Differential Equation for Logistic Growth

Try to solve $y' = ky$ for some constant $k$. This is a primitive model for population growth, where the rate of growth is proportional to the existing population.

This equation is easily solved using the technique of separation, as described earlier; we get $\ln(y) = kx + C_1$, hence $y=Ce^{kx}$. Hence in ideal cases population growth is exponential. But, as Malthus observed, population growth is not nearly that pretty. The logistic equation is a more realistic model. Feel free to look up what each of the constants represents.

$$y' = ry\frac{k-y}{k}$$

$$\begin{align*} \frac{dy}{dx} &= ry\frac{k-y}{k}\\ \int\frac{dy}{y(k-y)} &= \int\frac{r}{k}dx \end{align*}$$

We evaluate the integral on the left using partial fractions: $$\int\frac{dy}{y(k-y)} = \frac{1}{k}\int \frac{1}{y} + \frac{1}{k-y}dy = \frac{1}{k}(\ln(y)-\ln(k-y))$$ Hence $\frac{1}{k}(\ln(y)-\ln(k-y)) = \frac{rx}{k}+C_1$, so $$\frac{y}{k-y} = C_2e^{rx}$$ So $y\left(1+C_2e^{rx}\right) = kC_2e^{rx}$, or $$y = \frac{kC_2e^{rx}}{1+C_2e^{rx}}=\frac{k}{1+Ce^{-rx}},$$ where the last equality is obtained by dividing both numerator and denominator of the fraction by $C_2e^{rx}$ then setting $C = \frac{1}{C_2}$.

Some interesting properties of this solution:

After we wait a long time ($x\rightarrow\infty$), the population $y$ approaches $k$, which is the carrying capacity, as long as $C$ is finite. If we wanted to solve for $C$ given an initial condition $y(0) = y_0$, we would find $$y_0 = \frac{k}{1+C} \rightarrow C = \frac{k}{y_0}-1=\frac{k-y_0}{y_0}$$ so as long as $y_0\not=0$, the population will approach $k$.

The logistic equation is $$\frac{dy}{dt} = ky\left(1 - \frac{y}{L}\right)$$ where $k,L$ are constants. It is sometimes written with different constants, or in a different way, such as $y' = ry(L-y)$, where $r = k/L$. In either case, the constant $L$ is known as the carrying capacity limit, and the factor $1 - \frac{y}{L}$ represents growth inhibition. All solutions to the logistic equation are of the form $$y(t) = \frac{L}{1 + be^{-kt}}$$ for some constant $b$ (depending on the initial conditions or other information). In particular, regardless of the value of $b$, we see that $y(t) \to L$ as $t\to \infty$ (as long as $L,k,r$ are positive), so $L$ can also be thought of as the equilibrium value (as $t\to\infty$) in the logistic model.