Error in solution for a separable differential equation I "solved" the differential equation $x'=x^2-1$ a couple of months ago, now I checked the solution with wolfram and it seem I was wrong...
According to Wolfram the solution should be $x(t)=\displaystyle\frac{1-e^{2(t+C)}}{1+e^{2(t+C)}}$, but mine is $x(t)=\displaystyle\frac{1+e^{2(t+C)}}{1-e^{2(t+C)}}$.
What I did...
$$\frac{dx}{dt}=x^2-1\\t=\int\frac{1}{x^2-1}dx\\=\frac{1}{2}\int\frac{1}{x-1}dx-\frac{1}{2}\int\frac{1}{x+1}dx\\=\frac{1}{2}(\log(x-1)-\log(x+1))-C\\=\log\sqrt{\frac{x-1}{x+1}}-C\\ \implies \log\sqrt{\frac{x-1}{x+1}}=t+C\\ \implies \sqrt{\frac{x-1}{x+1}}=e^{t+C}\\ \implies \frac{x-1}{x+1}=e^{2(t+C)}\\ \implies x-1=xe^{2(t+C)}+e^{2(t+C)}\\ \implies  x(1-e^{2(t+C)})=1+e^{2(t+C)}\\ \implies x=\frac{1+e^{2(t+C)}}{1-e^{2(t+C)}}$$.
I can't spot the error.

The solutions $x(t)=\displaystyle\frac{1-e^{2(t+C_1)}}{1+e^{2(t+C_1)}}$ and $x(t)=\displaystyle\frac{1+e^{2(t+C_2)}}{1-e^{2(t+C_2)}}$ are the same in fact, but with different $C$. Note that $C$ can be complex. With this form of writing we have the relationship $C_2=C_1+i\pi$
Another equivalent form is $x(t)=\displaystyle\frac{1-c_1e^{2t}}{1+c_1e^{2t}}=\displaystyle\frac{1+c_2e^{2t}}{1-c_2e^{2t}}$ where $c_2=-c_1$
