基于$Y(t)=e^{tw}V$形式，求解$rac{d^2Y}{dt^2}=AY$的通解

Let's break down how to find the general solution using the given eigenvalues and eigenvectors of matrix $A$.

First, recall that for systems of the form $\frac{d^2Y}{dt2} = AY$, we look for solutions of the form $Y(t) = e^{rt}V$, where $r$ is a constant and $V$ is an eigenvector of $A$. Substituting this into the equation gives us $r^2 V = AV$, meaning $r^2$ must be an eigenvalue of $A$.

Step 1: Handle Eigenvalue $\lambda = 3$ #

For $\lambda = 3$, we have $r^2 = 3$, so $r = \sqrt{3}$ or $r = -\sqrt{3}$. Using the corresponding eigenvectors $\begin{bmatrix}1 \ 0 \ -1\end{bmatrix}$ and $\begin{bmatrix}0 \ 1 \ -1\end{bmatrix}$, we get four linearly independent solutions:

• $e^{\sqrt{3}t}\begin{bmatrix}1 \ 0 \ -1\end{bmatrix}$
• $e^{-\sqrt{3}t}\begin{bmatrix}1 \ 0 \ -1\end{bmatrix}$
• $e^{\sqrt{3}t}\begin{bmatrix}0 \ 1 \ -1\end{bmatrix}$
• $e^{-\sqrt{3}t}\begin{bmatrix}0 \ 1 \ -1\end{bmatrix}$

Step 2: Handle Eigenvalue $\lambda = 0$ #

For $\lambda = 0$, $r^2 = 0$, so $r = 0$ (a repeated root). When $r=0$, the basic solution is $V = \begin{bmatrix}1 \ 1 \ 1\end{bmatrix}$, but since it's a double root, we need a second linearly independent solution: $tV$. These two solutions are:

• $\begin{bmatrix}1 \ 1 \ 1\end{bmatrix}$
• $t\begin{bmatrix}1 \ 1 \ 1\end{bmatrix}$

General Solution #

Combining all linearly independent solutions, the general solution is a linear combination of all six solutions:

$$
Y(t) = C_1 e^{\sqrt{3}t}\begin{bmatrix}1 \ 0 \ -1\end{bmatrix} + C_2 e^{-\sqrt{3}t}\begin{bmatrix}1 \ 0 \ -1\end{bmatrix} + C_3 e^{\sqrt{3}t}\begin{bmatrix}0 \ 1 \ -1\end{bmatrix} + C_4 e^{-\sqrt{3}t}\begin{bmatrix}0 \ 1 \ -1\end{bmatrix} + C_5 \begin{bmatrix}1 \ 1 \ 1\end{bmatrix} + C_6 t\begin{bmatrix}1 \ 1 \ 1\end{bmatrix}
$$

Where $C_1, C_2, C_3, C_4, C_5, C_6$ are arbitrary constants.

内容的提问来源于stack exchange，提问作者M.Byrne