Solving a simple first order differential equation I've got the following first order differential equation that is doing something I can't quite figure out.  The question is to solve:
$$y' = 2x^2y^2$$
Here is my solution:
$$y' \frac{1}{y^2} = 2x^2$$
$$\int \frac{1}{y^2} y' dx = \int 2x^2 dx$$
$$\int \frac{1}{y^2}dy = \int 2x^2 dx$$
$$\ln|y^2| = \frac{2}{3}x^3 + c$$
$$y^2 = e^{\frac{2}{3}x^3 + c} = e^{\frac{2}{3}x^3}.e^c = ce^{\frac{2}{3}x^3}$$
$$y = \sqrt{ce^{\frac{2}{3}x^3}}$$
However, I know from the textbook that this answer is incorrect.  Please help.

Note that $$\int \frac 1{y^2}\,dy = \int y^{-2} \,dy = -y^{-1} + C$$
So you should obtain $$-\frac 1y = \frac{2}{3}x^3 + c \iff y = -\frac 3{2x^3}\; \text{provided } x\neq 0$$
