Consider the system $\dot{x}=4x^{2}-16$ find an analytical solution.

I am working through a text book by Strogatz Nonlinear dynamics and chaos . In chapter 2 question 2.2.1 , I am looking for an analytical solution. I have the question's answer but would like to ask how a certain step was performed.

Question

Consider the system $$\dot{x}=4x^{2}-16$$ Find an analytical solution to the problem.

$$$$\dot{x}=4x^{2}-16$$$$

$$$$\int \frac{1}{x^{2}-4} dx = \int 4 dt \\ \frac{1}{4} \ln(\frac{x-2}{x+2}) = 4t + C_{1} \\ x = 2 \frac{1 + C_{2}e^{16t}}{1 - C_{2}e^{16t}}$$$$

$$$$C_{2}(t=0) = \frac{x-2}{x+2}$$$$

where $$C_{1}$$ and $$C_{2}$$ are constants.

Summary

In the first step to get to $$\int \frac{1}{x^{2}-4} dx = \int 4 dt$$ how does this happen? There is an intermediary step/result that is not clear. Any help would be really appreciated.

Edit 1:

In other words, is this step okay? $$$$\frac{\dot{x}}{x^{2}-4} = 4\\ \int \frac{1}{x^{2}-4} dx = \int 4 dt$$$$

Edit 2:

Can I then denote my solution as:

$$x(t) = \frac{2(e^{4c_{1}+16t})}{(e^{4c_{1}-16t})}$$

The equation $${\dot{x}\over x^2-4}=4$$ is actually equivalent to $${{dx\over dt}\over x^2-4}=4$$which, by multiplying both sides in $$dt$$ leads to $${dx\over x^2-4}=4dt$$or$$\int{dx\over x^2-4}=\int 4dt$$
• and the integral is with regards to $t$ right? Jun 17 '20 at 9:11
• Not both of them. The right one is with respect to $t$ while the left one is with respect to $x$. Jun 17 '20 at 9:12
They divided both sides of the equation by the term $$x^2-4$$ and then took the integral.