Q:

Solving Piece wise ODE using heaviside function laplace transforms

Solve $y''(t)-2y'(t)+y(t)=g(t)$ where $g(t)= \begin{cases}e^{-t} ,&  t \in [0,1] \\0 , & t \notin [0,1] \end{cases}$.
Below is my attempt at solving. I get stuck after applying the Heaviside.

First, I will define some of the Laplacian transformations.
\begin{align}
\mathscr{L}\{y''(t)\}&=s^2 \hat{F}(s)-sy(0)-y'(0)
\\ 
\mathscr{L}\{y'(t)\}&=s \hat{F}(s)-y(0). 
\\\text{ So, }~~\mathscr{L}\{2y'(t)\}&=2s \hat{F}(s)-2y(0)
\\   
\mathscr{L}\{y(t)\}&=\hat{F}(s)
\\  
\mathscr{L}\{g(t)\}&= \mathscr{L}\{e^t-e^tu_1(t)\},
\end{align} where $u_1(t)$ denotes the Heaviside function, activated at $t+1$.
And, $$ \mathscr{L}\{e^t-e^tu_1(t)\}=\mathscr{L}\{e^t\}-\mathscr{L}\{e^tu_1(t)\}=\mathscr{L}\{e^t\}-e^{-s}\mathscr{L}\{e^{t+1}\}= \frac{1}{s-1}-e^{-s}\cdot\frac{e}{s-1}$$ via the Heaviside transformation steps. 
Then, brought all together, we have...
$$s^2\hat{F}(s)-sy(0)-y'(0)+2s\hat{F}(s)-2y(0)+\hat{F}(s)=\frac{1}{s-1}-e^{-s}\cdot\frac{e}{s-1}. $$ 
From here, the $\hat{F}(s)$ needs to be isolated to one side of the equation, leaving us with...
$$\hat{F}(s)=\frac{\frac{1+e^{-s+1}}{s-1}+(s+2)y(0)+y'(0)}{(s+1)^2}.$$
And it is here where I am stuck, and I believe that I am missing vital information in terms of the $y(0)$ and $y'(0)$ values or I am not understanding something quite right. 

A:

You changed the right side from the task with a factor $e^{-t}$ to the computation where you consistently used $e^t$.
In composing the Laplace transforms of the left side, you missed a sign so that you got $(s+1)^2$ instead of $(s-1)^2$.
In transcribing the Laplace transform of the right side, you missed a sign from $\frac{1}{s-1}-e^{-s}\frac{e}{s-1}$ to the wrong $\frac{1+e^{1-s}}{s-1}$.

To verify your solution, use the integrating factor $e^{-t}$ of the left side and set $z=e^{-t}y(t)$. This now satisfies the ODE of the original task
$$
z''=e^{-t}(y''-2y'+y)=(u(t)-u(t-1))e^{-2t}
$$
Now consider that $f''(t)=u(t)e^{at}$ has a general solution $$f(t)=ct+d+u(t)\frac{e^{at}-1-at}{a^2}$$ to get
$$z(t)=ct+d+u(t)\frac{e^{-2t}-1+2t}{4}-e^{-2}u(t-1)\frac{e^{-2(t-1)}-1+2(t-1)}{4}$$
It is quite usual for this type of tasks to demand that also the solution satisfies $y(t)=0$ for $t<0$ so that $c=d=0$ and
$$
y(t)=u(t)\frac{e^{-t}+(2t-1)e^t}{4}-u(t-1)\frac{e^{-t}+(2t-3)e^{t-2}}{4}.
$$

For the modified task as used in the computations you get
$$
z''=u(t)-u(t-1)\implies z(t)=cx+d+\frac12u(t)t^2-\frac12u(t-1)(t-1)^2
$$
and with the same assumptions on initial conditions,
$$
y(t)=\frac12e^t\bigl[u(t)t^2-u(t-1)(t-1)^2\bigr].
$$

