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Phase Line and long term behavior with initial value

goku900

New member
Nov 25, 2013
8
Ok so i have the equation y'=y^2(y-3)(y-5)^3

I found the equilibrium positions to be y=0, y=3, y=5.

For my phase diagram all the arrows are pointing up so the solutions are nodes?

The last part asks Describe the long term behavior of the solution to the above di fferential equation with initial condition y(0) = 4.

I'm not sure how to get started with this.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You have two equilibria of odd multiplicity so you cannot have all arrows in the phase diagram pointing up. Can you identify the region in the phase space where the slope is negative?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
you have [tex]y'= y^2(y-3)(y- 5)^3[/tex]. [tex]y^2[/tex] will always be positive but, as MarkFL said,those two odd powered factors mean the product cannot always be positive.


The equilibrium points are, of course, y= 0, y= 3, and y= 5. I would continue like this:
Because the "x" factor is squared, it will always be positive.
If y< 3, then y is also less than 5 so y- 3 and y- 5 are negative, odd powers are negative, and y' is the product of (+)(-)(-) which is positive.
If y< 3, then y- 3 is positive but y- 5 is still negative. Now y' is the product of (+)(-)(+) which is negative.
If y> 3, then all factors are positive so y' is the product of (+)(+)(+) which is positive.

y is decreasing for y between 3 and 5. Since 4 is between 3 and 5, y will decrease as x increases. Further, since y(x) cannot cross 3, it must approach 3 as a limit as for large x. As x decreases toward negative infinity, y(x) approaches 5.