$$y=\pm \frac{e^{\frac{x^{2}}{3}+C}-1}{2}=\pm\frac{e^{\frac{x^{2}}{3}+C}-1}{2}\mp \frac{1}{2}$$.
I checked my result using WolframAlpha, and the solution it came up with was
$$y=Ce^{\frac{x^{2}}{3}}-\frac{1}{2}$$.
That confuses me. I know that we can eliminate $\frac{e^{C}}{2}$ by choosing a constant $D$ such that $D=\frac{e^{C}}{2}$, and since $D$ can be both positive or negative, the $\pm$ in front of $\frac{e^{\frac{x^{2}}{3}+C}-1}{2}$ can be eliminated as well, but how…