# Partial-fractions problem

1. May 29, 2005

### tandoorichicken

Find the inverse Laplace transform of
$$\frac{s-2}{s^3+2s^2+2s}$$

I think once I get the expression simplified I can do the rest by myself. I started to separate this expression out by partial fractions and got as far as this:
$$\frac{s-2}{s^3+2s^2+2s}=\frac{s-2}{(s^2+2s+2)s}=\frac{As+B}{s^2+2s+2} +\frac{C}{s}$$

$$(As+B)s+C(s^2+2s+2)=s-2$$
From this expression I got C=-1, which I checked was correct by using my calculator, but I still don't know how to find A, or B. Can anyone help with this please?

Any ideas or hints on how to do the inverse would be also be appreciated.

2. May 29, 2005

### Corneo

Note that

$$s^2+2s+2 = (s+1)^2+1$$

Opps major brain laspe. That should be fixed

Last edited: May 29, 2005
3. May 29, 2005

Nope, it equals $(s+1)^2 + 1.[/tex] edit: tex 4. May 29, 2005 ### Corneo Or perhaps that wasn't too helpful. Here is an trick I learned when doing these partial fractions problems. Consider what you have $$(As+B)s+C(s^2+2s+2)=s-2$$ Multiply out and gather the terms. $$(A+C)s^2 + (2C+B)s + 2C = s-2$$ Is there a [itex]s^2$ term on the right hand side of the equation? What does this tell you about $A+C$? What about $2C$, what should that be equal to? And $2C+B$?

5. May 29, 2005

### tandoorichicken

Thanks, Corneo.