I'd like to calculate the solution $y : [0, \infty) \longrightarrow \mathbb{R}$ of the IVP
$$\ddot{y}(t) -4\dot{y}(t)+4y(t) = t\mathrm e^{2t} \chi_{[0,1]}(t), \quad y(0) = 1,\ \dot{y}(0) = 0,$$
where $\chi_{[0,1]}(t)$ is the indicator function on the interval $[0, 1].$
What I was able to do is to solve the homogenous equation $\ddot{y}(t) -4\dot{y}(t)+4y(t) = 0$. After this I wanted to find a particular solution of the inhomogeneous equation. My idea was, that I use the ansatz $y_{\mathrm p}(t) = f(t) \mathrm e^{2t}$, where $f$ is $\mathcal{C}^2$, plug this into the ODE and multiplicate the resulting solution with the characteristic function. That lead me to the particular solution $y_{\mathrm p}(t) = \frac{t^3}{6} \mathrm e^{2t} \chi_{[0,1]}(t)$, but I was told that this is wrong. What is my mistake and how can one solve this problem?
Best Answer
You can't have $y_p\in C^2$. At $t=1$, $\ddot y$ would have to abruptly change.
So the way to do this is to compute the solution $t\in[0,1)$ and $t\in[1,\infty)$ separately, and imposing a $C^1$ transition at $t=1$.
So for $t\in[0,1]$, the DE is $$ \ddot{y}-4\dot y+4y = te^{2t}, y(0)=1, \dot{y}(0)=0 $$ giving $y=\frac16\exp(2t)(t^3-12t+6)$. At $t=1$ this gives $y(1)=-5e^2/6$, $\dot y(1)=-19e^2/6$, so the solution to $\ddot{y}-4\dot y+4y=0$ on $[1,\infty)$ is $$ y=\frac16\exp(2t)(4-9t) $$