The course I am taking is using the text Differential Equations by Blanchard, Devaney, and Hall, and I want to solve the following problem:
Solve $$\frac{dx}{dt}=-2x-2y, \quad \frac{dy}{dt}=-2x+y$$ where the initial condition $(x(0), y(0))$ is $(1, 0)$.
I started by finding the eigenvalues $$\lambda_1 = 1, \quad \lambda_2 = -2.$$ I then found the eigenvectors $$-2y_1=x_1, \quad -2x_1 = 0.$$ But I'm not sure where to go from here.
On
Maybe it's best not to get so technical here.
You are given $\tag1x'=Ax;\ x(0)=x_0$ as defined above.
If we take a proposed solution to be $x(t)=ve^{\lambda t}$, then on substituting into $(1)$ we are led to
$\tag2Av=\lambda v$
and so solving the differential equation amounts to solving an eigenvalue-eigenvector problem, which is what you are in fact doing. If the eigenvalues are distinct, you will get distinct eigenvectors $v_1$ and $v_2$ such that
$\tag3x(t)=av_{1}e^{\lambda _{1}t}+bv_{2}e^{\lambda _{2}t}$
and to finish, substitute the initial condition to determine $a$ and $b$.
The solution to a linear ODE system $x'=Ax,x(0)=x_0$ (where $A$ is a constant square matrix) is given by $x(t)=e^{At}x_0$. This should not be particularly surprising in view of the 1D case, but it also is not very useful by itself, because it is not obvious how to compute $e^{At}$, which is defined as $\sum_{k=0}^\infty \frac{t^k A^k}{k!}$.
The most convenient way to compute $e^{At}$ is if we have diagonalized $A$. Specifically, if $A$ has eigenvalue matrix $D$, and eigenvector matrix $P$, then $A^k=PD^kP^{-1}$. Since $D$ is diagonal, it is easy to compute $D^k$: you just raise each diagonal entry $\lambda_i$ to the power of $k$. Plugging this into the definition of $e^{At}$, we get $Pe^{Dt}P^{-1}$. Again $e^{Dt}$ is very simple: each diagonal entry is $e^{\lambda_i t}$.
So $x(t)=Pe^{Dt}P^{-1}x_0$. One can alternately write this as $x(t)=\sum_{j=1}^n c_j e^{\lambda_j t} v_j$, where $v_j$ are the eigenvectors and $c_j$ are coefficients depending on $x_0$.