# Homework Help: Finding the inverse Laplace transform

1. Aug 21, 2011

### Rubik

1. The problem statement, all variables and given/known data
Find the inverse Laplace transform of the following function
1/[s(s2 + 4)2]

2. Relevant equations
1/ (s2 + $\omega$2)2 = (1/ 2$\omega$3) (sin$\omega$t -$\omega$t cos$\omega$t)

3. The attempt at a solution
L-1 = (1/s) (1/ s2 +22)2
= (1/16) (sin 2t - 2t cos 2t) as the Laplace transform of (1/s) = 1

Is this what I am meant to do?

2. Aug 21, 2011

### Screwdriver

The inverse Laplace transform of a product isn't the product of the inverse Laplace transforms in general. You can go for the convolution theorem (might be a mess), or use partial fractions to decompose that, which requires you to use a few other inverse identities once you get the decomposition.

3. Aug 21, 2011

### Rubik

Right.. how silly of me, so doing it by partial fractions I got 0=B=C and A= 1/16
giving me F(s) = (1/16)(1/s) does that seem a little more on track?

4. Aug 21, 2011

### Screwdriver

Well using partial fractions is definitely on track. But remember that when you have a quadratic term in the denominator, you have to have a linear term in the numerator (plus, it's squared, so there are going to be two.) The form for the expansion should be:

$$\frac{1}{(s)(s^2+4)^2} = {\frac{A}{s}} + \frac{Bs + C}{s^2 +4} + \frac{Ds + E}{(s^2+4)^2}$$

Give that a shot.

Last edited: Aug 21, 2011
5. Aug 21, 2011

### Rubik

Okay after that I am now getting A = 1/16, B = -1/16, C = 0, D = -1/4, E = 0.

(1/16)(1/s) - (1/16)(s/[s2 + 4]) - (1/4)(s/(s2 + 4)2

6. Aug 21, 2011

### Screwdriver

That looks good! Now that you have those, you can invert them term-by-term since the inverse Laplace transform is linear, so $\mathcal{L}^{-1}(af(t)+bg(t)) = a\mathcal{L}^{-1}(f(t)) + b\mathcal{L}^{-1}(g(t))$ for constants $a$ and $b$.

7. Aug 21, 2011

### Rubik

Thank you so much!!!!! :D

8. Aug 21, 2011

### Screwdriver

You're welcome!

9. Aug 22, 2011

### Ray Vickson

There is a standard result: if f(t) <---> g(s), then integral_{x=0..t} f(x) dx < ---> g(s)/s.

RGV

10. Aug 22, 2011