Houjun Liu

First-Order Linear Systems of ODEs

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}


\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}