Differential equation with (bilateral) Laplace transformation So I have this:
$y'(t) - \int_0^tf(t-x)*y(x)dx=f(t), t\geq0$
where $y(0)=1$ and $f(t)=e^{-3t}$
I try solving it by multiplying both sides with the $H(t)$ (<-- Heaviside) and performing the Laplace transformation. But after some cleaning up I get $\mathcal{L}(y*H)=\frac{s+4}{s^2+3s-1}$ which seems wrong because the denominater has the roots $s= \frac{-3}2 \pm \frac{\sqrt13}2$.
Have I done something wrong along the way? Please help!!!
 A: In the general case consider
$$y'(t) - \int_{0}^{t} e^{-a (t-x)} \, y(x) \, dx = e^{-at} \hspace{5mm} t\geq 0.$$
Since $t\geq 0$ consider applying the one sided Laplace transform for which
\begin{align}
s \, \overline{y} - y(0) - \frac{1}{s+a} \, \overline{y} &= \frac{1}{s+a} \\
\frac{s^2 + as -1}{s+a} \, \overline{y} &= y(0) + \frac{1}{s+a} \\
\overline{y} &= \frac{1}{s^2 + as -1} + y(0) \, \frac{s+a}{s^2 + as -1}  
\end{align} 
Now, $s^2 + as -1 = (s+\alpha)(s + \beta)$ with $2 \alpha = a + \theta$, $2 \beta = a - \theta$, $\theta = \sqrt{a^2 + 4}$ such that
\begin{align}
\overline{y} &= \frac{1}{\alpha - \beta} \, \left[ ((\alpha - a) \, y(0) -1) \, \frac{1}{s+\alpha} - ((\beta - a) \, y(0) -1) \, \frac{1}{s+\beta} \right] \\
&= \frac{1}{\alpha - \beta} \, \left[ (\alpha \, y(0) +1) \, \frac{1}{s + \beta} - (\beta \, y(0) +1) \, \frac{1}{s+\alpha}  \right]
\end{align}
and finally
\begin{align}
y(t) = \frac{e^{-a t/2}}{\sqrt{a^2+4}} \, \left[ (a \, y(0) + 2) \, \sinh\left(\frac{\sqrt{a^2 + 4} \, t}{2} \right) + y(0) \, \sqrt{a^2 + 4} \, \cosh\left(\frac{\sqrt{a^2 + 4} \, t}{2} \right) \right].
\end{align}
For this particular problem $a=3$ and $y(0) = 1$ and yields
$$y(t) = e^{-3 t/2} \, \left[ \frac{5}{\sqrt{13}} \, \sinh\left(\frac{\sqrt{13} \, t}{2}\right) + \cosh\left(\frac{\sqrt{13} \, t}{2}\right) \right].$$
Suppose the conditions were $a=4$ and $y(0) = 0$ then
$$y(t) = \frac{1}{\sqrt{5}} \,  e^{-2 t} \, \sinh(\sqrt{5} \, t).$$
