Closed form solution to this system of ODEs Does the following system of ODEs
$$y'=\left(\begin{matrix}0&100&1\\-100&0&0\\1&0&-1\\ \end{matrix} \right)y$$
with some initial conditions $y(0) = [a, b, c]^T$ have a closed-form analytical solution?
 A: Yes, there is.
Let's write up $y := [f,g,h]^T$. Then the matrix-equation can be transformed into
$$\begin{align}
\text{I.} \quad f' &= 100g+h \\
\text{II.} \quad g' &= -100f \\
\text{III.} \quad h' &= f-h
\end{align}$$
From $\text{II.}$, we get that
$$g = -100 \smallint f$$
Substituting this into $\text{I.}$, we get that
$$f' = -10^4 \smallint f + h$$
Adding the $10^4 f$ term to both sides and differentiating we get
$$\begin{align}f' + 10^4 \smallint f &= h \\
f'' + 10^4 f &= h'\end{align}$$
Finally, substituting this into $\text{III.}$ we obtain a 1-variable homogeneous ODE:
$$\begin{align}
f''+10^4f &= f-(f'+10^4 \smallint f) \\
f''+10^4f &= f-f'-10^4 \smallint f \quad /' \\
f'''+10^4f' &= f'-f''-10^4 f \\
f'''+f''+(10^4-1)f'+10^4 f &= 0 \quad (*)
\end{align}$$
These types of equations can be solved by $f(x) = ae^{\alpha x}$. Taking its derivatives we get:
$$\begin{align}
f(x) &= ae^{\alpha x} \\
f'(x) &= a\alpha e^{\alpha x} \\
f''(x) &= a\alpha^2e^{\alpha x} \\
f'''(x) &= a\alpha^3e^{\alpha x} \\
\end{align}$$
Substituting this back into $(*)$:
$$\begin{align}
\alpha^3ae^{\alpha x}+\alpha^2ae^{\alpha x}+(10^4-1)\alpha ae^{\alpha x}+10^4ae^{\alpha x} &= 0\\
ae^{\alpha x}\left(\alpha^3+\alpha^2+(10^4-1)\alpha+10^4\right)&= 0
\end{align}$$
Assuming the initial condition $a$ isn't $0$, and using that $e^x$ never takes on zero, we can reduce this to:
$$\alpha^3+\alpha^2+(10^4-1)\alpha+10^4=0$$
Which we can solve for $\alpha$. I've used WolframAlpha for this, and the only real solution for the above equation is:
$$\alpha = -1.0001$$
So we got the first solution of
$$f(x) = ae^{-1.0001 x}$$
We can simply substitute this back to obtain $g$ and $h$:
$$g(x) = -100 \smallint f(x)dx = \frac{100a}{1.0001} e^{-1.0001 x}$$
$$h(x) = f''(x)+10^4f'(x) = a\left(1.0001^2e^{-1.0001 x}-10001 e^{-1.0001 x}\right)$$
So the solution is
$$y(x) = \begin{bmatrix}f(x) \\ g(x) \\ h(x)\end{bmatrix} = a\begin{bmatrix}e^{-1.0001x} \\ \frac{100}{1.0001} e^{-1.0001x} \\ (1.0001^2-10001) e^{-1.0001x}\end{bmatrix}$$

Note: This is just $1$ solution. For example when integrating in the first step (after "From $\text{II.}$"), we get another constant that can be carried through.
