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By "equations" I meant the differential equation (definition of logistic) and its solution: $$\frac{dy}{dt}=ky(1-\frac{y}{L})$$ and $$y=\frac{y_0L}{y_0+(L-y_0)e^{-kt}}$$ given that $y(0)=y_0$

The first or the differential equation has the two constant solution: y=0,L (which I don't know how to find, appreciate if anyone can show me. Are there any more constant solution?)

The second equation also has these two solutions: $$y_0=0, L$$. I guess $y_0$ acts like x, or the independent variable, here.

But why? I can't make the connection.

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    $\begingroup$ Wait! The latter (the function) solves the former (the differential equation)... This is all there is to understand. $\endgroup$ – Did Apr 4 '16 at 13:52
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HINT

If $y$ is a constant solution then $y'=0$ and then $ky\left(1-\frac{y}{L}\right)=0$

Solving by separation of variables you will find $$ \int \frac{dy}{y\left(1-\frac{y}{L}\right)}=\int k\, dt $$ and after observing that $\frac{1}{y\left(1-\frac{y}{L}\right)}=\frac{1}{y}+\frac{1}{L-y}$ and integrating you'll find $$ \frac{L-y}{y}=Ae^{-kt} $$ and substituting $y(0)=y_0$ you'll find $ \frac{L-y_0}{y_0}=A $ and then your expression $$ y(t)=\frac{L}{1+A\,e^{-kt}}=\frac{y_0L}{y_0+(L-y_0)e^{-kt}} $$

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For the differential equation:

$y=0$ and $y=L$ give $\frac{dy}{dt}=0$ so these are constant solutions since the derivative is null.

For the explicit equation: you find these solutions for the initial conditions: $$ y_0=0 \qquad y_0=L $$

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