a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phas

A. Identify the equilibrium values. Which are stable and...

Key Concepts

Autonomous Differential Equations

Autonomous differential equations are those in which the derivative of the dependent variable depends only on the variable itself and not explicitly on the independent variable. This property allows for analysis through techniques such as phase line diagrams and qualitative methods, since the behavior of the system can be studied by looking solely at the function of the dependent variable.

Stability Analysis

Stability analysis involves determining whether small perturbations around an equilibrium solution grow or decay over time. An equilibrium is considered stable if nearby solutions tend to approach it, and unstable if they diverge away. Techniques such as linearization or the sign analysis of the derivative of the function at the equilibrium point are commonly used to assess stability.

Phase Line Analysis

Phase line analysis is a graphical technique used to summarize the qualitative behavior of an autonomous differential equation. It involves plotting equilibrium points on a line and indicating the direction of the solution trajectories based on the sign of the derivative. This method helps visualize the stability of equilibria and the overall dynamics of the system.

Qualitative Sketching of Solutions

Qualitative sketching of solutions refers to drawing approximate curves that represent the behavior of the solutions to a differential equation without solving it explicitly. By combining the information from the equilibria, stability analysis, and phase line analysis, one can illustrate how solutions evolve over time, indicating regions of increasing or decreasing behavior and capturing the general motion within the phase space.

Equilibrium Solutions

Equilibrium solutions, also known as steady states, are constant solutions of a differential equation where the derivative is zero. They are found by setting the right-hand side of the differential equation to zero and solving for the dependent variable. These solutions represent states where the system does not change in time.

Transcript

00:01 But this problem will be sketching the phase line and solution curve.

00:06 But first what we'll need to do is find the equilibrium point.

00:10 And we call that it where our derivative is equal to zero.

00:17 So here we can factor that as y times y plus one times y plus one.

00:29 And solving this gives us equilibrium points where y equals zero, y equals negative 1 and y equals positive 1 so to determine whether they're stable or unstable click sketch here when y is less than negative 1 you know that our derivative is less than 0 so that will tend to the left between negative 1 and 0 you know that the derivative is less than or greater than 0 so that will tend to the right between 0 and 1 if y is between 0 and 1 the derivative is negative so if our solution tends to the left and if it's greater than 1 then our derivative is positive so the solution will tend to the right towards infinity so from what we observe here the only stable point is where y equals 0 so this is our stable solution and at y equals negative one and one those would be unstable now if we go ahead and plot the baseline okay i'm going to add what we learned in part a to this baseline so this goes away one times less than zero one is greater than zero here what we need to add here is out the information about our second derivative.

02:47 So if y prime is equal to y -cuk minus y, then our second derivative is equal to three y -square times y prime minus y prime or three y squared minus one.

03:14 If it's a substitute y prime in here, it's one times y -cute minus y.

03:26 When we said that equal to zero, what we find is that our second derivative changes sign when y equal positive and negative 1 over the square of 3 negative 1 0 and 1 okay so i need to add positive and negative 1 divided by the score that's approximately close to a half so that would roughly be around here negative 1 over the square 3 positive 1 over the square with 3 now we can test for where our second derivative will be positive or negative.

04:18 At less than one, the second derivative is less than zero.

04:27 Here, the second derivative turns out to be greater than zero.

04:33 Here it turns out to be less than zero.

04:38 Here, greater than zero...

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