dy/dt = y^2(1-y)(y-3)
a) Determine whether the equilibrium solutions are stable , unstable or neither?
b) for which value of y(0) will y(t) approach to 0 as t approach to infinity?
Stable equilibrium solution problem PLEASE HELP?
2014-05-04 2:03 pm
回答 (1)
2014-05-04 4:45 pm
First of all, the equilibrium solutions occur when dy/dt = 0.
==> y^2 (1 - y)(y - 3) = 0
==> y = 0, 1, 3.
a) For y < 0, we have dy/dt < 0 (try y = -1 for instance).
For 0 < y < 1, we have dy/dt < 0 (try y = 1/2 for instance).
For 1 < y < 3, we have dy/dt > 0 (try y = 2 for instance).
For y > 3, we have dy/dt < 0 (try y = 4 for instance).
(Sketch isoclines from this information.)
Hence, y = 0 is neither, y = 1 is unstable, and y = 3 is stable.
b) From the information above, we know that dy/dt < 0 when 0 < y < 1,
but dy/dt > 0 when y > 1.
So, we need y(0) to be in [0, 1) for y(t) to approach 0 as t goes to infinity.
I hope this helps!
==> y^2 (1 - y)(y - 3) = 0
==> y = 0, 1, 3.
a) For y < 0, we have dy/dt < 0 (try y = -1 for instance).
For 0 < y < 1, we have dy/dt < 0 (try y = 1/2 for instance).
For 1 < y < 3, we have dy/dt > 0 (try y = 2 for instance).
For y > 3, we have dy/dt < 0 (try y = 4 for instance).
(Sketch isoclines from this information.)
Hence, y = 0 is neither, y = 1 is unstable, and y = 3 is stable.
b) From the information above, we know that dy/dt < 0 when 0 < y < 1,
but dy/dt > 0 when y > 1.
So, we need y(0) to be in [0, 1) for y(t) to approach 0 as t goes to infinity.
I hope this helps!
收錄日期: 2021-04-29 00:11:33
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