# Inverse Laplace Transform

1. May 15, 2005

### cyberdeathreaper

I am asked to find the inverse laplace transform of the following function:

$$\frac{ \left( s+3 \right) }{ \left( s+1 \right) \left( s+2 \right) }$$

Using tables, can anyone help me understand why the answer is:

$$2e^{-t} - e^{-2t}$$

I'm completely loss on this one, and yet the book assumes this is easily determined. Any ideas?

Note: I already realize that the bottom can be rewritten using partial fractions, but it seems to me that the book assumes that isn't even neccessary - which makes sense, since it doesn't seem to get me anywhere anyway.

Last edited: May 15, 2005
2. May 15, 2005

### OlderDan

Look's like it might be this Heaviside formula for the ratio of polynomials. Have you encountered this before?

http://www.plmsc.psu.edu/~www/matsc597/fourier/laplace/node10.html

Last edited by a moderator: Apr 21, 2017
3. May 15, 2005

### Gale

you have to do partial fractions thats all. $$\frac{\left( s+3 \right) }{ \left( s+1 \right) \left( s+2 \right) }= \frac{A}{s+1} + \frac{B}{s+2}$$

s+3= A(s+2) + B(s+1)= (A+B)s + (2A+B)
A=2 B=-1

so you get
$$\frac{s+3}{(s+1)(s+2)} = \frac{2}{s+1} - \frac{-1}{s+2}$$

from there you can see pretty simply from $$e^{at}= \frac {1}{s-a}$$ that you get the textbooks given answer.

4. May 16, 2005

### cyberdeathreaper

Good grief....

That's totally correct Gale17 - thanks. I was trying to use partial fractions on just this part:

$$\frac{ 1 }{ \left( s+1 \right) \left( s+2 \right) }$$

and then mutliply the answer by s+3. Obviously that was not getting me anywhere, because I still had an s term on the top to deal with.

Thanks again!

PS: OlderDan, thanks for the suggestion though. I don't think we have touched on that approach specifically yet, but based on your link, it does seem feasible.

5. May 16, 2005

### Gale

an interesting way to use partial fractions... also, when you have the right answer, try working backwards with it.

you're welcome! (i like feeling useful anyways... sides, i have my own diff eq final coming up.. woo... )

6. May 16, 2005

### OlderDan

The partial fraction approach is definitely the way to go. I thought you were looking for an alternative. That other thing might come in handy for higher order polynomials.