Inital value problem, Laplace transform

[saxen]

Homework Statement

Solve inital value problem: x''+2x'+x=g(t)

g(t)=
t 0<t<1
2-t 1<t<2
0 t>2

Homework Equations

Second shift theorem, Heaviside function and Laplace transforms. I denote Heaviside,functuon H(t-a), and Laplace transform with L

The Attempt at a Solution

I rewrote the problem with Heaviside functions:

g(t)=t+(2-t)H(t-1)
L(t+(2-t)H(t-1))= [itex]\frac{1}{s^{2}}[/itex]+[itex]\frac{2e^{-s}}{s^{2}}[/itex]

After this I transformed the original diff. equation and plugged in my transform of g(t). After some simplification I got:G(s)=(1-[itex]2e^{-s}[/itex])([itex]\frac{1}{s^{2}(s+1)^{2}}[/itex])
And after partial fraction decomp. I got:

G(s)=(1-[itex]2e^{-s}[/itex])([itex]\frac{2}{s}[/itex]+[itex]\frac{1}{s^{2}}[/itex]+[itex]\frac{1}{(s+1)^{2}}[/itex])

If I use inverse Laplace transform I get

x(t)=f(t)-2f(t-1)H(t-1)

where

f(t)=(t+2)e[itex]^{-t}[/itex]+t-2

According to my book, this is only about half of the answer. I am missing some step and I can't figure out what is it! All help is appreciated

Thanks in advance!

 

[LCKurtz]

I didn't check your work, but I notice that you haven't mentioned the initial conditions. Unless you were given x(0) = 0 and x'(0) = 0 you have left something out, which might explain why you don't have the whole answer.