Consider the case where there are two functions interacting with each other:
\begin{equation} y_1(t) \dots y_{2}(t) \end{equation}
So we have more than one dependent function, with functions \(y_1, y_1’, y_2, y_2’\) and so forth. To deal with this, we simply make it into a matrix system:
\begin{equation} y(t) = \mqty(y_1(t) \\ \dots \\ y_{n}(t)) \end{equation}
For instance, should we have:
\begin{equation} \begin{cases} y_1’ = 3y_1 - 2y_2 \\ y_2’ = -y_1 + 5y_2 \end{cases} \end{equation}
We can write this system in a matrix like such:
\begin{equation} y’(t) = \mqty(3 & -2 \\ -1 & 5) y(t) \end{equation}
Meaning:
\begin{equation} y’ = Ay \end{equation}
which is a single linear equation.
Recall that we had:
\begin{equation} y’ = Ay \end{equation}
Let \(v\) be an eigenvector of \(A\) with \(\lambda\) be an eigenvalue. Let us guess that \(y = e^{\lambda t} v\) is a solution.
Plugging this in, we have:
\begin{equation} y’ = Ay = A(e^{\lambda t} v) = e^{\lambda t} Av = \lambda e^{\lambda t} v \end{equation}
Of course, \(y’ = \lambda e^{\lambda t} v\).
Meaning this is a solution of our system. Recall finding eigenvalues with actual numbers, so we want some \(\lambda\) for which \(det(A-\lambda I)=0\).
Plugging the eigenvalues back, and recalling the superposition principle, we are left with some:
\begin{equation} y(t) = c_1 e^{\lambda_{1}} v_1 + \dots + c_{n} e^{\lambda_{n}} v_{n} \end{equation}
This is true if we have enough eigenvalues which forms a basis. Now, at \(y(0)\), we have some \(y_0 = c_1v_1 + … + c_{n}v_{n}\).
This yields a system \(y_{0} = \mqty[v_1 & \dots & v_{n}] \mqty[c_1 \\ \dots \\ c_{n}]\).
We call this matrix written in terms of eigenvectors \(E\), that is:
\begin{equation} E = \mqty[v_1 & \dots & v_{n}] \end{equation}
Finally, we have:
\begin{equation} \mqty[c_1 \\ \dots \\ c_{n}] = E^{-1} y_0 \end{equation}
This method works for cases where we have enough independent eigenvectors to admit enough initial conditions. Otherwise, matrix exponentiation.
Special Cases
2x2 with \(\lambda_{2} = \bar{\lambda_{1}}\)
For any two by two system, where there the eigenvalues are conjugates of each other, we can formulate a solution in the form:
\begin{equation} y(t) = c_1 Re(e^{\lambda t} v) + c_2 Im(e^{\lambda t}v) \end{equation}
if the matrix representing the system admits two eigenvalues, \(\lambda\) and \(\bar{\lambda}\). We can obtain this by rephrasing one solution as \(e^{\lambda t} = e^{a + ib} e^{t} = e^{a+t}(\cos b + i\sin b)\).
Tips and Tricks
Changing higher order system into lower orders
We can actually write higher order linear system this way too:
\begin{equation} y’’ + ay’ + by = 0 \end{equation}
we can actually construct:
\begin{align} & y_1(t) = y(t) \\ & y_2(t) = y’(t) \end{align}
And therefore, we can construct:
\begin{equation} \mqty(y_1 \\ y_2)’ = \mqty(y_2 \\ -by1 - ay2) = \mqty(0 & 1 \\ -b &-a) \mqty(y_1 \\ y_2) \end{equation}