Question

Computer Lab Assignments; Find phase portraits for the systems in Problems 22 and 23. For each system and any half-line trajectories and include these lines in your phase portrait.

Short answer

Viewing the graph without the parameter, we see $y=x,x<0$

Viewing the graph without the parameter, we see $y=x,x<0$

Step-by-step solution

Define half-line trajectories and include them in your facial portrait.

In the phase plane, a phase portrait is a geometric representation of a dynamical system's trajectories. A separate curve, or point, represents each set of initial conditions.

Phase portrait of the system

$x^1=\left(\begin{array}{cc}-6& 5\\ -5& 4\end{array}\right)x$is shown below. We previously found that the general solution of the system is

$x^1=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{-t}+c_2\left[\left(\begin{array}{c}1\\ 1\end{array}\right)t+\left(\begin{array}{c}0\\ 1/5\end{array}\right)\right]e^{-t}$

Let$c_1=1$and$c_2=0$we get$x=e^{-t},y=e^{-t}$

Eliminating the parameter, we find$y=x,x>0$

Let$c_1=-1$and$c_2=0$we get $x=-e^{-t},y=-e^{-t}$

Eliminating the parameter, we find $y=x,x<0$

Step 2:The system's phase portrait is depicted below. We previously discovered that the system's overall solution is

Phase portrait of the system

$x^1=\left(\begin{array}{cc}-1& 3\\ -3& 5\end{array}\right)x$is shown below. We previously found that the general solution of the system is

$x=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)e^{2t}+c_2\left[\left(\begin{array}{c}1\\ 1\end{array}\right)t+\left(\begin{array}{c}\frac{{\displaystyle -1}}{3}\\ 0\end{array}\right)\right]e^{2t}$

Let$c_1=1$and$c_2=0$we get $x=e^{2t},y=e^{2t}$

Eliminating the parameter, we find$y=x,x>0$

Let$c_1=-1$and$c_2=0$we get $x=-e^{2t},y=-e^{2t}$

Eliminating the parameter, we find $y=x,x<0$

As a result, the overall solution has been developed. View the half-line trajectories on a graph and add these lines in your phase portrait$y=x,x<0$