DE using Laplace transform

[manenbu]

Homework Statement

y'' -6y' + 9y = f(t)

f(t)=
0, 0<t<1
1, 1<t<3
0, t>3

Homework Equations

The Attempt at a Solution

Turning it to heaviside functions I get:
y'' -6y' + 9y = u₁ - u₃

and I solve.

in the answers it should be:
y(t) = 2te^t + u₁(1/9 - 1/9e^3(t-1)+ 1/3(t-1)e^3(t-1)).

where did u₃ go in the answers? why doesn't it appear? the u₁ I can get just like that, but I can't seem to figure out where the u₃ went and why there is a 2te^t in there.

 

[Mathnerdmo]

It looks like we're missing some information. Were there additional conditions given? (e.g. y(0) = ? and y'(0) = ?)

Also, what did the DE look like after you applied the Laplace transform?

 

[dacruick]

it doesn't look as though you changed anything by putting it in unit step form.

 

[manenbu]

Sorry, forgot.
y(0) = 1, y'(0) = 2.