Substituting back constants for system of differential equations Solve \begin{cases}
x''(t)+y'(t)-x(t)+y(t)&=-1 \\
x'(t)+y'(t)-x(t)&=t^2
\end{cases}
I know this is a system of differential equation and can get through most of the problem with the exception of substituting the constants back into each other so there is no extra constant values.
\begin{cases}
(D^2-1)x+&(D+1)y&=-1 \\
(D-1)x+&Dy&=t^2
\end{cases}
\begin{cases}
(-D^3+D^2+D-1)y=t^2-1& \text{eliminating x}\\
(-D^3+D^2+D-1)x=t^2+2t& \text{eliminating y}
\end{cases}
The complementary functions is =$$ c_1e^{-t}+c_2e^t+c_3te^t $$
After solving for the particular solution of both, I get 
\begin{cases}
y=&c_1e^{-t}&+c_2e^t&+c_3te^t&-t^2&-2t&-3\\
y'=&-c_1e^{-t}&+c_2e^t+c_3e^t&+c_3te^t&&-2t&-2\\
x=&c_4e^{-t}&+c_5e^t&+c_6te^t&-t^2&-4t&-6\\
x'=&-c_4e^{-t}&+c_5e^t+c_6e^t&+c_6te^t&&-2t&-4\\
x''=&c_4e^{-t}&+c_5e^t+2c_6e^t&+c_6te^t&&&-2
\end{cases}
Here is the where the problem is: notice $$c_5$$
If I sub back my x, x', x'', y, y' back into my original two equations, regardless of which one, my $$c_5$$ disappears, so how do I find what that is supposed to be to sub it back into x.
 A: If you compute a little more clever, you can see that $D^2-1=(D+1)(D-1)$ by binomial formula and thus the elimination of $x$ gives
\begin{align}
(D^2-1)x+(D+1)y&=-1\\
(D^2-1)x+D(D+1)y&=(D+1)t^2=t^2+2t\\[1em]\hline
(D-1)(D+1)y=(D^2-1)y&=t^2+2t+1
\end{align}

One could also start by partially integrating the first equation as there is a common factor $(D+1)$ on the left side, leading to
\begin{gather}
(D+1)z=-1\implies z=-1+ce^{-t}\\[1em]\hline
(D-1)x+y = -1 + c_1e^{-t}
\end{gather}
which also should simplify the further solution.
A: You don't, but I don't think it's that big of a deal. The constraint equations determining the solution to the homogeneous equation are
$$\begin{align}\begin{bmatrix}0 & 0 \\ -2 & 1\end{bmatrix}\begin{bmatrix}c_4\\c_1\end{bmatrix}&=\begin{bmatrix}0\\0\end{bmatrix}\\
\begin{bmatrix}2D &D+2\\ D & D+1\end{bmatrix}\begin{bmatrix}c_5+c_6t\\c_2+c_3 t\end{bmatrix}&=\begin{bmatrix}0\\0\end{bmatrix}\text{.}
\end{align}$$
That's $3$ relations among $6$ constants, leading to $3$ linearly dependent solutions to the homogeneous equation, which is what you might expect.
