# 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 $y=C_{2}e^{3t}$ . 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: $x'=2x+C_{2}e^{3t}$ . We can solve this equation by using an integrating factor to get that:

${\begin{cases}x(t)=C_{1}e^{2t}+C_{2}e^{3t}\\y(t)=C_{2}e^{3t}\end{cases}}$ 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: $y_{1}=C_{1}e^{6t}$ and $y_{2}=C_{2}e^{2t}$ . And so we find $x_{1}=C_{1}e^{6t}+C_{2}e^{2t}$ and $x_{1}=C_{1}e^{6t}-C_{2}e^{2t}$ . 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 ${\vec {x}}_{1}(t)=C_{1}e^{6t}{\begin{pmatrix}1\\1\end{pmatrix}}$ and ${\vec {x}}_{2}(t)=C_{2}e^{2t}{\begin{pmatrix}1\\-1\end{pmatrix}}$ 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{aligned}{\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{aligned}} 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: