### Author Topic: LEC5101 QUIZ6  (Read 2204 times)

#### fanqing2

• Newbie
• Posts: 2
• Karma: 0
##### LEC5101 QUIZ6
« on: November 15, 2019, 01:07:58 PM »
Question:

a. Find the general solution of the given system of equations and describe the behavior of the solution as $t \rightarrow \infty$.

b. Draw a direction field and plot a few trajectories of the system.

${x}^{\prime}=\left(\begin{array}{ll}{3} & {-2} \\ {2} & {-2}\end{array}\right) {x}$

$A=\left|\begin{array}{cc}{3} & {-2} \\ {2} & {-2}\end{array}\right|$

det$(A-r I)={\rm det}\left|\begin{array}{cc}{3-r} & {-2} \\ {2} & {-2-r}\end{array}\right|=(r+2)(r-3)+4=0$

$(r+2)(r-3)+4=0$

$r^{2}-3 r+2 r-6+4=0$

$r^{2}-r-2=0$

$(r-2) \cdot(r+1)=0$

$r_{1}=2 \quad, \quad r_{2}=-1$

For $r_{1}=2$

$A-r_{1} I=\left|\begin{array}{cc}{3-2} & {-2} \\ {2} & {-2-2}\end{array}\right|=\left|\begin{array}{cc}{1} & {-2} \\ {2} & {-4}\end{array}\right|$

$\left|\begin{array}{cc:cc}{1} & {-2} & {0} \\ {2} & {-4} & {0}\end{array}\right|$ ${R_{2}-2 R_{1}}$ $\left|\begin{array}{cc:cc}{1} & {-2} & {0} \\ {0} & {0} & {0}\end{array}\right|$

$x_{2}=t \quad x_{1}-2 x_{2}=0$

$x_{1}=2 t$

$\left|\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right|=\left|\begin{array}{l}{2} \\ {1}\end{array}\right|t$

For $r_{2}=-1$

$A-r I=\left|\begin{array}{cc}{3-(-1)} & {2} \\ {2} & {-2-(-1)}\end{array}\right|=\left|\begin{array}{cc}{4} & {-2} \\ {2} & {-1}\end{array}\right|$

$\left|\begin{array}{cc:cc}{4} & {-2} & {0} \\ {2} & {-1} & {0}\end{array}\right|$ ${R_{2} \times 2-R_{1}}$ $\left|\begin{array}{cc:cc}{4} & {-2} & {0} \\ {0} & {0} & {0}\end{array}\right|$

$x_{2}=t \quad 4 x_{1}-2 x_{2}=0$

$x_{1}=\frac{1}{2} t$

$\left|\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right|=\left|\begin{array}{c}{\frac{1}{2}} \\ {1}\end{array}\right|t$$=\left|\begin{array}{l}{1} \\ {2}\end{array}\right|t$

The general solution is

$x=c_{1} e^{-t}\left|\begin{array}{l}{1} \\ {2}\end{array}\right|+c_{2} e^{2 t}\left|\begin{array}{l}{2} \\ {1}\end{array}\right|$

$\left\{\begin{array}{l}{x_{1}(t)=c_{1} e^{-t}+2 c_{2} e^{2 t}} \\ {x_{2}(t)=2 c_{1} e^{-t}+c_{2} e^{2 t}}\end{array}\right.$

when $t \rightarrow+\infty$

when $c_{2}=0,\lim \limits_{t \rightarrow+\infty} x_{1}(t)=0$ and $\lim \limits_{t \rightarrow +\infty} x_{2}(t)=0$

Then $x$ approaches to zero as $t \rightarrow+\infty$

when $c_{2} \neq 0,\lim \limits_{t \rightarrow+\infty} x_{1}(t)=+\infty$ and $\lim \limits_{t \rightarrow+\infty} x_{2}(t)=+\infty$

Then $x$ approaches to $+\infty$ as $+\rightarrow +\infty$

$(b) \quad r_{1}=2>0 \quad\quad r_{2}=-1<0$
The graph is in the attachment