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

Question

I am solving the ODE above, it is a question from Strogatz Nonlinear dynamics and chaos, chapter 2 question 2.2.1.

Question

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

Answer

\begin{equation} \frac{\dot{x}}{4x^{2}-16} = 1\\ \frac{\dot{x}}{x^{2}-4} = 4\\ {{dx\over dt}\over x^2-4}=4 \\ {{dx\over dt}\over x^2-4}. dt=4. dt \\ {dx\over x^2-4}=4dt \\ \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}

Summary

I am looking to understand the intermediary step in the proof above. How do we get to this step $\frac{1}{4} \ln(\frac{x-2}{x+2}) = 4t + C_{1} $ from the previous step. Can we remove the constant $\frac{1}{4} $ then integrate the remaining portion?

Answer 1

$$I=\int \frac{dx}{x^{2}-4}$$ Partial fraction decomposition gives us: $$I=\int \left ( \frac A {x-2}-\dfrac B {x+2} \right)dx$$ $$I=\int \frac{x(A-B)+2(A+B)}{x^{2}-4}dx$$ $$\implies A=B=\dfrac 14$$ $$I=\dfrac 14\int \left ( \frac 1 {x-2}-\dfrac 1 {x+2} \right)dx$$ Then integrate with $\ln $ function . $$I=\dfrac 14 \ln |{x-2}|-\dfrac 14\ln |{x+2}|+C$$ $$I=\dfrac 14 \ln \left | \dfrac {x-2}{x+2} \right |+C$$

Answer 2

The author has used the method of "partial fractions" to write $$ \frac{1}{x^2-4} = \frac{A}{x-2} + \frac{B}{x+2}, $$ (solving for $A$ and $B$, which I'm not going to do), integrated each of the right hand items to get a "log" term, and then combined a difference-of-logs into a log-of-a-quotient. There's nothing subtle here except skipping steps from calculus class.