Solving linear system of differential equations of 2nd order

Question

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.

Answer 1

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.

Answer 2

In fact using the classical substitution method should be even simpler:

$\begin{cases}\ddot{x}=8x+4y\\\ddot{y}=-4x\end{cases}$

$\therefore\ddddot{x}=8\ddot{x}+4\ddot{y}=8\ddot{x}-16x$

$\ddddot{x}-8\ddot{x}+16x=0$

$x=C_1te^{2t}+C_2te^{-2t}+C_3e^{2t}+C_4e^{-2t}$

$\dot{x}=2C_1te^{2t}+C_1e^{2t}-2C_2te^{-2t}+C_2e^{-2t}+2C_3e^{2t}-2C_4e^{-2t}=2C_1te^{2t}-2C_2te^{-2t}+C_2e^{-2t}+(C_1+2C_3)e^{2t}+(C_2-2C_4)e^{-2t}$

$\ddot{x}=4C_1te^{2t}+2C_1e^{2t}+4C_2te^{-2t}-2C_2e^{-2t}+2(C_1+2C_3)e^{2t}-2(C_2-2C_4)e^{-2t}=4C_1te^{2t}+4C_2te^{-2t}+4(C_1+C_3)e^{2t}-4(C_2-C_4)e^{-2t}$

$\therefore y=\dfrac{\ddot{x}}{4}-2x=C_1te^{2t}+C_2te^{-2t}+(C_1+C_3)e^{2t}-(C_2-C_4)e^{-2t}-2C_1te^{2t}-2C_2te^{-2t}-2C_3e^{2t}-2C_4e^{-2t}=-C_1te^{2t}-C_2te^{-2t}+(C_1-C_3)e^{2t}-(C_2+C_4)e^{-2t}$