/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 The planar, autonomous system ha... [FREE SOLUTION] | 91Ó°ÊÓ

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The planar, autonomous system having the form $$ \begin{aligned} &x^{\prime}=a x+b y \\ &y^{\prime}=c x+d y \end{aligned} $$ where \(a, b, c\), and \(d\) are arbitrary real numbers, is called a linear system of first order differential equations. Exercises 1 \(-6\) each contain a solution of some linear system. Use MATLAB to create a plot of \(x\) versus \(t, y\) versus \(t\), and a plot of \(y\) versus \(x\) in the phase plane. Use the subplot command, as in subplot (221), plot (t, \(x)\), axis tight subplot(222), plot(t,y), axis tight subplot (221), plot \((t, x)\), axis tight subplot (222), plot (t,y), axis tight subplot (212), plot( \(x, y)\), axis equal subplot (212), plot \((x, y)\), axis equal to produce a plot containing all three plots. Use the suggested time interval. $$ \begin{aligned} &x=-2 e^{-2 t}+3 e^{-3 t} \\ &y=4 e^{-2 t}-3 e^{-3 t} \\ &{[-0.5,2]} \end{aligned} $$

Short Answer

Expert verified
Plot the given functions using MATLAB over the interval [-0.5, 2] with subplots for x vs. t, y vs. t, and y vs. x.

Step by step solution

01

Define the Time Interval

First, specify the time interval over which you want to plot the solutions. According to the exercise, the interval is from -0.5 to 2. We'll create a vector for time, \( t \), using MATLAB code: ```matlabt = linspace(-0.5, 2, 100);```This generates 100 points between -0.5 and 2.
02

Calculate x and y

Compute the values for \( x(t) \) and \( y(t) \) using the given functions:\[x = -2 e^{-2t} + 3 e^{-3t} \y = 4 e^{-2t} - 3 e^{-3t}\]In MATLAB, you would write:```matlabx = -2 * exp(-2 * t) + 3 * exp(-3 * t);y = 4 * exp(-2 * t) - 3 * exp(-3 * t);```
03

Plot x versus t

Use the `subplot` function to plot \( x(t) \) versus \( t \):```matlabsubplot(221), plot(t, x), axis tighttitle('x versus t')xlabel('t')ylabel('x')```
04

Plot y versus t

Create a plot for \( y(t) \) versus \( t \) using the `subplot` function:```matlabsubplot(222), plot(t, y), axis tighttitle('y versus t')xlabel('t')ylabel('y')```
05

Plot y versus x in the Phase Plane

In this step, plot \( y \) against \( x \) to produce the phase plane trajectory:```matlabsubplot(212), plot(x, y), axis equaltitle('Phase Plane: y versus x')xlabel('x')ylabel('y')```
06

Adjust the Subplots

Ensure that all plots are aligned properly using `axis tight` for time series plots, which automatically adjusts the axis limits to fit the data tightly and `axis equal` for the phase plane plot to maintain equal scaling of data on both axes. These ensure that the data visualization is accurate and comprehensible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

MATLAB plotting
MATLAB is a powerful tool for visualizing mathematical functions and their behaviors. When working with linear systems of differential equations, plotting can help understand how these systems evolve over time. In the exercise, MATLAB is used to plot solution curves of the system, like \(x(t)\), \(y(t)\), and their relation. This involves commands like `subplot`, which allows plotting multiple graphs in a single figure. This is essential for comparing different solutions side-by-side.
The `plot` function is another vital command. It creates a graph based on the data you provide. For instance, `plot(t, x)` graphs the value of \(x(t)\) over time. When adding labels or titles with `xlabel`, `ylabel`, and `title`, we make the graph easier to read and understand.
Additionally, formatting using `axis tight` and `axis equal` helps by adjusting the graph's layout so that the data fits neatly within the screen, giving accurate scale and comparison between the plotted lines. This is especially important when dealing with complex data like differential equations.
phase plane analysis
Phase plane analysis is a technique used to study autonomous systems by plotting their trajectories in a coordinate plane defined by the dependent variables, like \(x\) and \(y\). This gives a visual representation of the system's behavior over time, making it easier to see patterns or types of motion.
In our linear system, the phase plane plot of \(y\) versus \(x\) helps to visualize how these two variables interact. This is crucial in identifying key features such as equilibrium points or the nature of trajectories that the system follows. Understanding the phase plane gives insight into the stability and long-term behavior of the system. It can indicate whether solutions converge to a steady state, cycle, or diverge.
With the help of MATLAB’s `subplot` function, this analysis becomes straightforward. You can view how the solution behaves over time and across variables in a single figure, enhancing comprehension and interpretation.
autonomous systems
An autonomous system is one where the system of differential equations does not explicitly depend on the independent variable, usually time. This means that their behavior depends purely on the state of the system, such as the variables \(x\) and \(y\).
In the exercise's context, the given differential equation has no explicit \(t\) dependence within its structure. As a result, the evolution of \(x\) and \(y\) depends solely on how they influence each other through parameters \(a, b, c,\) and \(d\).
Autonomous systems often exhibit predictable patterns or behavior, which can be modeled and visualized more easily using phase plane analysis. Understanding these systems can be greatly beneficial because it allows for predictions on the equilibrium solutions or the stability of the dynamics over time.
Autonomous structures are common in real-world scenarios, like population dynamics or chemical reactions, making this concept applicable and valuable in various fields.
first order differential equations
First order differential equations involve derivatives of the first degree, which means they include terms like \(x'\) or \(y'\). In a linear system, each equation is linear in its dependent variable and its derivative.
The exercise presents a first-order system, which describes how the rate of change of \(x\) and \(y\) relies on their current values. The simplicity of first-order equations makes them easier to analyze, yet they can model complex systems and behaviors.
To solve first-order linear differential equations, one typically uses techniques like separation of variables or matrix methods, but MATLAB provides numerical solutions and direct plotting of these equations. This ease of computation is invaluable for checking analytical solutions or exploring the dynamics of models too complex for manual computations.
Understanding these equations is foundational in differential equations, offering insight into how systems behave, change, or stabilize over time. When combined with computational tools, even intricate systems become accessible for analysis.

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Most popular questions from this chapter

Enter the system $$ \begin{aligned} &x^{\prime}=-\cos y+2 y \cos y^{2} \cos 2 x \\ &y^{\prime}=-\sin x+2 \sin y^{2} \sin 2 x \end{aligned} $$ in the PPLANE6 Setup window. Set the display window to \(-10 \leq x \leq 10\) and \(-2 \leq y \leq 4\) and select None for the vector field. a) Select File \(\rightarrow\) Save the current system and save the system with the name teddybears.pps. b) Select Gallery \(\rightarrow\) linear system to load the linear system template. Select File \(\rightarrow\) Load a system and load the system teddybears.pps. c) Select Solutions \(\rightarrow\) Keyboard input and start a solution trajectory with initial condition \((\pi / 2,0)\) to create the "legs" of the Teddy bears, and a trajectory with initial condition \((-7.2,3.5)\) to create the "heads" of the Teddy bears. d) Experiment further with this "wild and wooly" example. Click the mouse in the phase plane to start other solution trajectories. Can you find the "eyes" of the Teddy bears?

Repeat the pendulum experiment of the previous problem. Displace the mass \(30^{\circ}\) counter-clockwise (positive \(\pi / 6\) radians) from the vertical, only this time do not release the mass from rest. Instead, push the mass in the clockwise (negative) direction with enough negative angular velocity so that it spins around in a circle exactly one time before settling into a motion decaying to a stable equilibrium. The tricky part of this experiment is the fact that the stable equilibrium point is now \(\theta=-2 \pi, \omega=0\). a) Without using any technology, sketch graphs of \(\theta\) versus \(t\) and \(\omega\) versus \(t\) approximating the motion of the pendulum. b) Without using any technology, sketch graphs of \(\omega\) versus \(\theta\). Place \(\omega\) on the vertical axis, \(\theta\) on the horizontal axis. Note: This is a lot harder than it looks. We suggest that you work with a partner or a group and compare solutions before moving on to part c). c) Select Gallery \(\rightarrow\) pendulum in the PPLANE6 Setup window. Adjust the damping parameter to \(D=0.1\), and set the display window so that \(-10 \leq \theta \leq 5\) and \(-4 \leq \omega \leq 4\). Select Options \(\rightarrow\) Solution direction \(\rightarrow\) Forward and use the Keyboard input window to start a solution trajectory with initial condition \(\theta(0)=\pi / 6 \mathrm{rad}\) and \(\omega(0)=-2.5 \mathrm{rad} / \mathrm{s}\). Compare this result with your hand-drawn solution in part b). Select Graph \(\rightarrow\) Both and click your solution trajectory in the phase plane to produce plots of \(\theta\) versus \(t\) and \(\omega\) versus \(t\). Compare these with your hand- drawn solutions in part a).

The equation \(m y^{\prime \prime}+d y^{\prime}+k y=0\) represents a damped, spring-mass system. In Exercises \(23-26\), values of the parameters \(m, d\), and \(k\) have been chosen to create a specific example of a damped, spring-mass system. We are looking for the solution \(y\) with the given initial position and velocity. Let $$ x_{1}=y, \quad \text { and } \quad x_{2}=y^{\prime}, $$ and use this change of variables to write each spring-mass system as a planar, autonomous system. Use pplane6 to obtain a printout of the graph of \(y\) versus \(t\). $$ \begin{aligned} &y^{\prime \prime}+3 y^{\prime}+2 y=0 \\ &y(0)=3, y^{\prime}(0)=2 \end{aligned} $$

Consider the predator-prey system $$ \begin{aligned} &R^{\prime}=R-R F, \\ &F^{\prime}=-F+R F, \end{aligned} $$ where \(R\) and \(F\) represent rabbit and fox populations, respectively. Enter the system in pplane6 and set the display window so that \(0 \leq R \leq 2\) and \(0 \leq F \leq 2\). Use Keyboard input to start a solution trajectory at \(R=0.5\) and \(F=1\). Note that the trajectory is periodic. Use pplane6 to estimate the time it takes to travel this periodic trajectory exactly once; i.e., find the period of the oscillation. Hint: Try cropping an \(x\) versus t plot.

The planar, autonomous system having the form $$ \begin{aligned} &x^{\prime}=a x+b y \\ &y^{\prime}=c x+d y \end{aligned} $$ where \(a, b, c\), and \(d\) are arbitrary real numbers, is called a linear system of first order differential equations. Exercises 1 \(-6\) each contain a solution of some linear system. Use MATLAB to create a plot of \(x\) versus \(t, y\) versus \(t\), and a plot of \(y\) versus \(x\) in the phase plane. Use the subplot command, as in subplot (221), plot (t, \(x)\), axis tight subplot(222), plot(t,y), axis tight subplot (221), plot \((t, x)\), axis tight subplot (222), plot (t,y), axis tight subplot (212), plot( \(x, y)\), axis equal subplot (212), plot \((x, y)\), axis equal to produce a plot containing all three plots. Use the suggested time interval. $$ \begin{aligned} &x=-2 e^{-2 t}-e^{3 t} \\ &y=e^{-2 t}+e^{3 t} \\ &{[-1.5,1.0]} \end{aligned} $$

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