# Ordinary Differential Equations/Linear Systems

A system of differential equations is a collection of two or more differential equations, which each ODE may depend upon the other unknown function.

For example consider the equations:

$\begin{cases}x'(t)=2x(t)+y(t)\\ y'(t)=3y(t)\end{cases}$

In this case the equation for differential equation for $x'(t)$ depends on both $x(t)$ and $y(t)$. In principle we could also allow $y'(t)$ to depend on both $x$ and $y$, but it is not necessary.

Notice in some cases we find a solution for a system of ODE's. For example in the case above, because $y'$ doesn't depend on $x$ we can solve the second equation (by separating variables or using an integrating factor) to get that $y=C_2e^{3t}$. Since there will be a second constant when we solve the first ODE, we choose to call the constant here $C_2$. Now we can plug this into the first equation to get that: $x'=2x+C_2e^{3t}$. We can solve this equation by using an integrating factor to get that:

$\begin{cases}x(t)=C_1e^{2t}+C_2e^{3t}\\y(t)=C_2e^{3t}\end{cases}$

In other cases a clever change of variables allows one to separate the two ODE's. Consider the system

$\begin{cases}x_1'(t)=4x_1(t)+2x_2(t)\\ x_2'(t)=2x_1(t)+4x_2(t)\end{cases}$.

If we let $y_1=x_1+x_2$ and $y_2=x_1-x_2$. Then we find that

$\begin{cases}y_1'(t)=6y_1(t)\\ y_2'(t)=2y_2(t)\end{cases}$

and each of these are easy to solve: $y_1=C_1e^{6t}$ and $y_2=C_2e^{2t}$. And so we find $x_1=C_1e^{6t}+C_2e^{2t}$ and $x_1=C_1e^{6t}-C_2e^{2t}$. It turns out to be helpful with systems to work with vectors and matrices so if we introduce $\textstyle\vec{x}(t)=\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix}.$ Then the above system can be re-written as:

$\frac{d}{dt}\vec{x}(t)=\begin{pmatrix}4 & 2 \\ 2 & 4\end{pmatrix}\vec{x}(t).$

And we have solutions $\vec{x}_1(t)=C_1e^{6t}\begin{pmatrix}1\\1\end{pmatrix}$ and $\vec{x}_2(t)=C_2e^{2t}\begin{pmatrix}1\\-1\end{pmatrix}$

Notice that the solutions we found were of the for $e^{\lambda t}\vec{\xi}$ for some constant vector $\vec{\xi}$. Using this as motivation we will investigate the question, when does $\vec{x}(t)=e^{\lambda t}\vec{\xi}$ solve the system:

$\frac{d}{dt}\vec{x}(t)=A\vec{x}(t).$

for some constant matrix $A$.

By substituting into the equation we see that:

\begin{align} \frac{d}{dt}(e^{\lambda t}\vec{\xi})&=A(e^{\lambda t}\vec{\xi})\\ \lambda e^{\lambda t}\vec{\xi}&=e^{\lambda t}A\vec{\xi}\\ e^{\lambda t}(A-\lambda I)\vec{\xi}&=\vec{0}. \end{align}

Since $e^{\lambda t}\neq 0$, the only way for the left hand side to be $\vec{0}$ is if $\lambda$ is an eigenvalue and $\vec{\xi}$ is a corresponding eigenvector.

This is not quite the end of the story. When the matrix is real we shall consider the following cases: