ODE using Laplace transform [
I got my Y(t) to be :
$$12 \, e^{-4} \, e^{-2s} \, [\frac{1}{12(s+2)} + \frac{1}{4(s-2)} - \frac{1}{3(s-1)}] + \frac{1}{(s-2)} - \frac{1}{(s-1)}.$$
so i assume I need to use t shifting for the laplace inverse/transformation (not sure of the term to use), but i'm not sure what to do with the first terms. Could someone help me here.
 A: I don't agree with your $Y(s)$. Here comes a suggestion on how to get $Y(s)$ and on how to get your function back in the end:
Taking Laplace transform of the left-hand side, invoking the initial value conditions, gives
$$
(s^2-3s+2)Y(s)-1.
$$
The right-hand side can be written
$$
12\theta(t-3)e^{-2t}.
$$
The Laplace transform of $12\theta(t)$ is $12/s$, and so the Laplace transform of $12\theta(t-3)$ is $12e^{-3s}/s$. Using the shift rule, we find that the Laplace transform of the right-hand side is
$$
\frac{12e^{-3(s+2)}}{s+2}.
$$
Thus, your equation for $Y(s)$ should be (since $s^2-3s+2=(s-1)(s-2)$)
$$
(s-1)(s-2)Y(s)-1=\frac{12e^{-3(s+2)}}{s+2}.
$$
Thus,
$$
Y(s)=\frac{1}{(s-1)(s-2)}+\frac{12e^{-3(s+2)}}{(s-1)(s-2)(s+2)}
$$
If I understood you correct, the first part of this expression is not a problem, so let us concentrate on how to calculate the inverse Laplace transform of the second one. We first do a partial fraction decomposition, to write
$$
\frac{12}{(s-1)(s-2)(s+2)}=-\frac{4}{s-1}+\frac{3}{s-2}+\frac{1}{s+2}.
$$
I show you how to calculate the inverse Laplace transform of
$$
-\frac{4e^{-3(s+2)}}{s-1},
$$
and leave the other terms to you. This is like working with the right-hand side, but backwards. We write
$$
-\frac{4e^{-3(s+2)}}{s-1}=-4e^{-9}\frac{e^{-3(s-1)}}{s-1}.
$$
This is the $1$-shift of
$$
-4e^{-9}\frac{e^{-3s}}{s}
$$
which should give a $e^t$ factor in the end. Now, the term $e^{-3s}$ implies that we should have the inverse Laplace transform of
$$
-4e^{-9}\frac{1}{s}
$$
shifted $3$ steps. But this we recognize as the Laplace transform of $\theta(t)$. Hence, we get
$$
-4e^{-9}\theta(t-3)e^t,
$$
or, maybe a bit cleaner
$$
-4e^{t-9}\theta(t-3).
$$
You do the same with the other terms.
A: I would suggest doing it one time interval at a time to keep your work neat. It gets quite messy with all the partial fractions sometimes.
Start with the interval $0<t<3$, with $y(0)=0$ and $y'(0)=1$ as initial conditions.
$\frac{d^2y}{dt^2}-3\frac{dy}{dt}+2y=0$
$(s^2Y-sy(0)-y'(0))-3(sY-y(0))+2Y=0$
$Y={1 \over s^2-3s+2}=\frac{1}{s-2}-\frac{1}{s-1}$
$y(t)=e^{2t}-e^t \ \ \ | \ \  0<t<3$
Now do the part for $t>3$. Initial conditions are $y(3)=e^6-e^3=c_1$ and $y'(3)=2e^6-e^3=c_2$. I used $c_1$ and $c_2$ to make the work neater.
$(s^2Y-sc_1-c_2)-3(sY-c_1)+2Y=\frac{12}{s+2}$
$(s-1)(s-2)Y=\frac{12}{s+2}+(c_1s+c_2+3c_1)$
$Y=\frac{12}{(s+2)(s-1)(s-2)}+\frac{c_1s+c_2+3c_1}{(s-1)(s-2)}$
$Y=\frac{1}{s+2}+\frac{3}{s-2}-\frac{4}{s-1}+\frac{-4c_1-c_2}{s-1}+\frac{5c_1+c_2}{s-2}$
$Y=\frac{1}{s+2}+\frac{3+5c_1+c_2}{s-2}-\frac{4-4c_1-c_2}{s-1}$
$y(t)=e^{-2(t-3)}+(3-6e^3+7e^6)e^{(t-3)}+(4+5e^3-6e^6)e^{2(t-3)} \ \ \ | \ \  t>3$
