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!!!
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 $$y(t) = \frac{1}{\sqrt{5}} \, e^{-2 t} \, \sinh(\sqrt{5} \, t).$$