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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 $0<y<\frac{b}{a}$, 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. – Reader Nov 14 '12 at 4:27
up vote 1 down vote accepted

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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Thanks. one thing how do you get that the slope is $b^2/4a$.But my question is also that I dont understand how can I have a function y such that $y>b/a?$ Can you give an example of this? – Reader Nov 14 '12 at 5:49
@Reader: If $0<c<1$, then your formula for $y(t)$ will give a value greater than $b/a$. – Hans Lundmark Nov 14 '12 at 10:23
@Reader: Maybe you're thinking that $y>b/a$ is to be true for all $t$. What one usually does is to pick some initial value for $t$, say $t=0$, and then specify the value of $y(0)$. This then determines a specific point $(0,y(0)$ in the plane [horizontal axis as $t$, vertical as $y$]. Then depending only on $y(0)$ and the slope field, you get the various solution curves. Of course you can definitely start with $y(0)>b/a$, and in that case get a curve which asymptotically approaches $y=b/a$ from above, as $t \to \infty$. – coffeemath Nov 14 '12 at 13:52
Thanks Lundmark, and coffeemath. That is what I was thinking about. – Reader Nov 14 '12 at 16:11

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