Ordinary differential equations: Separable differential equations
Separable ODEs in chemical reaction kinetics
A second-order chemical reaction with one reactant The ODE that describes the reaction rate of the chemical reaction \[2\,\text{A}\longrightarrow \textit{B}\] is \[-\frac{1}{2}\frac{\dd [\text{A}]}{\dd t} = k\cdot [\text{A}]^2\] when elementary chemical kinetics laws are applicable. Using symbol \(y\) instead of the concentration \([\text{A}]\) we can rewrrite the so-called rate equation in differential form: \[\dd y=-2\,k\,y^2\,\dd t\] By separating variables we get: \[-\frac{1}{y^2}\,\dd y=2\,k\,\dd t\] Integrating both sides \[-\int\frac{1}{y^2}\,\dd y=2\int k\,\dd t\] gives \[\frac{1}{y}= 2\,k\,t+c\] where the integration constant \(c\) can be determined from the initial value \(y(0)=y_0\) as \[\frac{1}{y_0}= c\] So: \[\frac{1}{y}=\frac{1}{y_0} +2\,k\,t\] This can be rewritten (check this!) as \[y=\frac{y_0}{1+2\,k\,y_0t}\] Thus: \[[\text{A}]=\frac{[\text{A}]_0}{1+2\,k\,[\text{A}]_{{}_0}\,t}\] where \([\text{A}]_{{}_0}\) is the initial concentration at time \(t=0\). When we look at the implicit solution, i.e., the relationship \[\frac{1}{[\text{A}]}=\frac{1}{[\text{A}]_{{}_0}}+2\,k\,t\] then we can understand that a plot of \(\dfrac{1}{[\text{A}]}\) against time \(t\) is a straight line whose slope is \(2k\) and whose intercept with the vertical axis is \(\dfrac{1}{[\text{A}]_{{}_0}}\).
A second-order chemical reaction of two reagents The rate equation of the chemical reaction \[\text{A}+\text{B}\longrightarrow \textit{C}\] is: \[-\frac{\dd [\text{A}]}{\dd t} = -\frac{\dd [\text{B}]}{\dd t} = k\cdot [\text{A}]\cdot [\text{B}]\] when elementary chemical kinetics laws are applicable. Let the initial concentrations of \(\text{A}\) and \(\text{B}\) be \(a\) and \(b\), respectively, and let the symbol \(y\) be defined as \([\text{A}]=a-y\). Then: \([\text{B}]=b-y\) and the ODE for \(y\) is (check this!) as follows: \[\frac{\dd y}{\dd t}=k\,(a-y)\,(b-y)\] Two cases are distinguished.
a = b In this case, we can write the rate equation in differential form as \[\dd y=k\,(a-y)^2\,\dd t \] Separation of variables gives: \[\frac{1}{(a-y)^2}\,\dd y=k\,\dd t\] Integrating both sides \[\int\frac{1}{(a-y)^2}\,\dd y=\int k\,\dd t\] gives \[\frac{1}{a-y}= k\,t+c\] where the integration constant \(c\) can be determined from the initial value \(y(0)=0\) as \[\frac{1}{a}= c\] So: \[\frac{1}{a-y}= \frac{1}{a}+ k\,t\] This can be rewritten (check this!) as \[[\text{A}]=\frac{[\text{A}]_{{}_0}}{1+k[\text{A}]_{{}_0}t}\] where \([\text{A}]_{{}_0}\) is the initial concentration at time \(t=0\). When we look at the implicit solution, i.e., the relationship \[\frac{1}{[\text{A}]}=\frac{1}{[\text{A}]_{{}_0}}+k\,t\] then we can understand that a plot of \(\dfrac{1}{[\text{A}]}\) against time \(t\) is a straight line whose slope is \(k\) and whose intercept with the vertical axis is \(\dfrac{1}{[\text{A}]_{{}_0}}\).
a ≠ b In this case, one of the two reagents remains. In this case we must apply partial fraction decomposition to solve separable ODE \[\frac{1}{(a-y)(b-y)}\,\dd y=k\,\dd t\] This leads (check this!) to \[\frac{1}{(a-y)(b-y)}=\frac{1}{b-a}\left(\frac{1}{a-y}-\frac{1}{b-y}\right)\] So \[\begin{aligned}\int \frac{1}{(a-y)(b-y)}\,\dd y &= \frac{1}{b-a}\int\left(\frac{1}{a-y}-\frac{1}{b-y}\right)\dd y \\ \\ &= \frac{1}{b-a}\bigl(-\ln(a-y)+\ln(b-y)\bigr)+c\\ \\ &= \frac{1}{b-a}\ln\left(\frac{b-y}{a-y}\right)+c\end{aligned}\] with some constant \(c\), and this should be equal to the integration result \(k\,t\) of the right-hand side of the rate equation in differential form. At the beginning of the reaction, \(y(0)=0\) and therefore \[c= -\frac{1}{b-a}\ln\left(\frac{b}{a}\right)\] We can finally write (check this!) \[\frac{1}{b-a}\ln\left(\frac{a\,(b-y)}{b\,(a-y)}\right)=k\,t\] In terms of concentrations, it can be rewritten as: \[\frac{1}{[\text{B}]_0-[\text{A}]_0}\ln\left(\frac{[\text{A}]_{{}_0}\cdot [\text{B}]}{[\text{B}]_0\cdot [\text{A}]}\right)=k\,t\] A plot of \(\displaystyle\ln\left(\frac{ [\text{B}]}{ [\text{A}]}\right)\) expand against time \(t\) is a straight line whose slope is \(k\).
An autocatalytic chemical reaction We consider the kinetics of the following autocatalytic chemical reaction \[\text{A}+\text{B}\longrightarrow 2\,\text{B}\] in which the reaction rate is proportional to the product of the concentrations of substances A and B, say \[\frac{\dd [\text{A}]}{\dd t}=-k\, [\text{A}]\, [\text{B}]\] for some reaction rate constant \(k\). The formation rate of substance B satisfies the differential equation \[\frac{\dd [\text{B}]}{\dd t}=k\, [\text{A}]\, [\text{B}]\] Because of the stoichiometry of the reaction it is true that the sum of the concentrations of the substances A and B is constant, say \[[\text{A}]+[\text{B}]=K\] Then we can rewrite the differential equation for the formation rate of substance B as \[\frac{\dd\,[\text{B}]}{\dd t}=k\, [\text{B}]\bigl(K-[\text{B}]\bigr)\] The concentration of substance B is herewith a solution of a logistic differential equation and is explicitly described by a logistic function.
