3.3: Solution of the Logistic Equation
- Page ID
- 106213
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We have seen that one does not need an explicit solution of the logistic equation (3.2) in order to study the behavior of its solutions. However, the logistic equation is an example of a nonlinear first order equation that is solvable. It is an example of a Riccati equation.
The general form of the Riccati equation is
\[\dfrac{d y}{d t}=a(t)+b(t) y+c(t) y^{2} \label{3.4} \]
As long as \(c(t) \neq 0\), this equation can be reduced to a second order linear differential equation through the transformation
\[y(t)=-\dfrac{1}{c(t)} \dfrac{\dot{x}(t)}{x(t)} \nonumber \]
We will demonstrate this using the simple case of the logistic equation,
\[\dfrac{d y}{d t}=k y-c y^{2} \label{3.5} \]
We let
\[y(t)=\dfrac{1}{c} \dfrac{\dot{x}}{x} \nonumber \]
Then
\[\begin{aligned}
\dfrac{d y}{d t} &=\dfrac{1}{c}\left[\dfrac{\ddot{x}}{x}-\left(\dfrac{\dot{x}}{x}\right)^{2}\right] \\
&=\dfrac{1}{c}\left[\dfrac{\ddot{x}}{x}-(c y)^{2}\right] \\
&=\dfrac{1}{c} \dfrac{\ddot{x}}{x}-c y^{2}
\end{aligned} \label{3.6} \]
Inserting this into the logistic equation (3.5), we have
\(\dfrac{1}{c} \dfrac{\ddot{x}}{x}-c y^{2}=k \dfrac{1}{c}\left(\dfrac{\dot{x}}{x}\right)-c y^{2}\),
or
\[\ddot{x}=k \dot{x} \nonumber \]
This equation is readily solved to give
\[x(t)=A+B e^{k t} \nonumber \]
Therefore, we have the solution to the logistic equation is
\[y(t)=\dfrac{1}{c} \dfrac{\dot{x}}{x}=\dfrac{k B e^{k t}}{c\left(A+B e^{k t}\right)} \nonumber \]
It appears that we have two arbitrary constants. But, we started out with a first order differential equation and expect only one arbitrary constant. However, we can resolve this by dividing the numerator and denominator by \(k B e^{k t}\) and defining \(C=\dfrac{A}{B}\). Then we have
\[y(t)=\dfrac{k / c}{1+C e^{-k t}} \label{3.7} \]
showing that there really is only one arbitrary constant in the solution.
We should note that this is not the only way to obtain the solution to the logistic equation, though it does provide an introduction to Riccati equations. A more direct approach would be to use separation of variables on the logistic equation. The reader should verify this.