Consider the IVP given by
y' = (1-y)^2 y(0) = yo
For each equilibrium solution, determine the basin of attraction.
I found the only equilibrium solution, y=1, but can't find the basis of attraction.
Separating the variables we have...
If we impose the condition we obtain...
If then and ...
... and the 'basis of attraction' is .
If however the situation is quite different and becomes 'basis of repulsion'...
The question defines the basis of attraction as follows...
"Basins of Attraction:
Suppose that y=c is an equilibrium solution or constant solution of the first-order DE y'=f(y).
The basin of attraction is the set of initial conditions y(0)=yo so that the solutions satisfying the IVP y'=f(y), y(0)=yo tend to c as t-->infinity."
So in other words, the basis of attraction is the set of initial conditions such that the function will be attracted to a given equilibrium solution c.
I approached this problem by graphing the DE y' = (1-y)^2 using this online app DiffEqu.
I noted the equilibrium solution y=1, which my online homework says is correct.
I then inputted the basis of attraction as (-inf,1), because all points contained in this basis of attraction tend towards y=1 as t approaches infinity.
I'm pretty sure that this is the correct answer, but my online homework says it's incorrect.
Perhaps I am entering it in the wrong format.
From my homework...
"Rules for inputting answer:
(1) Start with the smallest equilibrium solution and input these solutions in increasing order.
(2) If the basin of attraction is the point c, input only the number c without any spaces.
(3) If the basin of attraction is an interval, input the interval without any spaces. For example, input can look like (-5,7) for an interval. Use -inf for negative infinity and inf for positive infinity.
(4) If any answer fields are unused, type only an upper-case N in each of these."
I inputted (-inf,1), but this is incorrect, still unsure where my error lies.