Particular solution of a system of second order ODE I have the following system of two ODEs:
$$\begin{bmatrix}
-k & k & \\
k & -k & \\
\end{bmatrix}\begin{bmatrix}x_1(t)\\x_2(t)\end{bmatrix}+\begin{bmatrix}F-k\times a\\k\times a\end{bmatrix} =\frac{d^2}{dt^2}\begin{bmatrix}x_1(t)\\x_2(t)\end{bmatrix}$$
where $F, k$ and $a$ are constants. How can I find a particular solution for the system? I have been trying to guess a form that may work but with no success.
UPDATE:
If I change the system to first order ODE:
$$\begin{bmatrix}
\mathbf{0} & \mathbf{I} & \\
\mathbf{K} & \mathbf{0} & \\\end{bmatrix}
\begin{bmatrix}\mathbf{x}(t)\\\mathbf{y}(t)\end{bmatrix}+
\begin{bmatrix}0\\0\\F-k\times a\\k\times a\end{bmatrix} =
\frac{d}{dt}\begin{bmatrix}\mathbf{x}(t)\\\mathbf{y}(t)\end{bmatrix}$$
The problem here is that the matrix $\begin{bmatrix}
\mathbf{0} & \mathbf{I} & \\
\mathbf{K} & \mathbf{0} & \\\end{bmatrix}$ is not invertible and I can't solve for unknown coefficients in an assumed form.
 A: Writing your system as $$x'' = k(y-x) + F - a k,$$ and $$y'' = k(x-y) + a k,$$ leads me to change coordinates to $u = x+y$, $v=x-y$, in which case your equations decouple to $$u''=F$$ and $$v''=-2k v + F -2 a k.$$ I would expect both of these ODEs to be covered in a first course on differential equations. You can use the linear transformation that I used in my other answer to remove the constant term from the second equation.
A: I will write your system as $$x'' = M x + b,$$ where $x\in \mathbb{R}^2$, $M = \begin{pmatrix} -k & k & \\ k & -k & \\ \end{pmatrix}$, and $b =\begin{pmatrix}F-k\times a\\k\times a\end{pmatrix}$, assuming that $\times$ is multiplication of scalars.
Now do a linear transformation to $y = x-M^{-1}b$, giving that $$y'' = M y.$$ Then I believe this equation has solutions of the form $$y(t) = y_1 \exp(N t) + y_2 \exp(-N t),$$ where $y_i \in \mathbb{R}^2 $, $\exp$ is the matrix exponential,  and $N$ is any matrix square root of $M$, satisfying $N^2 =M$.
It suffices to find square roots of your $M$, which may or may not exist, I didn't check.
Edit;
I realised that your matrix is not invertible, so my solution doesn't work. I will leave it up in case this solution for invertible $M$ is still useful for someone.
