Can someone help me find the Inverse Laplace Transform of

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

Question

Can someone help me find the Inverse Laplace Transform of

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

anonymous · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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: $$\frac{1}{(s+2)}*...$$

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

right

anonymous · Answer

Wait, made a mistake there.

anonymous · Answer

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

anonymous · Answer

second one

anonymous · Answer

Then worry about the e afterwards?

anonymous · Answer

yes

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

But your approach is good.