# Solving linear system of differential equations of 2nd order

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.

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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.

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Thank you! I've came across similar formulas when looking for an answer, but never quite fully understood them. They basically stated the same process as you did, but the formula for determing the second eigenvector sometimes had an exponent, like $(A-\lambda I)^p v_2 = v1$ where $\lambda$ is a $p$th eigenvector. According to your solution, I don't need this exponent? –  StringerBell Apr 17 '13 at 23:04
That does come up if you want even more solutions because you don't have enough linearly independent (genuine) eigenvectors. Then you have to solve the same problem linear equation for successively higher powers $p$ in order to get $v_3, v_4$, etc, until you've exhaustively found a full set of solutions. –  Christopher A. Wong Apr 18 '13 at 0:13

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

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

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

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

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

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

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

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

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This would probably work in this specific case, but actually I wanted to understand the most generic approach (if such exists at all) - but of course any solution is welcome, so thank you! Just one question: in your solution, how did you figure out that the exponent of the fundamental system is $2$ resp. $-2$? –  StringerBell Apr 18 '13 at 18:01