<s>[INST] Solving $\frac{dy}{dx} = xy^2$ This problem appears to be pretty simple to me but my book gets a different answer.
$$\frac{dy}{dx} = xy^2$$
For when y is not 0
$$\frac{dy}{y^2} = x \, dx$$
$$\int \frac{dy}{y^2} = \int x \, dx$$
$$\frac{-1}{y^1} = \frac{x^2}{2}$$ 
$$\frac{-2}{x^2} = y$$ 
Is there anything wrong with this solution? It is not what my book gets but it is similar to how they do it in the example.
 [/INST] Continuing from what you did, we have
$$\int \frac{dy}{y^2} = \int x\,dx$$
$$-\frac{1}{y} + C_2= \frac{x^2}{2}+C_1$$
Where $C_1$ and $C_2$ are constants of integration.  Letting $C_3=C_1-C_2$
$$-\frac{1}{y} + C_2= \frac{x^2}{2}+C_1=-\frac{1}{y}= \frac{x^2}{2}+C_3$$
Solving for $y$, we have:
$$-\frac{1}{y}= \frac{x^2}{2}+C_3= \frac{x^2+2C_3}{2} \implies -y=\frac{2}{x^2+2C_3} \\
\implies y=\frac{2}{2C_3-x^2}$$
Letting $2C_3=K$
$$y=\frac{2}{2C_3-x^2}=\frac{2}{K-x^2}$$
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