Inverse Laplace Transform, I need help

Question

What is the ILT of $H(s)=\frac{7(3s+1)}{(s-3)(s^2+10s-13)}$

Also, if you kindly want to help with this another inverse transform, I'd really appreciate it: $H(s)=\frac{6(s+2)}{s^3(s-5)}$

Thanks!

Answer 1

Here is a hint:

For the first: $H(s) = \frac{35}{13 ( s-3\ ) }-\frac{35 s+182}{13 ( {s}^{2}+10 s-13 ) }$, and you can write $\frac{35 s+182}{13 ( {s}^{2}+10 s-13 ) } = \frac{35 ( s+5 ) +7}{13 { ( s+5 ) }^{2}-494} $.

If you let $\hat{h}_3(s) = {35 \over 13} {s \over x^2-38}$, and $\hat{h}_4(s) = {7 \over 13} {1 \over x^2-38}$, then the corresponding time functions are $h_3(t) = {35 \over 13} \cosh(\sqrt{38}t)$, $h_4(t) = {35 \over 13} {1 \over \sqrt{38}} \sinh(\sqrt{38}t)$ (I am assuming $t>0$ here).

Note that the last term in $H(s)$ can be written as $-(\hat{h}_3(s+5)+\hat{h}_4(s+5))$, hence the corresponding time function is $-e^{-5t} (h_3(t)+h_4(t))$.

For the second: $H(s) = \frac{42}{125 ( s-5 ) } -\frac{42}{125 s}-\frac{42}{25 {s}^{2}}-\frac{12}{5 {s}^{3}} $.