ODE system

We can express a linear system of differential equations of the form

\begin{align*} \dot x_1 &= a_{1,1} x_1 + a_{1,2} x_2 + \cdots + a_{1,n} x_n \\ \dot x_2 &= a_{2,1} x_1 + a_{2,2} x_2 + \cdots + a_{2,n} x_n \\ & ~~\vdots \\ \dot x_n &= a_{n,1} x_1 + a_{n,2} x_2 + \cdots + a_{n,n} x_n \end{align*}

using a matrix such that $\dot{\mathbf x} = A \mathbf x$ :

\begin{align*} \frac{\mathrm d}{\mathrm dt} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} &= \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \cdots & a_{n,n} \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}. \end{align*}

To solve an ODE system of this form we guess the solution $\mathbf x = \mathbf C e^{\lambda t}$ where $\mathbf C$ is a constant vector.

\begin{align*} \dot{\mathbf x} &= \lambda e^{\lambda t} \mathbf C \\ \lambda \mathbf C &= A \mathbf C. \end{align*}

We see that $\lambda$ must be an eigenvalue of $A$ , and $\mathbf C$ its corresponding eigenvector. The general solution is then

\mathbf x(t) = c_1 \mathbf C_1 e^{\lambda_1 t} + c_2 \mathbf C_2 e^{\lambda_2 t} + \cdots + c_n \mathbf C_n e^{\lambda_n t}

where $c_i$ is determined by initial conditions.