> 
> Solve $\frac{dy}{dx} = y(y-1)$.
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I got

$$\frac{dy}{y(y-1)} = dx \implies \ln|y-1| - \ln|y| = x + c$$

$$\implies \ln \bigg| \frac{y - 1}{y}\bigg| = x + c$$

$$\implies 1 - \frac{1}{y} = Ce^{x}$$

$$\implies y = \frac{1}{1 - Ce^{x}}$$

However, [Wolfram|Alpha](http://m.wolframalpha.com/input/?i=solve%20dy%2Fdx%20%3D%20%28y-1%29y&x=0&y=0) seems to disagree with this. Can anyone point out my mistake?