# Thread: Basis of Attraction (differential eq'ns)

1. ## Basis of Attraction (differential eq'ns)

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.

2. Separating the variables we have...

$\displaystyle \int \frac{dy}{(1-y)^{2}} = \int dx \rightarrow \frac {1}{1-y} = x + c \rightarrow y= \frac{x+c-1}{x+c}$

If we impose the condition $\displaystyle y(0)= y_{0}$ we obtain...

$\displaystyle c=\frac{1}{1-y_{0}}$

If $\displaystyle y_{0}<1$ then $\displaystyle c>0$ and ...

$\displaystyle \lim_{x \rightarrow \infty} y(x) = 1$

... and the 'basis of attraction' is $\displaystyle y=1$ .

If $\displaystyle y_{0}>1$ however the situation is quite different and $\displaystyle y=1$ becomes 'basis of repulsion'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Thanks chisigma.

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...
(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.

4. Originally Posted by FZ44
Thanks chisigma.

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...