Im trying to find the inverse laplace transformation of $\frac{(2s^2 + 9s + 5)}{((s^2 - 16s + 73)(s^2 + 25))}$. when $s \gt 8$.
Here is my work: $$\begin{align} &\frac{(2s^2 + 9s + 5)}{((s^2 - 16s + 73)(s^2 + 25))} \\ &= \frac{\left(\frac{1}{272}\right) * (9s + 580)}{(s^2 - 16s + 73) - \frac{(9s + 180)}{(s^2 + 25)}} \\ &= \left(\frac{1}{272}\right) * \frac{(9s + 580)}{((s - 8)^2 + 9) - \frac{(9s + 180)}{(s^2 + 25)}}, \text{completing the square} \\ &= \left(\frac{1}{272}\right) * \frac{(9(s - 8) + 508)}{((s - 8)^2 + 9) - \frac{(9s + 180)}{(s^2 + 25)}} \end{align}$$
Inverting now yields $$\left(\frac{1}{272}\right) * e^{(8t)} (9 \cos(3t) + \left(\frac{508}{3}\right) \sin(3t)) - (9 \cos(5t) + \left(\frac{180}{5}\right) \sin(5t))$$
Please let me know what is the correct answer because this is incorrect. Appreciate it.
We are are asked to find:
$\tag 1 \displaystyle \mathcal{L^{-1}} \frac{2s^2 + 9s + 5}{(s^2 - 16s + 73)(s^2 + 25)}, \text{when} ~~ s \gt 8.$
We start off by doing the partial fraction expansion.
$\tag 2 \displaystyle \frac{2s^2 + 9s + 5}{(s^2 - 16s + 73)(s^2 + 25)} = \frac{1}{272}\left(\frac{9 s+580}{s^2-16 s+73}-\frac{9 (s+20)}{s^2+25}\right)$
Now, we are looking for having specific forms in order to do the ILT, and for this problem, we are going to use the following four forms:
$\displaystyle \mathcal{L^{-1}} \frac{a}{s^2+a^2}, s \gt 0 \rightarrow \sin (at)$
$\displaystyle \mathcal{L^{-1}} \frac{s}{s^2+a^2}, s \gt 0 \rightarrow \cos (at)$
$\displaystyle \mathcal{L^{-1}} \frac{a}{(s-b)^2 + a^2}, s \gt b \rightarrow e^{bt}\sin (at)$
$\displaystyle \mathcal{L^{-1}} \frac{s-b}{(s-b)^2 + a^2}, s \gt b \rightarrow e^{bt}\cos (at)$
So, our task now is to take the partial fraction expansion and write so it looks like the four forms above. We get:
$ \tag 3 \displaystyle \frac{1}{272}\left( 9 \frac{s-8}{(s-8)^2+3^2} + \frac{652}{3} \frac{3}{(s-8)^2 + 3^2} -9 \frac{s}{s^2 + 5^2} -36 \frac{5}{s^2+5^2} \right)$
Do you understand where your error is now? Expand $(3)$ and make sure it is exactly what we started with in $(2)$.
Now, since these are in the exact form we need them to be in, we can use the four $\mathcal{L^{-1}}$ above to write:
$$\displaystyle \frac{1}{272}\left(9 e^{8t}\cos (3t) + \frac{652}{3} \left(e^{8t} \sin(3t)\right) -9 \cos(5t) -36 \sin(5t)\right)$$