Using Laplace Transforms to Solve ODE with Piecewise Forcing Function

[nobodyuknow]

Homework Statement

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).

Homework Equations

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

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!

 

[SithsNGiggles]

For visibility:

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

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

 

[nobodyuknow]

Thanks for that.

 

[SteamKing]

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

 

[HallsofIvy]

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

 

[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?