# Inital value problem, Laplace transform

1. Sep 5, 2011

### saxen

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

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

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

2. Relevant equations

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

3. 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))= $\frac{1}{s^{2}}$+$\frac{2e^{-s}}{s^{2}}$

After this I transformed the original diff. equation and plugged in my transform of g(t). After some simplification I got:

G(s)=(1-$2e^{-s}$)($\frac{1}{s^{2}(s+1)^{2}}$)
And after partial fraction decomp. I got:

G(s)=(1-$2e^{-s}$)($\frac{2}{s}$+$\frac{1}{s^{2}}$+$\frac{1}{(s+1)^{2}}$)

If I use inverse Laplace transform I get

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

where

f(t)=(t+2)e$^{-t}$+t-2

According to my book, this is only about half of the answer. I am missing some step and I cant figure out what is it! All help is appreciated