# Inverse Laplace transforms with quadratic factors

1. Nov 4, 2012

### Exocet

1. The problem statement, all variables and given/known data

This is a practice problem for a test on Laplace transforms

Find L^-1[ (9s^3+17s^2+66s+45) / (s^2+9)(s+2)^2 ]

(L^-1 = inverse laplace transform)

2. Relevant equations

From Laplace transform tables:

L^-1[ 1 / s-α ] = e^αt

L^-1[ s / s^2+ω^2 ] = cos(ωt)

L^-1[ ω / (s-α)^2+ω^2 ] = (e^αt)sin(ωt)

L^-1[ s-α / (s-α)^2+ω^2 ] = (e^αt)cos(ωt)

3. The attempt at a solution

First I expanded the function using partial fractions and found that it is equivalent to:

6/(s+2) - 7/(s+2)^2 + 3s/(s^2+9)

From the first equation above, the first fraction can be inverted to give

6e^-2t

From the second equation, the third fraction can be inverted to give

3cos(3t)

Usually at this point I would invert the second fraction using the two remaining equations above. This involves completing the square for the denominator. However, in this case the denominator is already in completed square form and ω is equal to zero.
I might be missing something blindingly obvious but I haven't got a clue how to solve this problem and I have a test in a few days so any help would be greatly appreciated

2. Nov 4, 2012

### LCKurtz

Can you invert $-\frac 7 {s^2}$? Then use if $\mathcal L^{-1}F(s) = f(t)$ then $\mathcal L^{-1}F(s+a) = e^{-at}f(t)$.

3. Nov 5, 2012

### Exocet

Aha! This definitely looks like it should work... but how do I go about inverting -7 / s^2 ???

4. Nov 5, 2012

### Exocet

Ah, obviously we use L-1[ n! / s^n+1 ] = t^n.... so f(t) is -7t which mean F(s-a) = -7te^-2t.

Thanks for the tip in the right direction :)