.