# An Epilogue of “Qualitative Analysis of a Simple Chemical Reaction”

In “Qualitative Analysis of a Simple Chemical Reaction“, we deduced from

$\begin{cases}\frac{dx}{dt}=2 k_2 y - 2 k_1 x^2 \quad\quad(0-1)\\ \frac{dy}{dt}=k_1 x^2-k_2 y\quad\quad\quad(0-2)\end{cases}$

a 1st order differential equation

$\frac{dx}{dt}=-2k_1x^2-k_2x+c_0.$

This differential equation can be solved to obtain $x$ first, then by (0-1),

$y = \frac{1}{2k_2}(\frac{dx}{dt}+2k_1x^2).$

As expected,

$\lim\limits_{t\rightarrow \infty} x(t) = \frac{\sqrt{k_2^2+8c_0k_1}-k_2}{4k_1}=x_*$

and

$\lim\limits_{t\rightarrow \infty} y(t) = -\frac{k_2\sqrt{k_2^2+8c0k_1}-k_2^2-4c_0k_1}{8k_1k_2}=\frac{k_1}{k_2}x_*^2=y_*.$

Exercise-1 Solve differential equation $\frac{dx}{dt}=-2k_1x^2-k_2x+c_0$ without CAS.

Hint: rewrite $\frac{dx}{dt}=-2k_1x^2-k_2x+c_0$ as $\frac{1}{2k_1x^2+k_2x-c_0}\cdot\frac{dx}{dt}=-1.$