Differential equation $\frac{dy}{dt}=y(1-y)$ I'm starting with
$$\frac{dy}{dt}=y(1-y)$$
Then I take the obvious steps.
$$\int\left(\frac1y+\frac1{1-y}\right)dy=t+C\\\ln|y|+\ln|1-y|=t+C\\\ln(|y||1-y|)=t+C\\\sqrt{y^2}\sqrt{(1-y)^2}=Ce^t\\\sqrt{y^2(1-y)^2}=Ce^t\\y^2(1-y)^2=Ce^{2t}\\y^2-2y^3+y^4=Ce^{2t}$$
I am stuck here for finding an explicit solution.

Going from line 2 to line 3, you should have a minus sign, so it should be $$\ln|y|-\ln|1-y|=t+C \iff \ln \left| \frac{y}{1-y}\right|=t+C \iff \frac{y}{1-y}=e^{t+C}$$
$$\iff\frac{y}{1-y}=Ae^t$$ (letting $A=e^C$)
$$\iff y=Ae^t-Aye^t \iff y+Aye^t=Ae^t \iff y[1+Ae^t]=Ae^t$$
$$\iff \boxed{y=\frac{Ae^t}{1+Ae^t}}$$

Also, you may divide through by $A$ to obtain $y=\frac{e^t}{\left(\frac{1}{A}\right)+e^t}$ and let $B=\frac{1}{A}$ to give 
$$\boxed{y=\frac{e^t}{B+e^t}},$$ but this is just a matter of personal preference.
