Question

Inverse Laplace transform:

\[\mathcal{L}^{-1} \left( \frac{5}{s(s^2+4s+5)} \right) \]

Answer 1

Since we have that grouping of two factors in the bottom, I think a good first step is to break it up by partial fractions and then use the fact that:
\(\mathcal{L}^{-1} \left[F + G\right] = \mathcal{L}^{-1} \left[F\right] + \mathcal{L}^{-1} \left]G \right]\)

Answer 2

\[ 5*\mathcal{L}^{-1} \left( \frac{1/5}{s}-\frac{s}{s^2+4s+5}-\frac{4}{s^2+4s+5} \right) \]

Answer 3

oops \[ 5*\mathcal{L}^{-1} \left( \frac{1/5}{s}-\frac{-s/5}{s^2+4s+5}-\frac{4/5}{s^2+4s+5} \right) \]

Answer 4

\[ \mathcal{L}^{-1} \left( \frac{1}{s} \right) - \mathcal{L}^{-1} \left( \frac{s}{(s+2)^2+1} \right)- 4*\mathcal{L}^{-1} \left( \frac{1}{(s+2)^2+1} \right) \]

Answer 5

\[ 1 - e^{-2t}cos(t)- 4e^{-2t}sin(t) \] ??

Answer 6

Hmm... I think the second term is not correct
\( \displaystyle \mathcal{L}^{-1} \left[ \frac{s-a}{(s-a)^2+ b^2} \right] = e^{at} \cos bt \)
We need b=1 and a=-2, but the numerator is only s and not s+2.
Instead, we could pull out the extra +2 we need from that last term. The last term won't be affected besides a smaller coefficient. (2 instead of 4).

Answer 7

\( \displaystyle -\mathcal{L}^{-1} \left[\frac{s+4}{(s + 2)^2 + 1} \right] = -\mathcal{L}^{-1} \left[ \frac{s+2}{(s+2)^2 + 1} + \frac{2}{(s + 2)^2 + 1} \right] \)

Answer 8

oh that's interesting. I thought I could use the simpler cosine transform , \[\mathcal{L}[cos(bt)]=\frac{s}{s^2+b^2}\] and use the shift theorem.

Answer 9

oh because there's an "s" on top as well.. -_-

Answer 10

You could, although we have to be careful b/c the shift changes all s \(\to\) s-a, so that'd be both the s in the numerator and denominaotr becoming \(s-a\)
But that will also work. :)

Answer 11

I get it now :)

Answer 12

lol sorry for making you write all that I just noticed it now

Answer 13

Oh, sorry, I think my internet just played a trick on me. lol
I didnt see that last msg. :)

It's fine, and glad you got it now! :D

Answer 14

thank you soo much :)!!

Answer 15

You're welcome!