# Math Help - Laplace Transform

1. ## Laplace Transform

x'' + 4x' -8x = 0
x(0) = 0 x'(0) = 1

So I set up the transform, and got it to (s^2)L(x) +4sL(x)-8L(x) = 1
From here, x(t) = L^(-1) of 1/(s^2+4s-8)
I did a sort of completing the square, to get 1/((s+2)^2-12), which almost fits the equation for a laplace transform of e^(at)cos(bt), but now this b value would be the square root of a negative number...where did I go wrong? Any clarification would be great, thanks!

2. You have constant coefficients. I would substitute $x(t)=e^{mt}$ instead.

Then you only need to solve $m^2+4m-8=0$.

3. Originally Posted by mistykz
x'' + 4x' -8x = 0
x(0) = 0 x'(0) = 1

So I set up the transform, and got it to (s^2)L(x) +4sL(x)-8L(x) = 1
From here, x(t) = L^(-1) of 1/(s^2+4s-8)
I did a sort of completing the square, to get 1/((s+2)^2-12), which almost fits the equation for a laplace transform of e^(at)cos(bt), but now this b value would be the square root of a negative number...where did I go wrong? Any clarification would be great, thanks!
Factorise the denominator and then express in partial fraction form.

4. How would I factor the bottom? I don't see anything that works without getting complex numbers..

5. Originally Posted by mistykz
How would I factor the bottom? I don't see anything that works without getting complex numbers..
Last time I checked, $(s+2)^2-12$ could be factorised using the difference of two squares formula.

6. Ah, I didn't think of using that...got it! Thanks!