How to verify my solution to an separable differential equation? I have this question: 
Find the general solution to the separable differential equation
$$
\frac{dy}{dx} = y(1-y).
$$
My attempt is :
$$
\frac{dy}{y(1-y)} = dx
$$
$$
\frac{1}{y(1-y)} = \frac{A}{y}+\frac{B}{(1-y)}
$$
$$
1=A(1-y)+ B y
$$
$$
A=1, B-A=0, so B=1 
$$
$$
\frac{dy}{y} + \frac{dy}{1-y} = dx
$$
$$
ln(y)-ln(1-y) =x+k 
$$
$$
\frac{y}{(1-y)}=K^x 
$$
$$
\frac{(1-y)}{y}=c^{-x}
$$
$$
\frac{1}{y-1}=c^{-x}
$$
$$
y=\frac{1}{1+c^{-x}}
$$
Firstly is my attempt correct is or is there a better/simpler/easier way of solving this? 
And secondly how would I formally verify my solution?
 A: For $y\ne0$, you can linearize by rewriting
$$\frac1{y^2}\frac{dy}{dx}=\frac1y-1,$$
i.e.
$$-\frac{dz}{dx}=z-1.$$
The solution of the homogenous equation is 
$$z=Ce^{-x}$$and a particular solution is
$$z=1.$$
So
$$y=\frac1{Ce^{-x}+1}.$$
A: Starting from
$$\ln y-\ln(y-1)=x+c$$
We find
$$\ln \left( \frac{y}{y-1}\right)=x+c$$
taking the exponent of both sides
$$\frac{y}{y-1}=Ae^x$$
or
$$\frac{y-1}{y}=Be^{-x}$$
From which we see
$$1-\frac{1}{y}=Be^{-x}$$
Hence
$$y=\frac{1}{1-Be^{-x}}$$
Subject to initial conditions and bounds on $x$.
A: $$
\frac{dy}{y} + \frac{dy}{1-y} = dx\Longrightarrow \frac{dy}{y} - \frac{dy}{y-1} = dx
$$
But now
$$
\ln|y| - \ln|y-1| = x + C\Longrightarrow \ln \left|\frac{y}{y-1}\right| = x + C\Longrightarrow \left|\frac{y}{y-1}\right| = e^{x + C} = C_1e^x
$$
$C_1$ must be positive, but
$$
\frac{y}{y-1} = \pm C_1e^x = De^x,
$$
and $D$ may be negative or positive constant. Now
$$
\frac{y-1}{y} = \frac{1}{De^x}=1 - \frac1y \Longrightarrow y = \frac{1}{1-\frac{1}{De^x}} = \frac{De^x}{De^x - 1}
$$
And don't forget that $y\equiv 0$ and $y\equiv 1$ are solutions too.
