# Solving D.E with Laplace

1. Dec 5, 2005

### Larsson

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

### Larsson

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
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