Author Topic: Q6--T0501  (Read 1470 times)

Meng Wu

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Q6--T0501
« on: March 16, 2018, 07:19:49 PM »
For the given equation:$\\$
    a. Express the general solution of the given system of equations in terms of real-valued functions. $\\$
    b. Also draw a direction field, sketch a few of the trajectories, and describe the behaviour of solutions as $t \rightarrow \infty$.
$$\textbf{x}'=\begin{pmatrix}2&-5\\1&-2\end{pmatrix}\textbf{x}$$

Meng Wu

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Re: Q6--T0501
« Reply #1 on: March 16, 2018, 07:20:06 PM »
(a)$\\$
Find eigenvalues with $\det(\boldsymbol{A}-\lambda\boldsymbol{I_2})=0:$
$$\begin{aligned}\begin{array}{|c c|}2-\lambda&-5\\1&-2-\lambda\end{array}=0\\(2- \lambda)(-2-\lambda)-(-5)(1)=0\\\lambda^2+1=0\end{aligned}$$
We have $\cases{\lambda_1=i\\\lambda_2=-i}$ $\\$
Find egienvectors with $(\boldsymbol{A}-\lambda\boldsymbol{I_2})\textbf{x}=\boldsymbol{0}$, where $\textbf{x}$ represents the eigenvectors.$\\$
When $\lambda=i,$ $$\begin{pmatrix}2-i&-5\\1&-2-i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2-i&-5\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}5\\2-i\end{pmatrix}\text{where the eigenvector is}\space \boldsymbol{\xi}^{(1)}=\begin{pmatrix}5\\2-i\end{pmatrix}.$$ Thus one solution of the given system is $\textbf{x}^{(1)}(t)=\begin{pmatrix}5\\2-i\end{pmatrix}e^{it}.$ $\\$
When $\lambda=-i,$ $$\begin{pmatrix}2+i&-5\\1&-2+i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2+i&-5\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}5\\2+i\end{pmatrix}\text{where the eigenvector is}\space \boldsymbol{\xi}^{(2)}=\begin{pmatrix}5\\2+i\end{pmatrix}.$$ Thus one solution of the given system is $\textbf{x}^{(2)}(t)=\begin{pmatrix}5\\2+i\end{pmatrix}e^{it}.$ $\\$
To obtain a set of real-valued solution:
$$\begin{align}\textbf{x}^{(1)}(t)&=\begin{pmatrix}5\\2-i\end{pmatrix}e^{0}(\cos t+i\sin t)\\&=\begin{pmatrix}5\cos t\\ 2\cos t+\sin t\end{pmatrix}+i\begin{pmatrix}5\sin t\\2\sin t-\cos t\end{pmatrix} \\&=\boldsymbol{u}(t)+i\boldsymbol{v}(t)\end{align}$$
By the Theorem$\space 7.4.5$, we know that $\boldsymbol{u}(t)\space\text{and}\space\boldsymbol{v}(t)$ are also solutions of given equation. $\\$
Therefore, the general solution of the given system equation expressed in terms of real-valued functions is $$\textbf{x}(t)=c_1\begin{pmatrix}5\cos t\\ 2\cos t+\sin t\end{pmatrix}+c_2 \begin{pmatrix}5\sin t\\2\sin t-\cos t\end{pmatrix}$$
(b)$\\$
All solutions are periodic as $t\rightarrow \infty$. The origin $\textbf{x}=\boldsymbol{0}$ is a center.
« Last Edit: March 16, 2018, 07:22:55 PM by Meng Wu »