Laplace Transform of ODE

1. Mar 9, 2013

nobodyuknow

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

ODE: y'' + 4y' + 3y = f(t)
f(t) = (?? HELP - What's the mathematical term to describe these? I can't seem t o find it in my notes )
1, 0 ≤ t < 2
t², 2 ≤ t < 3
0, t ≥ 3

Write a brief description on how you would solve this ODE using Laplace transforms.

Also use the integral definition of Laplace Transforms to calculate the Laplace Transform of f(t).

2. Relevant equations

We know the Laplace Transform Definition is:
F(s) = ∫$^{∞}_{0}$f(t)e-st

3. The attempt at a solution

After integrating the f(t)'s I get...

0 + 1/s - ((e^-2s)/s) + (((4e^-2s) - (9e^-3s))/s) + (((4e^-2s) - (6e^-3s))/s^2) + (((2e^-2s) - (2e^-3s))/s^3)

This looks completely wrong, but, I really have little idea on how to do this. I don't have tutorials for at least a couple of days so I cannot ask my tutorer about it quite yet.

All help appreciated, thanks!

Last edited: Mar 9, 2013
2. Mar 9, 2013

SithsNGiggles

For visibility:

$y''+4y'+3y=f(t),$

$f(t)=\begin{cases}1&0\le t<2\\ t^2&2\le t<3\\ 0&t\ge3\end{cases}$

3. Mar 10, 2013

nobodyuknow

Thanks for that.

4. Mar 10, 2013

SteamKing

Staff Emeritus
The forcing function f(t) is composed of several piecewise functions.

5. Mar 10, 2013

HallsofIvy

Staff Emeritus
I would say rather that f(t) is one piecewise defined function, that function having several pieces!

6. Mar 11, 2013

nobodyuknow

Could you possibly do this...
y'' + 4y' + 3y = f(t)
y'' + 3y' + y' + 3y = f(t)
d/dx [y' + 3y] + [y' + 3y] = f(t)
let u = y' + 3y
u' + u = f(t)

Would that be correct?