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.


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


\begin{equation} \dot{x}=4x^{2}-16 \end{equation}

\begin{equation} \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}} \end{equation}

\begin{equation} C_{2}(t=0) = \frac{x-2}{x+2} \end{equation}

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


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? \begin{equation} \frac{\dot{x}}{x^{2}-4} = 4\\ \int \frac{1}{x^{2}-4} dx = \int 4 dt \end{equation}

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$$\int{dx\over x^2-4}=\int 4dt$$

  • $\begingroup$ and the integral is with regards to $t$ right? $\endgroup$
    – user4933
    Jun 17 '20 at 9:11
  • $\begingroup$ Not both of them. The right one is with respect to $t$ while the left one is with respect to $x$. $\endgroup$ Jun 17 '20 at 9:12
  • $\begingroup$ @ThamuMnyulwa: See also math.stackexchange.com/questions/27425/…. $\endgroup$ Jun 17 '20 at 9:15
  • $\begingroup$ Can I interpret the solution as specified in my second edit? $\endgroup$
    – user4933
    Jun 17 '20 at 9:33
  • $\begingroup$ Your Edit 1 sounds fine to me. $\endgroup$ Jun 17 '20 at 10:55

They divided both sides of the equation by the term $x^2-4$ and then took the integral.

  • $\begingroup$ @ GSofer , thank you for your answer. Is the integration step performed as I stated above in the edit? $\endgroup$
    – user4933
    Jun 17 '20 at 9:02

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