Direction field of a first orden variable separable differential equation

Question

I solved the equation

$y'=y(1-y)$

and I found

$ln(y)-ln(1-y)=x+c$

i.e. $y=\frac{c_{1}e^{x}}{c_{2}e^{x}-1}$

However, as I try to plot the direction field (using isoclines) I get

$y-y^{2}-y'=0$

i.e. $y=\frac{1\pm\sqrt{1-4y'}}{2}$

Does this means that the solutions must have a slope $y'<\frac{1}{4}$ at all points?

The thing is, when I plot the solution in my computer, it spits out some positive hiperbolae. Does this means that the solutions are only defined for values of $x$ to the left of the asymptote?

Note:

Wolframalpha gives me a solution $y=\frac{e^x}{c+e^x}$ wich seems a lot more elegant and easy to work with. Any idea on how to get this result?

Answer 1

For the wolfram solution, here are some steps

$$y'=y(1-y)$$ $$\int \frac {dy}{y(1-y)}=x+K$$ $$\int \frac {dy}{y}+ \int \frac{dy}{1-y}=x+K$$ $$\ln (y)-\ln {(y-1)}=x+K$$ $$\ln (\frac y{y-1})=x+K$$ $$\frac y{y-1}=Ce^x$$ $$ y=-\frac {Ce^x }{1-Ce^x} \implies y=\frac {Ke^x }{1+Ke^x}$$ $$ y=\frac {e^x }{R+e^x}$$

Answer 2

Taking the derivative of the equation $y'=y(1-y)$ and make it equal to $0$ gives us:

$y''=y'(1-2y)$

$y'(1-2y)=0$

$\Rightarrow y'(1-2y)=0$

$\Rightarrow y=\frac{1}{2}$

Using this on the original equation gives us $y'=1/4$

$\frac{1}{4} $ is a maximum for the slope!

The slope is indeed always less than $\frac{1}{4}$. In fact, if we study the direction field outside the region $0<y<1$, the slope will be always negative.