Inverse Laplace transform calculation with steps

Question

I need to calculate with steps this inverse laplace transform

$\mathcal{L} [(3s+2)/(4s^2+12s+29)*e^{-2s}]$

Unfortunately,I cant factor the dividor and I tried different ways to spit it up (creating ($3s+2)^{2}$ in the dividor) but that didnt get me anywhere. I am really stuck so I would appreciate any tips

Answer 1

Start with $4s^2 + 12s + 29 = 4\left(s^2+3s+\frac{29}{4}\right) = 4\left(s^2+3s+\frac{9}{4} + 5\right) = 4\left(s+\frac{3}{2}\right)^2 + 20$. This polynomial has roots $-\frac{3}{2} \pm i\sqrt{5}$, so $4s^2 + 12s + 29 = 4\left(s - (-\frac{3}{2} + i\sqrt{5}) \right)\left(s - (-\frac{3}{2} - i\sqrt{5}) \right)$ and $$ \frac{3s+2}{s^2+3s+\frac{29}{4}} = \frac{3s+2}{\left(s - (-\frac{3}{2} + i\sqrt{5}) \right)\left(s - (-\frac{3}{2} - i\sqrt{5}) \right)} = \frac{A}{s - (-\frac{3}{2} - i\sqrt{5})} + \frac{B}{s - (-\frac{3}{2} + i\sqrt{5})} = \\ \\ = \frac{A\left(s - (-\frac{3}{2} - i\sqrt{5})\right) + B\left(s - (-\frac{3}{2} + i\sqrt{5})\right)}{s^2+3s+\frac{29}{4}} \Rightarrow \begin{cases} A + B = 3 \\ A(-\frac{3}{2} + i\sqrt{5}) + B (-\frac{3}{2} - i\sqrt{5}) = -2 \end{cases} $$ After doing some algebra, $A = \frac{3}{2} + i \frac{\sqrt{5}}{4}$ and $B= \frac{3}{2} - i \frac{\sqrt{5}}{4}$.

Now, using $\mathcal{L}^{-1} \left(\frac{1}{s - (-\frac{3}{2} \pm i\sqrt{5})}\right) = \exp\left\{ (-\frac{3}{2} \pm i\sqrt{5}) t\right\} = e^{-\frac{3}{2}t}\left( \cos(\sqrt{5}t) \pm i \sin\sqrt{5}t)\right)$, you can get $$ \mathcal{L}^{-1} \left(\frac{3s+2}{4s^2 + 12s + 29}\right) = e^{-\frac{3}{2}t}\cdot \frac{6\sqrt{5} \cos (\sqrt{5}t) - 5 \sin(\sqrt{5}t)}{8\sqrt{5}} \cdot\theta(t) $$ Finally, because of $e^{-2s}$, apply shift theorem and change argument from $t$ to $t-2$.

Answer 2

Mathematica says:

$$\frac{e^{-\frac{3}{2} (x-2)} \theta (x-2) \sin \left(\sqrt{5} (x-2)\right)}{2 \sqrt{5}}+\frac{3 e^{-\frac{3}{2} (x-2)} \theta (x-2) \left(2 \sqrt{5} \cos \left(\sqrt{5} (x-2)\right)-3 \sin \left(\sqrt{5} (x-2)\right)\right)}{8 \sqrt{5}}$$