Bifurcation Diagram <Diff. Eqns>

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SUMMARY

The discussion centers on constructing a bifurcation diagram for the differential equation dP/dt = P(4-P) - h. The critical points, or equilibrium solutions, are identified as P(t) = 0 and P(t) = 4 when h equals 0. However, the equilibrium solutions depend on the parameter h, leading to a quadratic equation P² - 4P + h = 0. The bifurcation diagram is characterized by plotting h on the horizontal axis and the equilibrium values of P on the vertical axis, resulting in a sideways parabola representing the relationship between h and P.

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dalarev
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Homework Statement



Draw the bifurcation diagram for the following equation.

:attached:

Homework Equations



I believe this is a population D.E.

The Attempt at a Solution



I'm ashamed to say I'm at a complete loss with this problem. The only step I'm familiar with is finding the critical points, which would be p=0 and p=4 in this case. My professor gave is this problem and told us to study it for a final tomorrow. Help would be GREATLY appreciated.
 

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Your differential equation is dP/dt= P(4-P)- h. (Why make a gif and attach it for something that easy to type?) No, you did not find the critical points (more correctly "equilbrium solutions" since those are constant functions that satisfy the equation) . Equillibrium solutions are such that dP/dt= P(4-P)-h= 0. They would be P(t)= 0 and P(t)= 4 only if h were equal to 0. As it is, you have the quadratic equation P^2- 4P+ h= 0 so the equilibrium solutions depend upon h. You can use the quadratic formula to write them. You will note that for some values of h, there is no (real) solution. For some values of h, there are two distinct solutions and for precisely one value of h, there is a single solution.

The "bifurcation diagram" is a graph with h on the horizontal axis, the equilibrium values of P on the vertical axis. For this problem that should be a parabola lieing on it's side: precisely the graph of h= P(4-P).
 

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