Answer:
The critical points are 0,2,4
0 is semi-stable, 2 is stable, 4 is unstable.
Check the phase line and the solution curves in the files attached below
Step-by-step explanation:
[tex]dy/dx = y^{4} - 6y^{3} + 8y^{2}[/tex]
To get the critical points, dy/dx = 0
[tex]y^{4} - 6y^{3} + 8y^{2} = 0\\y^{2} ( y^{2} - 6y + 8) = 0\\y^{2} ( y-2)(y-4) = 0\\y = 0, 2, 4[/tex]
To classify the stability, the interval [tex](-\infty, \infty)[/tex] is divided into [tex](-\infty, 0), (0,2), (2,4),(4, \infty)[/tex]
[tex]f(y) = y^{4} - 6y^{3} +8y^{2}[/tex]
If y = -1, f(-1) = 15 > 0
If y = 1, f(1) = 3 > 0
If y = 3, f(3) = -9 < 0
If y = 5, f(5) = 75 > 0
The phase line is drawn in the file attached.
By critically observing the phase line:
The point y = 4 is unstable because the arrows are moving away from it
The point y = 2 is stable because the arrows are moving towards it
The point y = 0 is semi-stable because one arrow is moving towards it while the other moves away from it.
The typical solution curves are sketched in the file attached below.
