22–25. Equilibrium solutions Find the equilibrium solutions of...

Question

22–25. Equilibrium solutions Find the equilibrium solutions of the following equations and determine whether each solution is stable or unstable.

y′(t) = y(3+y)(y-5)

Accepted Answer

Welcome back, everyone. Find the equilibrium solutions of the equation C T equals C multiplied by Z minus 1 multiplied by C + 6, and classify each as stable or unstable. First of all, let's begin by finding the equilibrium solutions, and this means that we're going to set Z T equal to 0. In other words, Z multiplied by Z minus 1, multiplied by Z + 6. is equal to 0. We have a product in this case, and we can easily find our solutions. First of all, we can either have Z equal to 0. Then, considering our second factor, C minus 1 can be equal to 0, which means that Z is going to be 1, right? So our second solution is Z equals 1. And for the third factor, if Z + 6 is equal to 0, then Z is equal to -6. Well done, so now we have 3 equilibrium solutions, 01, and -6. We want to classify each as stable or unstable. What we're going to do is simply identify the second derivative. So we're going to evaluate Z of T. We are going to expand the factor form, and we're going to get Z cubed plus 5 Z2 minus 6 Z. And we're going to perform differentiation to get the 2nd derivative. The derivative is going to be 3 Z2 plus 10Z minus 6. And now if our second derivative at an equilibrium solution is greater than 0, we have an unstable solution. Otherwise, if our 2nd derivative is less than 0, it is a stable solution. So let's evaluate Z at each equilibrium point starting with 0. That would be 3 multiplied by. 00 squared. Plus 10 multiplied by 0 minus 6, and this is equal to -6. Now because it is less than 0, it is going to be a stable equilibrium solution. This is how we're going to proceed. Now let's valid C at 1. Which is 3 multiplied by 1 squared, plus 10 multiplied by 1 minus 6. So this is 13 minus 6. Which is 7, and now this is greater than 0, meaning it is unstable, right? And finally, we are going to valid C at -6. Which is 3 multiplied by -6 squared plus. Then multiplied by -6 minus 6. We got 36 multiplied by 3 minus 60 minus 6, and that's 42. Which is greater than 0, and because it's greater than 0 again, it's unstable. Now let's summarize the results. So we have our equilibrium solutions. Z1 is equal to 0. Which is stable. Followed by Z2, which is 1. And it is unstable. And finally, C3 is equal to -6, which is also unstable. That would be our final answer for this problem. Thank you for watching.

22–25. Equilibrium solutions Find the equilibrium solutions of the following equations and determine whether each solution is stable or unstable.
y′(t) = y(3+y)(y-5)

Key Concepts

Equilibrium Solutions

Stability of Equilibrium Solutions

Sign Analysis of the Derivative Function

Watch next