Why Is the Inverse Laplace Transform of This Function 2e^{-t} - e^{-2t}?

[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 necessary - which makes sense, since it doesn't seem to get me anywhere anyway.

 

[OlderDan]

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 necessary - which makes sense, since it doesn't seem to get me anywhere anyway.

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

 

[Gale]

you have to do partial fractions that's 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.

 

[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.

 

[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... )

 

[OlderDan]

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.

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.