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: