OpenStudy (anonymous):

Find the inverse transform using partial fractions and the shifting theorem: F(s) = 1/[(s+3)^2(s-2)^2]

OpenStudy (anonymous):

Laplace transform*

OpenStudy (anonymous):

Well, quick, concise, right to the point -- these are the Laplace transform problems I like.$$\frac1{(s+3)^2(s-2)^2}=\frac{A}{s+3}+\frac{B}{(s+3)^2}+\frac{C}{s-2}+\frac{D}{(s-2)^2}$$

OpenStudy (anonymous):

hold on I hate typing math. I'm doing it on a separate sheet of paper.

OpenStudy (anonymous):

I'm counting on you man. Anything you can help me with would be appreciated.

OpenStudy (anonymous):

\[\frac{1}{(s+3)^2(s-2)^2}=\frac{2}{125(s+3)}+\frac{1}{25(s+3)^2}-\frac{2}{125(s-2)}+\frac{1}{25(s-2)^2}\]

OpenStudy (anonymous):

@SithsAndGiggles well yes all of us can use WolframAlpha do to it for us...

OpenStudy (anonymous):

Yep, I prefer leaving the work to the askers.

OpenStudy (anonymous):

I think the reason it hasn't been working for me for the past hour is that I'm collecting my coefficients incorrectly... the system I've ended up with I have correctly solved but they aren't the coefficients.

OpenStudy (anonymous):

\(1=A(s+3)(s-2)^2+B(s-2)^2+C(s+3)^2(s-2)+D(s+3)^2\) Letting \(s=-3\), you'll find that \(B=\dfrac{1}{25}\), and letting \(s=2\) gives you \(D=\dfrac{1}{25}\) as well. Next, I'd try letting \(s=0\): \(1=12A+\dfrac{4}{25}-18C+\dfrac{9}{25}\\ \color{red}{\dfrac{2}{25}=2A-3C}\) Letting \(s=1\): \(1=4A+\dfrac{1}{25}-16C+\dfrac{16}{25}\\ \color{blue}{\dfrac{2}{25}=A-4C}\) Which tells you \(A-4C=2A-3C~\Rightarrow~ A=-C\). Substituting into the red equation: \(\dfrac{2}{25}=5A\\ A=\dfrac{2}{125}~\Rightarrow~C=-\dfrac{2}{125}\) And thus the result I got from WA.

OpenStudy (anonymous):

@SithsAndGiggles for some reason I convinced myself that wouldn't work in this case... ah, I did eventually finish it using Heaviside's cover-up method.

OpenStudy (anonymous):

From here, it looks like the inverse Laplace transform has two purely exponential terms and two exponential/trig terms. @oldrin.bataku, I usually use the method that involves matching up coefficients of like-powered terms, but I don't really like multiplying polynomials.