Author Topic: TT2--P4  (Read 2553 times)

Victor Ivrii

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TT2--P4
« on: March 21, 2018, 02:58:11 PM »
Find the general real solution to
$$
\mathbf{x}'=\begin{pmatrix}
2 & 3\\
-3 &2\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

Siyuan Tao

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Re: TT2--P4
« Reply #1 on: March 21, 2018, 03:01:36 PM »
Find eigenvalues with det(A−λI)=0: det(X-λI)=

$$\begin{pmatrix}2-\lambda&3\\-3&2-\lambda\end{pmatrix}=0\\(2-\lambda)(2-\lambda)-(-3)(3)=0\\\lambda^2-4\lambda+13=0\\$$
Since $\lambda={4\pm\sqrt{(-4)^2-4\cdot1\cdot13}\over2},$ we have $\cases{\lambda_1=2+3i\\\lambda_2=2-3i}$ $\\$
Find egienvectors with (A−λI)x=0, where x represents the eigenvectors:
when $\lambda=2+3i,$
$$\begin{pmatrix}-3i&3\\-3&-3i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Due to elementary row operation:
$$\begin{pmatrix}1&i\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies\begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\i\end{pmatrix}\text{where the eigenvector is}\space
\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\i\end{pmatrix}.$$
So, one of the solutions for the system is
$\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\i\end{pmatrix}e^{(2+3i)t}.$ $\\$
to obtain a set of real-valued solution:
$$=\begin{pmatrix}1\\i\end{pmatrix}e^2t(\cos3t+isin3t)\\=\begin{pmatrix}\cos3t\\-sin3t\end{pmatrix}e^{2t}+ie^{2t}\begin{pmatrix}sin3t\\cos3t\end{pmatrix}$$
Therefore, the general solution of the given system equation expressed in terms of real-valued functions is
$$\textbf{x}(t)=c_1e^{2t}\begin{pmatrix}\cos3t\\-sin3t\end{pmatrix}+c_2e^{2t}\begin{pmatrix}sin3t\\cos3t\end{pmatrix}$$

Victor Ivrii

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Re: TT2--P4
« Reply #2 on: March 24, 2018, 10:37:17 AM »
OK