# Laplace transforms and partial fractions

1. Feb 3, 2006

### morry

Hey guys, Im supposed to find the Laplace transform of a set of ODEs.

Ive broken it down a bit and Im left with finding the Laplace transform of:

(-2e^-s)/(s(s+4)(s+1))

Is this something I have to use partial fractions for? Or is there another way? Im a bit confused.

2. Feb 3, 2006

### benorin

Do you mean the inverse Laplace transform of (-2e^-s)/(s(s+4)(s+1))?

3. Feb 3, 2006

### morry

Ahh, yeah that would be correct. oops

4. Feb 3, 2006

### benorin

If so, factor out the exponential and just deal with the rational part:

$$-\frac{2e^{-s}}{s(s+4)(s+1)}=-2e^{-s}\frac{1}{s(s+4)(s+1)}$$

So $$\frac{1}{s(s+4)(s+1)}=\frac{A}{s}+\frac{B}{s+4}+\frac{C}{s+1}$$

5. Feb 3, 2006

### morry

Ok, so after finding values for A,B,C, do I just use the tables in order to work out the individual bits?
ie ((-2e^-s)*A)/s etc?

6. Feb 3, 2006

### benorin

Cross-multiply to get:

$$1=A(s+4)(s+1)+Bs(s+1)+Cs(s+4)$$

then plug, say, s=0,-4,-1 into the above to solve for A,B, and C

7. Feb 3, 2006

### benorin

yep, use tables if ya got 'em.

8. Feb 3, 2006

### benorin

s=0 gives 1=4A or $$A=\frac{1}{4}$$

s=-4 gives 1=12B or $$B=\frac{1}{12}$$

and s=-1 gives 1=-3C or $$C=-\frac{1}{3}$$

9. Feb 3, 2006

### morry

Cheers Benorin, partial fractions are no worries, just that exponential that was confusing me, seems pretty simple now.