ODE with time dependent jumps of coefficients

Question

Consider the following ODE \begin{align} x'(t) = -x(t)k(t)r \end{align} where $r > 0$ and \begin{align} k(t) = \begin{cases} 1 \quad t\in [0,\bar t)\\ \frac{3}{2} \quad t\in [\bar t,\bar t + 1)\\ 1\quad t\in [\bar t + 1, \infty) \end{cases} \end{align} We thus get \begin{align} x(t) = \begin{cases} x(0)e^{-rt} \quad &t\in [0,\bar t)\\ x(\bar t)e^{-r\frac{3}{2}(t-\bar t)} \quad &t\in [\bar t,\bar t + 1)\\ x(\bar t + 1)e^{-r(t-\bar t-1)}\quad &t\in [\bar t + 1, \infty) \end{cases} \end{align} Is it possible to get rid of the cases and give a general solution for $x(t)$ over $t \in [0,\infty)$?

Answer 1

Hint.

With the Laplace transform

$$ \mathcal{L}(x(t)k(t)) = \int_0^{\bar t}x(t) e^{-st}dt + \frac 32\int_{\bar t}^{\bar t +1}x(t) e^{-st}dt +\int_{\bar t + 1}^{\infty}x(t) e^{-st}dt $$

NOTE

$$ \int_a^b = \int_a^c - \int_b^c $$