Consider the following differential equation:
$$y''+2\omega_0 y'+\omega_0^2 y=A_0I(t)$$
where $I$ is the indicator function for the interval $(-\tau,\tau)\subset\mathbb{R}$. Suppose we seek solutions which are continuous $L^1(\mathbb{R})$ functions. Then we can solve this via Fourier transform and then invert the transformation via contour integration. I did arrive at a solution, but not the one that I expected. The solution that I obtained has a discontinuity at $t=\tau$. How can this be possible despite the prior assumption that $y$ be a continuous $L^1(\mathbb{R})$ function? For reference, the solution that I obtained is shown below:
$$y(t)=\begin{cases} \frac{A_0}{\omega_0}\big(t+\tau+\frac{1}{\omega_0}\big)\left[e^{-\omega_0(t+\tau)}-e^{-\omega_0(t-\tau)}\right]&,t\geq\tau \\ \frac{A_0}{\omega_0^2}-\frac{A_0}{\omega_0}\left(t+\tau+\frac{1}{\omega_0}\right)e^{-\omega_0(t+\tau)}&,|t|<\tau\\ 0&,t\leq-\tau \end{cases}$$