<s>[INST] 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.
 [/INST] Instead of keeping $y'$ as $y'$, you should change it into $\frac{dy}{dx}$. Then,$$\frac{dy}{dx} = 2x^2y^2$$ $$y^2 dy = 2x^2 dx$$ $$\int y^2\, dy = \int2x^2\, dx$$ $$\frac{y^3}{3} = \frac{2}{3}x^3 + C_1$$ $$y = \sqrt[3]{2x^3+C}$$
</s>