Direction field of a first orden variable separable differential equation 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?

For the wolfram solution, here are some steps
$$y'=y(1-y)$$
$$\int \frac {dy}{y(1-y)}=x+K$$
$$\int \frac {dy}{y}+ \int \frac{dy}{1-y}=x+K$$
$$\ln (y)-\ln {(y-1)}=x+K$$
$$\ln (\frac y{y-1})=x+K$$
$$\frac y{y-1}=Ce^x$$
$$ y=-\frac {Ce^x }{1-Ce^x}
\implies y=\frac {Ke^x }{1+Ke^x}$$
$$ y=\frac {e^x }{R+e^x}$$
