To solve this convolution problem, we need to find the convolution of two signals given as:

1. $x(t) = e^{-3t} u(t)$
2. $h(t) = e^{-2t} u(t - 2)$

where $u(t)$ is the unit step function.

Step 1: Set Up the Convolution Integral

The convolution of two signals $x(t)$ and $h(t)$ is defined as: $y(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau) h(t - \tau) \, d\tau.$

Given $x(t) = e^{-3t} u(t)$, this signal is zero for $t < 0$. So: $x(\tau) = e^{-3\tau} u(\tau).$

The second signal $h(t) = e^{-2t} u(t - 2)$ also involves a delay by 2, which shifts the function to start at $t = 2$. Thus: $h(t - \tau) = e^{-2(t - \tau)} u(t - \tau - 2).$

Step 2: Define the Limits of Integration

Since both functions are causal (involving unit steps), the integral limits will depend on the time $t$ and the delay introduced by the unit step function in $h(t)$. This will lead to piecewise definitions based on the ranges of $t$ where the product $x(\tau) h(t - \tau)$ is non-zero.

Step 3: Solve the Convolution Integral

To proceed further, let's apply the convolution in two cases based on time intervals.

1. For $t < 2$: Here, $h(t - \tau) = 0$ for all $\tau$ since $t - \tau - 2 < 0$. Hence, $y(t) = 0$ in this interval.

2. For $t \ge 2$: Both $x(\tau)$ and $h(t - \tau)$ contribute to the integral: $y(t) = \int_0^t e^{-3\tau} e^{-2(t - \tau)} u(\tau) u(t - \tau - 2) \, d\tau.$ The step functions imply that $\tau$ ranges from $2$ to $t$: $y(t) = \int_2^t e^{-3\tau} e^{-2(t - \tau)} \, d\tau.$ Simplifying inside the integral: $y(t) = \int_2^t e^{-3\tau} e^{-2t + 2\tau} \, d\tau = e^{-2t} \int_2^t e^{-\tau} \, d\tau.$ Integrate $e^{-\tau}$ from $2$ to $t$: $y(t) = e^{-2t} \left[ -e^{-\tau} \right]_2^t = e^{-2t} \left( -e^{-t} + e^{-2} \right).$ Simplifying further: $y(t) = e^{-2t} \cdot e^{-2} - e^{-3t} = e^{-2(t+1)} - e^{-3t}.$

So the final solution is: