Solving a homogenous system of linear ODE with Pauli matrices I was asked to solve find a general solution to $\overrightarrow{x'}=P\overrightarrow x$ where $P=\begin{pmatrix} -1 & 2 \\-1 & 1\end{pmatrix}$.
Using the "regular" method of finding the eigenvalues and eigenvectors of $P$, I arrived at the solution
$\overrightarrow x=c_1 \begin{pmatrix}e^{it} \\ \frac{1}{2}e^{it}+\frac{1}{2}ie^{it}\end{pmatrix}+c_2\begin{pmatrix} e^{-it}\\\frac{1}{2}e^{-it}-\frac{1}{2}ie^{-it}\end{pmatrix}$
This is the most general solution, we were not given initial conditions on the problem.
I showed a friend my solution, and he said he solved it using pauli matrices, according to him, we can solve a linear system of ODE if we somehow decompose $P$ as a sum of pauli matrices.
I googled and youtubed but I found no mention of pauli matrices being relevant to differential equations. I would like if someone could explain to me or refer me to a guide of some sort on how to use pauli matrices to solve system of ODEs. Thank you.
Also - is my solution correct?
 A: The eigenvalues of $P$ are $\pm i$, corresponding to the vectors $\vec v_\pm=(2,1\pm i)$ (or $(1\mp i,1)$, whichever is more convenient).
The general solution is $\vec x(t)=c_+e^{it}\vec v_++c_-e^{-it}\vec v_-$ for coefficients $c_\pm$.
This is also your solution, and it is correct.
It is also possible to solve such problems using Pauli matrices.
First write your matrix $P$ as a (complex) linear combination of Pauli matrices and the unit matrix – this is always possible since the four matrices form a basis for the space of $2\times2$ matrices.
In this case we have for $\vec a=(-i/2,3/2,i)$ the identity
$$
P
=
i\vec a\cdot\vec\sigma
=
\frac12\sigma_1+\frac{3i}2\sigma_2-\sigma_3.
$$
Finding the coefficients is a simple exercise in linear algebra; in this case it happens that we do not get the identity matrix (this happens if and only if $\operatorname{tr} P=0$).
The general solution to $\vec x'(t)=P\vec x(t)$ is, just like in the one-dimensional case, $\vec x(t)=e^{tP}\vec x(0)$.
For a vector $\vec a$ with $a_1^2+a_2^2+a_3^2=1$ (no complex conjugates!), the Pauli matrices satisfy
$$
\exp(it\vec a\cdot\vec\sigma)
=
I\cos(t)+i\vec a\cdot\vec\sigma\sin(t).
$$
Our vector $\vec a$ satisfies this condition (otherwise you would just have to scale it), so our identity $P=i\vec a\cdot\vec\sigma$ yields
$$
e^{tP}
=
I\cos(t)+P\sin(t).
$$
Thus the solution to the ODE is
$$
\vec x(t)=\cos(t)\vec x(0)+\sin(t)P\vec x(0).
$$
If you plug in the initial condition $\vec x(0)=\vec v_\pm$ (which form a basis for the space of all possible initial conditions), you will quickly see that this gives the same result as the other method.
