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?