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The logistic differential equation $y'=y(b-ay) \, \textrm{with}\, a\neq 0, b\neq 0$ has the non-trivial solution $y(t) = \frac{\frac{b}{a}}{1+c\cdot e^{-bt}}, \quad (1)$ where $c$ is a constant.

My question is: If $0, then we get the s-form curve. If $y>\frac{b}{a}$ then we get the decreasing graph which is above it. In which situations do we get this curve. Can you give an example. We say that $y>0$, but when do we have that $y$ is also a solution?

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    I have seen but this has not been asked before.2012-11-14

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You can see the behavior of the solutions even without using the explicit solution. The equation is $y'=y(b-a y)$. I'll assume in fact that $a,b>0$ rather than just that they are nonzero; easy variation if not.

Now make a "slope field" by doing the following: At each point $(x,y)$ in the plane (or really onoly at a lot of points) make a tiny line segment having the slope of $y(b-ay)$. Since this expression has no $x$ in it, all the little segments at a given level $y=c$ have the same slope. The solution curves to the differential equation then "go along the slope field", that is, each solution is tangent to the slope curve at each point on the curve.

The slope $y(b-ay)$ is zero when $y=0$ and when $y=b/a$, so this means all the segments along the $x$ axis and along the line $y=b/a$ are horizontal. This corresponds to the fact that $y(t)=0$ and $y(t)=b/a$ are "trivial" solutions to the diffEQ.

The greatest (positive) slope occurs when $y=b/(2a)$ where the slope is $b^2/4a$. This is halfway between the upper and lower bounds of the "S shaped solutions.

Once the slope field is drawn you'll see that if you start anywhere above the upper line the solution curve will head downward toward the upper line, and if you start between the upper line and the x axis the solution will approach the upper line, and if you start below the x axis the solution will move down toward minus infinity. A similar thing happens if you move toward minus infinity, using the slope field as a guide to where the solution curves are going.

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    @HansLundmark Thank you for the answer.2018-11-19