# 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:

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

In this case the equation for differential equation for ${\displaystyle x'(t)}$ depends on both ${\displaystyle x(t)}$ and ${\displaystyle y(t)}$. In principle we could also allow ${\displaystyle y'(t)}$ to depend on both ${\displaystyle x}$ and ${\displaystyle y}$, 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 ${\displaystyle y'}$ doesn't depend on ${\displaystyle x}$ we can solve the second equation (by separating variables or using an integrating factor) to get that ${\displaystyle y=C_{2}e^{3t}}$. Since there will be a second constant when we solve the first ODE, we choose to call the constant here ${\displaystyle C_{2}}$. Now we can plug this into the first equation to get that: ${\displaystyle x'=2x+C_{2}e^{3t}}$. We can solve this equation by using an integrating factor to get that:

${\displaystyle {\begin{cases}x(t)=C_{1}e^{2t}+C_{2}e^{3t}\\y(t)=C_{2}e^{3t}\end{cases}}}$

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

${\displaystyle {\begin{cases}x_{1}'(t)=4x_{1}(t)+2x_{2}(t)\\x_{2}'(t)=2x_{1}(t)+4x_{2}(t)\end{cases}}}$.

If we let ${\displaystyle y_{1}=x_{1}+x_{2}}$ and ${\displaystyle y_{2}=x_{1}-x_{2}}$. Then we find that

${\displaystyle {\begin{cases}y_{1}'(t)=6y_{1}(t)\\y_{2}'(t)=2y_{2}(t)\end{cases}}}$

and each of these are easy to solve: ${\displaystyle y_{1}=C_{1}e^{6t}}$ and ${\displaystyle y_{2}=C_{2}e^{2t}}$. And so we find ${\displaystyle x_{1}=C_{1}e^{6t}+C_{2}e^{2t}}$ and ${\displaystyle x_{1}=C_{1}e^{6t}-C_{2}e^{2t}}$. It turns out to be helpful with systems to work with vectors and matrices so if we introduce ${\displaystyle \textstyle {\vec {x}}(t)={\begin{pmatrix}x_{1}(t)\\x_{2}(t)\end{pmatrix}}.}$ Then the above system can be re-written as:

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

And we have solutions ${\displaystyle {\vec {x}}_{1}(t)=C_{1}e^{6t}{\begin{pmatrix}1\\1\end{pmatrix}}}$ and ${\displaystyle {\vec {x}}_{2}(t)=C_{2}e^{2t}{\begin{pmatrix}1\\-1\end{pmatrix}}}$

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

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

for some constant matrix ${\displaystyle A}$.

By substituting into the equation we see that:

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

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