Solution of $x'(t) = - a (x(t)^2 - b^2)$ I am trying to reproduce the results of a paper and this differential equation 
$$x'(t) = - a(x(t)^2 - b^2)$$ 
is at the heart of the paper where a and b are positive numbers greater than zero. 
I cannot see how they arrive at a solution of $x(t) = -b \coth(b(a t+ C_1))$ just by integration. 
I have tried solving this differential equation using Mathematica and it gives me a solution of 
$$x(t) = b \tanh[b (a t + C[1])]$$
which I don't think is the same. I would like to know how they arrive at this result. Many thanks.
 A: This is a Riccati equation as was said by @YoTengoUnLCD in the comments.
The more general Riccati equation is:
$$x'=Ax^2+Bt^n$$
It is solved by first using the substitution:
$$x(t)=-\frac{1}{A} \frac{y(t)'}{y(t)}$$
Directly substituting, we obtain:
$$-\frac{1}{A} \frac{y''}{y}+\frac{1}{A} \frac{y'^2}{y^2}=\frac{1}{A} \frac{y'^2}{y^2}+Bt^n$$
$$y''+ABt^ny=0$$
In our case $AB=-(ab)^2$ and $n=0$, thus we obtain:
$$y''-(ab)^2y=0$$
But this is just the (imaginary) harmonic oscillator equation. The general solution is:
$$y=C_1 e^{abt}+C_2 e^{-abt}$$
Now getting back to $x(t)$:

$$x=b \frac{C_1e^{abt}-C_2 e^{-abt}}{C_1 e^{abt}+C_2 e^{-abt}}=b \frac{e^{2ab(t+t_0)}-1}{e^{2ab(t+t_0)}+1}=b \tanh(ab(t+t_0))$$

Here we have:
$$\frac{C_1}{C_2}=e^{2abt_0}$$
This is just the same solution as the one given by Mathematica. If you want to obtain another form of solution, used in the paper, you need to set:
$$\frac{C_1}{C_2}=-e^{2abt_0}$$
Then you will have:

$$x=b \frac{-e^{2ab(t+t_0)}-1}{-e^{2ab(t+t_0)}+1}=-b \coth(ab(t+t_0))$$

