Stability of Equilibrium solutions to ODE

[depre87]

Homework Statement

y'=(1-y)(3-y)(5-t)

Homework Equations

find equilibrium solutions of ODE and determine their stability

The Attempt at a Solution

equilibrium solutions are y = 1 and y = 3, I'm not sure how to determine their stability without some form of a slope field, is it possible to do so without one if so how? thanks!

 

[HallsofIvy]

There are three equilibrium solutions: y= 1, y= 3, and y= 5 because dy/dx is 0 at each of those three.

And you can determine stability just by looking at the sign of y' in each interval. I am going to write this as dy/dx= (-1)(x- 1)(x- 3)(x- 5) to get all factor in the form "x- a". If y< 1, all three of y-1, y- 3, and y- 5 will be negative. The product of four negative numbers (including (-1)) is positive which means that dy/dx= (-1)(y- 1)(y- 3)(y- 5) will be positive. Positive derivative means y is increasing toward from y= 1. If y is beween 1 and 3, y-1 is now positive while y- 3 and y- 5 are still negative. dy/dx is now the product of one positive and three negative numbers and so is now negative. That means y is decreasing- toward from y= 1, away from y= 3. For y between 3 and 5, both y-1 and y- 3 are positive, now y- 5 is negative so dy/dx is the product of two positive and two negative numbers and is positive. That means...

 

[jackmell]

Got a t in there Hall.

May I suggest a non-standard approach which I think would be helpful? Ok, draw it first, then interpret the slop field algebraically. You can draw the slope field easily by converting it to a coupled system:

$$\frac{dt}{dt}=1$$

$$\frac{dy}{dt}=(1-y)(3-y)(5-t)$$

Now just do a StreamPlot in Mathematica:

StreamPlot[{1,(1-y)(3-y)(5-t)j},{t,0,10},{y,-10,10}]

Note what's happening around t=5. The derivative can change from positive to negative without y doing so.