How to Sketch a Phaseline for Differential Equations?

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SUMMARY

The discussion focuses on sketching the phaseline for the differential equation x' = f(x) + a, where f(x) = (x+1)²(1-x). Participants explore how to qualitatively depict the phaseline for various values of 'a' (-2, -1, 0, 1) and identify conditions for three equilibrium points. It is established that for three equilibrium points to exist, the value of -a must lie between the maxima and minima of the function f(x).

PREREQUISITES
  • Understanding of differential equations and their graphical representations
  • Familiarity with the concept of equilibrium points in dynamical systems
  • Knowledge of maxima and minima in calculus
  • Ability to sketch qualitative graphs of functions
NEXT STEPS
  • Study the properties of equilibrium points in differential equations
  • Learn how to find maxima and minima of functions using calculus
  • Explore the concept of phase portraits in dynamical systems
  • Practice sketching phaselines for various differential equations
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Students and educators in mathematics, particularly those focusing on differential equations and dynamical systems, as well as anyone looking to improve their skills in qualitative analysis of functions.

SSJVegetto
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Hello all,

I have the following differential equation: [itex]x' = f(x) + a[/itex] with [itex]f(x) = (x+1)^{2}.(1-x)[/itex]

Now i have the following questions:
1. Sketch the phaseline for [itex]x' = f(x) + a[/itex] with [itex]a = -2, -1, 0, 1[/itex]
Don't calculate the exact intersections but make a qualitative correct picture.

2. Give the values for a where there are 3 equilibrium points. Hint: What are the maxima and
minima of f.

I really am trying but i don't understand how to solve this question and i really need some help on how to do this one.
 
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For (2), you need to solve x'=0= f(x)+a. Hence, (-a) must lie within the maxima & minima of f, so that the line y=-a cuts the graph of f in 3 distinct points.
 

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