I saw a question on MS about how to solve: $$(D^2+\operatorname{Id})y=4\cos{x}$$ (here: How do you solve $(D^2+1)y=4\cos{x}$?)
I thought, why not be overkill and solve this using some Linear Algebra. Define the vector space $$V = \{a\sin{x} + b\cos{x} + cx\sin{x} + dx\cos{x} \ |a,b,c,d\in\mathbb{R} \} $$
This is a vector space and an obvious basis is $$E = \{\cos x, \sin x, x\cos x, x\sin x\}$$
I glare over the details but the matrix representation of D wrt to E on the right and E on the left is then:
$$M(D)=\begin{bmatrix}0 & 1 & 1 & 0\\-1 & 0 & 0 & 1\\ 0 & 0 &0 &1\\ 0 & 0 &-1 & 0\end{bmatrix}$$
and so
$$M(D^2)=\begin{bmatrix}-1 & 0 & 0 & 2\\0 & -1 & -2 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\end{bmatrix}$$
and $M(D^2) + I_4$ has zeros on the diagonal and is obvisouly not invertible.
But, the solution to the ODE is obviously a linear combination of my basis vectors, so I don't understand why we can't solve the problem this way (i.e. by multiplying the coordinate vector of the function on the RHS by the inverse of $M(D^2) + I_4$).
The system $$ \pmatrix{0&0&0&2\\0&0&-2&0\\0&0&0&0&\\0&0&0&0} \pmatrix{a\\b\\c\\d} =\pmatrix{4\\0\\0\\0} $$ is solvable, even if the system matrix only has rank $2$. In fact, the degrees of freedom correspond to the parameters of the homogeneous solution.