Having some trouble with inverse Laplace tranform

Question

How to solve this using inverse Laplace transform?

1/[($s$+1)($s$+2)$^4$]

I though of this solution which is $A$/($s$+1) + $B$/($s$+2) + $C$/($s$+2)$^2$ + $D$/($s$+2)$^3$ + $E$/($s$+2)$^4$

Then I can start by finding $A$ when $s$ = -1 and $E$ when $s$ = -2. I guess finding $E$ also means finding $B$, $C$ & $D$ using some sort of an equation?

Answer 1

HINT:

$$\mathcal{L}_{s}^{-1}\left[\frac{1}{(s+1)(s+2)^4}\right]_{(t)}=$$ $$\mathcal{L}_{s}^{-1}\left[\frac{1}{s+1}-\frac{s^3+7s^2+17s+15}{(s+2)^4}\right]_{(t)}=$$ $$\mathcal{L}_{s}^{-1}\left[\frac{1}{1+s}-\frac{1}{(s+2)^4}-\frac{1}{(s+2)^3}-\frac{1}{(s+2)^2}-\frac{1}{s+2}\right]_{(t)}=$$ $$\mathcal{L}_{s}^{-1}\left[\frac{1}{1+s}-\frac{1}{(s+2)^4}-\frac{1}{(s+2)^3}-\frac{1}{(s+2)^2}-\frac{1}{s+2}\right]_{(t)}=$$

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Now, use:

• $$\mathcal{L}_{s}^{-1}\left[\frac{1}{s+1}\right]_{(t)}=e^{-t}$$
• $$\mathcal{L}_{s}^{-1}\left[\frac{1}{s+2}\right]_{(t)}=e^{-2t}$$

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$$\mathcal{L}_{s}^{-1}\left[-\frac{1}{(s+2)^4}-\frac{1}{(s+2)^3}-\frac{1}{(s+2)^2}\right]_{(t)}+e^{-t}-e^{-2t}=$$

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Now, use:

• $$\mathcal{L}_{s}^{-1}\left[\frac{1}{(s+x)^n}\right]_{(t)}=\frac{t^{n-1}}{e^{tx}\Gamma[n]}$$

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$$e^{-t}-e^{-2t}-\frac{t^{4-1}}{e^{2t}\Gamma[4]}-\frac{t^{3-1}}{e^{2t}\Gamma[3]}-\frac{t^{2-1}}{e^{2t}\Gamma[2]}=$$ $$e^{-t}-e^{-2t}-\frac{t^3}{e^{2t}\Gamma[4]}-\frac{t^2}{e^{2t}\Gamma[3]}-\frac{t}{e^{2t}\Gamma[2]}=$$ $$e^{-t}-e^{-2t}-\frac{t^3}{6e^{2t}}-\frac{t^2}{2e^{2t}}-\frac{t}{e^{2t}}$$