Inverse Laplace Transform partial fraction $\frac{8}{(S+2)^2 (S+2)} + \frac{S}{(S+2)^2} + \frac{5}{(S+2)^2}$

Question

Can anybody help me with the answer of this question?

Find the inverse Laplace transform of $$\frac{8}{(S+2)^2 (S+2)} + \frac{S}{(S+2)^2} + \frac{5}{(S+2)^2}$$

Answer 1

For the last term, $\displaystyle \frac{5}{(S+2)^2}\to 5te^{-2t}$.

For the second to last term, $\displaystyle \frac{S}{(S+2)^2}=\frac{1}{(S+2)}-\frac{2}{(S+2)^2}\to e^{-2t}-2te^{-2t}$.

For the first term, $\displaystyle \frac{8}{(S+2)^3}\to 4t^2e^{-2t}$.

Add these three terms up to get your answer, $e^{-2t}(4t^2+3t+1)$.

Answer 2

$$(1): \frac{8}{(s+2)^2+(s+2)}$$ Now we can say that: $$(s+2)^2+(s+2)=(s^2+4s+4)+(s+2)=s^2+5s+6=(s+3)(s+2)$$ Which means we can rewrite then use partial fractions: $$(1)\Rightarrow\frac{8}{(s+3)(s+2)}=\frac{8}{s+2}-\frac{8}{s+3}$$

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$$(2):\frac{s}{(s+2)^2}=\frac{1}{s+2}-\frac{2}{(s+2)^2}$$

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$$(3):\frac{5}{(s+2)^2}$$

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Now we can say that: $$\frac{8}{(s+2)^2+(s+2)}+\frac{s}{(s+2)^2}+\frac{5}{(s+2)^2}=\frac{8}{s+2}-\frac{8}{s+3}+\frac{1}{s+2}-\frac{2}{(s+2)^2}+\frac{5}{(s+2)^2}$$ $$=\frac{9}{s+2}-\frac{8}{s+3}+\frac{3}{(s+2)^2}$$

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$$\mathfrak{L}^{-1}\left(\frac{9}{s+2}-\frac{8}{s+3}+\frac{3}{(s+2)^2}\right)=9e^{-2t}-8e^{-3t}+3te^{-2t}=3\left(3+t\right)e^{-2t}-8e^{-3t}$$