My question is taken from: Finding the solution to a non-homogeneous matrix exponential.
Rather than ask all my questions I considered to post them as a new question. $$\frac{d}{du} \begin{bmatrix}a\\b \end{bmatrix} = \begin{bmatrix}-x& y\\ -y&-x\end{bmatrix} \begin{bmatrix}a\\ b\end{bmatrix} + \begin{bmatrix}\cos(zu)\\ -\sin(zu)\end{bmatrix}$$ where $P = \begin{bmatrix}-x& y\\ -y&-x\end{bmatrix}$. And I have it for $u=0:$ $$\begin{bmatrix}a\\b \end{bmatrix} = \begin{bmatrix}0\\0 \end{bmatrix}$$
Having proved that $\exp\left[Pu\right] = \exp\left[-xu\right]$
$$\begin{bmatrix}\cos(yu)& \sin(yu)\\ -\sin(yu) & \cos(yu)\end{bmatrix}.$$
The solution to this was provided by the user 'Hyperplane':
$$ c(u) = \frac{1}{x^2 + (y-z)^2}e^{xu} \begin{bmatrix} x\cos((y-z)u) + (y-z)\sin((y-z) u)\\ x\sin((y-z) u) - (y-z)\cos((y-z) u) \end{bmatrix} + \text{const.} $$
And to satify the initial condition $c(0)=0$ we need
$$ \text{const.} = -\frac{1}{x^2 + (y-z)^2}\begin{bmatrix} x \\- (y-z) \end{bmatrix}$$
The question that I have is whether the solution obtained would work for when x or y=0, or if x or y = z?
Thanks.
Surely you would have to say that the solution is valid unless x=0 or y=z since the fraction would become 1/0; hope this is useful :)