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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 $$y(t) = \frac{\frac{b}{a}}{1+e^{-bt}}, \quad (1)$$ where $c$ is a constant.

My questions are: 1) Why are we assuming that both $a$ and $b$ are nonzero. 2) Is $y(t)=0$ also a solution to the differential equation. 3) I want to show that $(1)$ is a solution to the differential equation. My idea is that I want to find $y'$ and then calculate $y(b-ay)$ where I insert (1). Then show that they are equal. When calculation $y'$ I dont want to use differentiation rule of quotient, I want to use rule of composition function. How can I do that?

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There is no reason to assume that both $a$ and $b$ are non-zero. –  Artem Nov 12 '12 at 17:34
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For question (1): If $a=0$ the equation is $y'=by$, a simple exponential function, and if $b=0$ the equation is $y'=-ay^2$, which I don't know offhand how to solve, but probably doesn't give a logistic function.

For question (2): Yes, $y=0$ is a solution; so is $y=b/a$. These are on a graph the lower and upper bounds of the logistic curves in between, and are asymptotes, for the typical initial values used.

For question (3) $y(t)$ doesn't look like a composition of functions to me, except in some trivial ways which likely won't help in taking the derivative, and maybe the quotient rule has to be dragged in at some point.

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Solving $y'=-ay^2$ is not hard, as the equation is separable. The solution satisfies $-\int y^{-2}\,dy=a\int dt=a(t-t_0)$ for some $t_0$, so that $y=1/(a(t_0-t))$. (Note that the solution blows up at $t=t_0$.) –  Harald Hanche-Olsen Nov 12 '12 at 12:10
    
Is it that if either a=0 or b=0 then the equation is not logistic? –  Reader Nov 12 '12 at 12:18
    
Harald: yes, I should have noticed it's separable for $y'=-ay^2$. That's not logistic, since a look at the line field shows all solutions go off to infinity or minus infinity. As you note from the explicit solution, this happens in finite time. –  coffeemath Nov 12 '12 at 13:02
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