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I tried solving this differential equation:

$$y'=y+y^2$$ I tried the substitution $$z=y^{-1}\\z'=-y^{-2}y'$$ and the differential equation becomes $z'+z=-1$ and now I have as solution $$z=ce^{-x}-1$$ and $$y=\frac{1}{ce^{-x}-1}$$ But the solution is not correct. Does someone can find my error?

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    "But the solution is not correct" Your book probably gives an equivalent formulation of your solution, for example, $$y=\frac{ce^x}{1-ce^x}$$ – 2017-02-16
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    @SpamIAm No. $ $ – 2017-02-16

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Simply differentiate $y$:

$$y'=-1(ce^{-x}-1)^{-2}(-ce^{-x})=\frac{ce^{-x}}{(ce^{-x}-1)^2}=y+y^2$$

You're right, as Did mentioned in the comments the book probably wrote:

$$y=\frac{ce^x}{1-ce^x}$$

Another way of solving this is:

$$\frac{dy}{y+y^2}=dx$$

$$\ln\left|\frac{y}{y-1}\right|=x+C$$

Solving for $y$ gives the same expression.

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    Oh. thank you. I did not notice it. That's really embarrassing. – 2017-02-16