# Solving D.E with Laplace

1. Dec 5, 2005

I have

y'' + 2y' - 3y = Dirac(t)

I use the laplacetransformation and get

s^2Y + 2sY - 3Y = 1

Y = 1/4 * 1/(s-1) - 1/4 * 1/(s+3) (skipped some steps)

I try to use inverselaplace and recieve

y = 1/4*exp(t) * H(t) - 1/4 * exp(-3t)*H(t)

Where H(t) is the heaviside function.
the correct answear should be 1/4*exp(t) *( H(t) -1) - 1/4 * exp(-3t)*H(t).

What am I doing wrong. They ask for "the limited solution" does that mean anything particulairy?

Thnx for any help

2. Dec 5, 2005

### Tide

You left out the initial values when you performed the Laplace transform on the original DE.

3. Dec 5, 2005

There were no initial values if you mean like y(0) och y'(0). But I think I understand it anyhow. the key lies in the "limited solution". A limited solution cant eb allowed to grow infinite, and therefor we have to compensate for the 1/4 exp(t)

4. Dec 5, 2005

### saltydog

Hello guys. Are you referring to the following IVP:

$$y^{''}+2y^{'}-3y=\delta_0(t);\quad y(0)=0\quad y^{'}(0)=0^-$$

(You'll need initial conditions to solve via Laplace transforms.)

It's a bit awkward that one initial condition is given "prior" to t=0. Perhaps this is the "limiting" case the author refers to. Anyway, I don't get all those Heaviside's. I get one and it has to be inserted manually:

$$y(t)=H(t)\left(\frac{e^t}{4}-\frac{e^{-3t}}{4}\right)$$

Last edited: Dec 5, 2005