In Simple complex ODE's in matrix form? we discussed how we can solve certain complex ODE's in matrix form. As an example:
$$y(x)^2+y'(x)^2=0$$
can be factored as $$(y(x)+ iy'(x))(y(x)-iy'(x))=0$$
and if we insist we could solve the two above equations by representing each term in a matrix form, i.e. $$y(x)\equiv \begin{pmatrix}y_r(x)&-y_i(x)\\y_i(x)& y_r(x)\end{pmatrix}\\ y'(x)\equiv \begin{pmatrix}y'_r(x)&-y'_i(x)\\y'_i(x)& y'_r(x)\end{pmatrix}\\ i \equiv \begin{pmatrix}0&-1\\1& 0\end{pmatrix}\\ $$ Multiplying the matrices out we get a system of coupled ODE's and the solution is found.
However, I was trying to apply this technique for a different problem:
$$y(x)^2+y'(x)^2-1=0$$
This expression cannot be factored as a product of two complex numbers (except the trivial factoring of 1). However, I tried to factor it over different types of numbers and in this post we discussed, that split-quaternions allow the following factorisation:
$$(y(x)+i y'(x)+j)(y(x)-i y'(x)-j)=0$$
Just like regular complex numbers allow a representation by 2x2 real matrices, split-quaternions can be written as 2x2 complex matrices. Wikipedia. The true solution is: $${y(x) = c_1\sin(x)}\\ {y(x) = c_2\cos(x)}$$ and what we get from the method I developed is the following $$y(x)= c_1 \cos(x)-c_2 \sin(x)+\\(c_1 \sin(x)+c_2 \cos(x))i+ \\(c_4 \sin(x)+c_3 \cos(x)-1)j+ \\(c_4 \cos(x)-c_3 \sin(x))k$$ So that $\mathcal{R}\{y(x)\}=c_1 \cos(x)-c_2 \sin(x)$ and the results agree.
My question is: when can I use this technique? Is it legitimate or is it just a lucky coincidence?