Question

Find the inverse place transform

Answer 1

As you mentioned in your other question, I don't think this needs to be done with the convolution relation of the transform.

Note that you can rewrite the transform as

$\dfrac{7s+66-\frac{9(s+5)}{(s+5)^2+64}}{s^2+10s+74}$
$=\dfrac{7(s+5)}{(s+5)^2+49}+\dfrac{31}{(s+5)^2+49}-\dfrac{9(s+5)}{((s^2+49)(s^2+64)}$
$=7\dfrac{s+5}{(s+5)^2+7^2}+\dfrac{31}7\dfrac7{(s+5)^2+7^2}+\dfrac35\dfrac{s+5}{(s+5)^2+8^2}-\dfrac35\dfrac{s+5}{(s+5)^2+7^2}$

Now, recall the following transforms:

$\mathcal L_s\{\cos at\}=\dfrac s{s^2+a^2}$
$\mathcal L_s\{\sin at\}=\dfrac a{s^2+a^2}$
$\mathcal L_s\{e^{ct}f(t)\}=F(s-c)$

With $c=-5$, you can see that

$\mathcal L_s\{f(t)\}=7\dfrac{s+5}{(s+5)^2+7^2}+\dfrac{31}7\dfrac7{(s+5)^2+7^2}+\dfrac35\dfrac{s+5}{(s+5)^2+8^2}-\dfrac35\dfrac{s+5}{(s+5)^2+7^2}$
$\implies\mathcal L_s\{e^{-5t}f(t)\}=7\dfrac s{s^2+7^2}+\dfrac{31}7\dfrac7{s^2+7^2}+\dfrac35\dfrac s{s^2+8^2}-\dfrac35\dfrac s{s^2+7^2}$

Taking the inverse transform of both sides gives

$e^{-5t}f(t)=7\cos7t+\dfrac{31}7\sin7t+\dfrac35\cos8t-\dfrac35\cos7t$
$\implies f(t)=e^{5t}\left(\dfrac{32}5\cos7t+\dfrac35\cos8t+\dfrac{31}7\sin7t\right)$