I need to solve the following system of differential equations:
$$ \ddot{x} = 8x + 4y \\
\ddot{y} = -4x$$
Here's what I've done so far: I have reduced this system to a first order system, by saying $x_1 := x, \ x_2 := \dot{x}, \ x_3 := y, \ x_4 := \dot{y}$. This yields the system $\dot{X} = A \cdot X$ with
$$ A = \begin{pmatrix} 0 & 1 & 0 & 0\\ 8 & 0 & 4 &0\\0 & 0 & 0 &1\\ -4 & 0 & 0 &0\end{pmatrix} \ \ \ X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\x_4 \end{pmatrix}$$
Then I've determined the eigenvalues $\lambda_1 = 2, \ \lambda_2 = -2$, with the corresponding eigenvectors $v_1 = \begin{pmatrix}1 & 2 & -1 & -2\end{pmatrix}^{T}$ and $v_2 = \begin{pmatrix}1 & -2 & -1 & 2\end{pmatrix}^{T}$.
Now what I'm struggling with is: how do I determine my set of fundamental solutions? I know that the terms $c_1e^{2t}$ and $c_2e^{-2t}$ are part of it for sure, but since I have two double eigenvalues, I also should have a solution somewhat like $te^{2t}$ resp. $te^{-2t}$. But I just don't see how they alle come together.
Best Answer
You have an eigenvalue $\lambda$ and its eigenvector $v_1$. So one of your solutions will be $$ x(t) = e^{\lambda t} v_1$$ As you've noticed however, since you only have two eigenvalues (each with one eigenvector), you only have two solutions total, and you need four to form a fundamental solution set. For each eigenvalue $\lambda$, you will calculate what's called a generalized eigenvector $v_2$, which is the solution to $$ (A - \lambda I)v_2 = v_1, \quad \text{where } (A - \lambda I)v_1 = 0;$$ in other words, $v_1$ was the first eigenvector. Then this contributes a new solution $$ x(t) = e^{\lambda t} v_2 + te^{\lambda t} v_1 $$ Now you have two linearly independent solutions corresponding to one eigenvalue. Now repeat the process for the second eigenvalue to get all four elements of your fundamental solution set.
Note: What I'm calling $v_2$ is NOT the same as what you're calling $v_2$ in your question.