Calculating an inverse Laplace transform

by George Dimitriou   Last Updated January 14, 2018 13:20 PM

I have a pain for a problem... need to find the inverse Laplace transform of: $$\frac{s^2+s+1}{(s+1)(s+2)^2(s^2+4s+9)}$$ Now, I get that we have to expand the partial fraction, I managed to get to this: $$\frac{A}{(s+1)}+\frac{B}{(s+2)}+\frac{C}{(s+2)^2}+\frac{Ds+E}{(s^2+4s+9)}$$ and after doing some expanding and simultaneous equations, I got: $A=\frac{5}{6}$; $B=\frac{64}{5}$; $C=\frac{-67}{5}$; $D=\frac{-289}{30}$; $C=\frac{-347}{10}$ So now, I have $$\mathcal{L}[f(t)]=\frac{5}{6}\frac{1}{(s+1)}+\frac{64}{5}\frac{1}{(s+2)}-\frac{67}{5}\frac{1}{(s+2)^2}-\frac{289}{30}\frac{s}{(s^2+4s+9)}-\frac{347}{10}\frac{1}{(s^2+4s+9)}$$ from here I use the table of inverses in the textbook to get: $F(s) = \frac{5}{6}e^{-t}+\frac{64}{5}e^{-2t}-\frac{67}{5}te^{-2t}+\dots$ and then I'm stuck on that last polynomial... I have looked and looked but found nothing... and I'm not even sure I did the first part correctly.

Please help!

Answers 1

Hint. Note that $$s^2+4s+9=(s+2)^2+(\sqrt{5})^2$$ and take a look at "exponentially decaying sine/cosine wave" in this Table.

P.S. Check your partial fraction decomposition. It should be $$\frac{1/6}{(s+1)}-\frac{3/5}{(s+2)^2}-\frac{s/6-1/10}{(s^2+4s+9)} =\frac{1/6}{(s+1)}-\frac{3/5}{(s+2)^2}-\frac{(s+2)/6-2/6-1/10}{(s+2)^2+(\sqrt{5})^2}.$$

Robert Z
Robert Z
January 14, 2018 12:47 PM

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