What method to use to reach the correct general solution in this differential equation? Some context to rule out previous mistakes; I started with a couple of DEs:
\begin{align}
(D-1)x+(D^2+1)y&=1 \\
(D^2-1)x+(D+1)y&=2
\end{align}
Then I multiplied in order to eliminate:
\begin{align}
(D+1)(D-1)x+(D+1)(D^2+1)y&=(D+1)1  \\
-(D^2-1)x-(D+1)y&=-2
\end{align}
Added both and got:
\begin{align}
(D+1)(D^2+1)y-(D+1)y&=-1    \\
(D^3+D^2+D+1)y-(D+1)y&=-1    \\
y'''+y''+y'+y-y'-y&=-1   \\
y'''+y''&=-1    \\  \\
m^3+m^2&=0    \\
m^2(m+1)&=0
\end{align}
So the zeros are $m=-1$, $m=0$, and $m=0$; thus:
$$Y_h=C_1+C_2t+C_3e^{-t}$$
At this point a couple people told me that $Y_p=-\frac{1}{2}t^2$ because allegedly you can eyeball it with experience (it is the right $Y_p$ according to my book though); however, I wanted to be able to reach to it with some procedure and I figured I would try the Wronskian. I don't really know how to format a matrix, but my Wronskian goes like this:
\begin{align}
Y_1&=1&
Y_2&=t&
Y_3&=e^{-t}  \\
Y'_1&=0&
Y'_2&=1&
Y'_3&=-e^{-t}  \\
Y''_1&=0&
Y''_2&=0&
Y''_3&=e^{-t}
\end{align}
From where I get:
$W=e^{-t}$
$W_1=te^{-t}$
$W_2=e^{-t}$
$W_3=-1$
$U_1=\frac{1}{2}t^2+t$
$U_2=t$
$U_3=-e^t$
And finally $Y_p=\frac{3}{2}t^2+t-1$, but since $t-1$ are already in $Y_h$ the final solution should be $X=C_1+C_2t+C_3e^{-t}+\frac{3}{2}t^2$ and then I would go on to get $Y$ and solve the DE.
However, the actual answer in my book is $$X=C_1+C_2t+C_3e^{-t}+C_4e^t-\frac{1}{2}t^2,$$ so I can only assume that my $Y_h$ is wrong, and that should be the reason why my $Y_p$ is wrong too. I have two questions:
What went wrong with $Y_h$?
How can people eyeball the right $Y_p$ so easily even though $Y_h$ is wrong?
 A: I will take a different approach to see if I arrive at your result.
We have the system
$$x' - x + y'' + y = 1 \\ x'' - x  + y' = y = 2$$
Taking the derivative of the first equation
$$x'' - x' + y''' + y' = 0 \\ x'' - x  + y' = y = 2$$
Subtracting the second equation from the first
$$-x'+ x + y''' - y = -2$$
Adding the first equation to this result
$$y''' + y'' = -1$$
Lets use Undetermined Coefficients
$$y''' + y'' = -1 \implies m^3 + m^2 = 0 \implies m = 0, 0, -1$$
Because of the double root, we write $(c_1 + c_2 t)e^{0 t}$ and for the single root, $c_3 e^{-t}$, for a homogeneous solution of
$$y_h(t) = c_1 + c_2 t + c_3 e^{-t}$$
For the particular solution, we see that we already have $c_1 + c_2 t$ in the homogeneous solution, so we will multiply by $t$ and try
$$y_p(t) = t( a + b t)$$
Substituting this in to $y''' + y'' = -1$ and solving for the constants gives
$$a = 0, b = -\dfrac{1}{2}$$
The solution is
$$y(t) = y_h(t) + y_p(t) = c_1 + c_2 t + c_3 e^{-t} -\dfrac{1}{2} t^2$$
Substituting this result into the original first equation
$$x' + x = -(c_1 + c_2 t + 2 c_3 e^{-t}-\dfrac{t^2}{2}-1) - 1$$
Solving for $x(t)$
$$x(t) = c_4 e^t + c_1 + c_2 (t+1)+ c_3 e^{-t} - \dfrac{t^2}{2} - t -1 = c_1 + c_2 t + c_3 e^{-t} + c_4 e^t - \dfrac{1}{2}t^2$$
In that last equality, we just combine like terms into common constants. 
A: Hint:
When the RHS is a constant, set the derivative of the lowest order (possibly $0$) to be a constant. Then all other derivatives vanish.
$$y'''+y''=-1$$ indeed gives $y''=-1$, which is $$y=-\frac{t^2}2+Ct+C'.$$
As you only need one particular solution, you can drop $Ct+C'$.
