I'm actually on the way to solve a little bit complicated differential-equation. Therefore I used the Laplace transformation. I've already solved it but I am actually not sure, whether my solution (especially the convolution integral) is correct. That's why, I want to show my way of "back-transformation".
Given is the following equation (in Laplace; $y_1$ and $p_i$ and $q_i$ are known constants; furthermore $y(t) = y_1\cdot t$):
$$ \mathcal{L}\{x(t)\}(s)=\mathcal{L}\{y(t)\}(s)\cdot\frac{q_0+q_1\cdot s+q_2\cdot s^2+q_3\cdot s^3+q_4\cdot s^4}{p_0+p_1\cdot s+p_2\cdot s^2+p_3\cdot s^3+p_4\cdot s^4} + (-y_1)\cdot\frac{q_2+q_3\cdot s+q_4\cdot s^2}{p_0+p_1\cdot s+p_2\cdot s^2+p_3\cdot s^3+p_4\cdot s^4}$$
Now, I expand the fractions to partial fractions. This gives:
$$ \mathcal{L}\{x(t)\}(s) = \mathcal{L}\{y(t)\}(s)\cdot\biggl[a_0+\frac{a_1}{s-b_1}+\frac{a_2}{s-b_2}+\frac{a_3}{s-b_3}+\frac{a_4}{s-b_4}\biggr] + (-y_1)\cdot \biggl[\frac{d_1}{s-f_1}+\frac{d_2}{s-f_2}+\frac{d_3}{s-f_3}+\frac{d_4}{s-f_4}\biggr]$$
Now I do $\mathcal{L}^{-1}\{\dots\}$ what gives ($\star$ marks a convolution):
$$x(t) = \delta\cdot a_0\star y(t) + a_1\cdot e^{b_1\cdot t}\star y(t)+ a_2\cdot e^{b_2\cdot t}\star y(t)+ a_3\cdot e^{b_3\cdot t}\star y(t)+ a_4\cdot e^{b_4\cdot t}\star y(t) + \bigl[d_1\cdot e^{f_1\cdot t}+d_2\cdot e^{f_2\cdot t}+d_3\cdot e^{f_3\cdot t}+d_4\cdot e^{f_4\cdot t}\bigr]\cdot(-y_1)$$
And if I now want to get the solution for any point $t_i$, I have to integrate $\int_0^{t_1}x(t)\text{d}t$ which leads me to:
$$x(t_1) = a_0\cdot y(t_1) + \int_0^{t_1}a_1\cdot e^{b_1\cdot\tau}\cdot y(t_1 - \tau)\text{d}\tau + \cdots + (-y_1)\cdot\int_0^{t_1}\bigl[d_1\cdot e^{f_1\cdot t} + \cdots \bigr]\text{d} t$$
When I replace $y(t_1)$ by $y(t_1) = y_1\cdot t_1$ as well as $y(t_1-\tau)$ by $y(t_1-\tau) = y_1\cdot (t_1-\tau)$ while performing the integral, I finally get
$$x(t_1) = a_0\cdot y_1\cdot t_1 + y_1\cdot\biggl[ y_1\cdot \frac{a_1}{b_1^2}\cdot\bigl(-b_1\cdot t_1 + e^{b_1\cdot t_1} - 1\bigr)+ \cdots\biggr] + (-y_1)\cdot\biggl[\frac{d_1}{f_1}\cdot\bigl(e^{f_1\cdot t_1}-1\bigr) + \cdots\biggr]$$
($\cdots$ marks the other parts which have the same form as the one denoted.)
So ... is my way of doing the Laplace backtransformation correct? Or did I something wrong while solving?
Thanks a lot in advance!