Question

I have an inverse laplace transform and I have no clue how to solve it with the allowed methods (no convolution, no heaviside):

Answer 1

\[L^{-1} \left\{\vphantom{} \frac{2s-1}{s^2(s+1)^3} \right\}\]

Answer 2

My first instinct was to factor out the one and add to the whole thing as such: (For some reason the eqn editor isn't working, I'm waiting on it to load)

Answer 3

*Factor out the two

Answer 4

I would suggest breaking up the argument into partial fractions first.

Answer 5

That sounds like a better idea and for some reason I totally overlooked it, one sec.

Answer 6

Eghh, I'll just write it by hand rather than using the editor, one minute

Answer 7

Got this so far, moving forward from here momentarily: http://i.imgur.com/48mggGC.png

Answer 8

http://www.wolframalpha.com/input/?i=partial+fraction+%282s-1%29%2F%28s%5E2%28s%2B1%29%5E3%29

Answer 9

Recall exponential shift formula and you may use below for the terms :\[\mathcal{L}^{-1} \dfrac{1}{(s-a)^n} = \dfrac{e^{at}t^{n-1}}{(n-1)!}\]

Answer 10

Yep, though you can reduce it to three eqs with three unknowns if you set \(s=0\) and \(s=-1\).

Answer 11

Am I doing this right? Is this right so far? I'm running into problems which suggest otherwise but: http://i.imgur.com/48mggGC.png

Answer 12

Yes, your set up is correct.

Answer 13

\[\begin{align*}
\frac{2s-1}{s^2(s+1)^3}&=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s+1}+\frac{D}{(s+1)^2}+\frac{E}{(s+1)^3}\\\\
2s-1&=As(s+1)^3+B(s+1)^3+Cs^2(s+1)^2+Ds^2(s+1)+Es^2\\\\
&=A(s^4+3s^3+3s^2+s)+B(s^3+3s^2+3s+1)\\&\quad +C(s^4+2s^3+s^2)+D(s^3+s^2)+Es^2\\\\
&=(A+C)s^4+(3A+B+2C+D)s^3\\&\quad+(3A+3B+C+D+E)s^2+(A+3B)s+B
\end{align*}\]
which gives
\[\begin{cases}
\begin{align*}
A+C&=0\\
3A+B+2C+D&=0\\
3A+3B+C+D+E&=0\\
A+3B&=2\\
B&=-1
\end{align*}
\end{cases}\]
You can find each constant sequentially starting with \(B\).

Answer 14

Wait, I'm getting it: http://i.imgur.com/JWwJJK0.png

Answer 15

(Just now saw your reply, for some reason I didn't get the notification, sorry about that.)

Answer 16

No worries. Yep, those are the right coefficients.

Answer 17

http://i.imgur.com/Q3idWcg.png

Answer 18

Got it. Thank you.