Question

Can someone help me find the Inverse Laplace Transform of

L^-1{(e^-2s)/(s^2+s-2)}

Answer 1

I'll try, give me a minute to review laplace transforms, been a couple of semesters since diffeq. I can tell you right off the bat though, that you will need a step function, and will need to split the bottom up into partial fractions.

Answer 2

hmm Okay... the e term on the top is the part that confuses me.

Answer 3

See it like this:
\[e^{-2s}\frac{1}{s^{2}+s-2}\]

Answer 4

Ok, give me a minute, I am going to attempt to solve it to make sure I understand it correctly. Thomas is correct, that is how you have to look at it. Then you have a function like: \[\frac{1}{(s+2)}*...\]

Answer 5

\[e^{-2s}\frac{1}{s^{2}+s-2}=e^{-2s}\frac{1}{(s+2)(s-2)}\]

Answer 6

@Thomas, slight typo, should be (s+2)(s-1)

Answer 7

right

Answer 8

Wait, made a mistake there.

Answer 9

\[A/(s-2)+b/(s-1)=e ^{-2s}\]
or
\[A /(s-2)+b/(s-1)=1\]

Answer 10

second one

Answer 11

Then worry about the e afterwards?

Answer 12

yes

Answer 13

The power of e is just a shift in the domain, so that's easiest to do last.

Answer 14

so
\[A(s-2)+B(s-1)=1\]
s=2 => B=1
s=1=> A=-1

Which will make it:

\[L ^{-1}\left\{ e ^{-2s}\left(( -1/(s-1))+(1/(s-2) \right) \right\}\]

and then just solve like this

Answer 15

it's A(s+2)+B(s-1)=1

Answer 16

But your approach is good.