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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.2012-11-12

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