Two common models lead to first-order differential equations with stable equilibrium values. A logistic population approaches its carrying capacity, while a falling object's velocity approaches its terminal velocity.

Logistic population growth

Let $N(t)$ be a population at time $t$. A population-dependent growth model has the form

$$\frac{dN}{dt}=R(N)N,$$

where $R(N)$ is the per-capita growth rate. To model limited resources, introduce a carrying capacity $K>0$. At zero population, the per-capita rate should equal the positive constant $r$; at capacity, it should vanish:

$$R(0)=r>0, \qquad R(K)=0.$$

It should also have the sign pattern

$$R(N)>0\quad\text{when }0<N<K,$$

so the population grows below capacity, and

$$R(N)<0\quad\text{when }N>K,$$

so it decreases above capacity. At capacity, $R(K)=0$.

The simplest function with this sign pattern is the decreasing line

$$R(N)=r\left(1-\frac{N}{K}\right) =\frac{r}{K}(K-N), \qquad r>0.$$

Substituting it into the population model gives the logistic equation:

$$\boxed{\frac{dN}{dt}=\frac{r}{K}(K-N)N}.$$

It is first order, nonlinear because of the $N^2$ term, and separable.

Per-capita growth rate

Growth is positive below the carrying capacity K, zero at K, and negative above K.

Population trajectories

K is a stable equilibrium: every positive trajectory shown approaches the carrying capacity.

Solving the logistic equation

For a nonconstant solution with $N\ne0$ and $N\ne K$, separate the variables:

$$\frac{K}{N(K-N)}\,dN=r\,dt.$$

The partial-fraction decomposition is

$$\frac{K}{N(K-N)} =\frac1N+\frac1{K-N}.$$

The second antiderivative has a minus sign because the derivative of $K-N$ is $-1$:

$$\begin{aligned} \int\left(\frac1N+\frac1{K-N}\right)dN &=\ln|N|-\ln|K-N|\\ &=\ln\left|\frac{N}{K-N}\right|. \end{aligned}$$

Therefore,

$$\ln\left|\frac{N}{K-N}\right|=rt+C,$$

and the sign can be absorbed into one arbitrary constant $C_1$:

$$\frac{N}{K-N}=C_1e^{rt}.$$

If $N(0)=N_0$ and $N_0\ne K$, then

$$C_1=\frac{N_0}{K-N_0}.$$

Solving for $N$ gives

$$\boxed{ N(t)=\frac{KN_0}{N_0+(K-N_0)e^{-rt}} }.$$

This formula describes the nonconstant positive solutions. The values excluded during separation must be checked separately. They give the two equilibrium solutions

$$N(t)\equiv0 \qquad\text{and}\qquad N(t)\equiv K.$$

For $r>0$:

• if $0<N_0<K$, then $N(t)$ increases toward $K$;
• if $N_0=K$, then $N(t)$ stays at $K$;
• if $N_0>K$, then $N(t)$ decreases toward $K$.

Thus $K$ is both the carrying capacity and the stable positive equilibrium.

A falling object with linear drag

Take downward as the positive direction. Let:

• $v(t)$ be the downward velocity;
• $m>0$ be the object's mass;
• $g>0$ be gravitational acceleration;
• $\gamma>0$ be the linear drag coefficient.

Gravity contributes $+mg$, while drag opposes the velocity and contributes $-\gamma v$. Newton's second law gives

$$mv'(t)=mg-\gamma v(t),$$

or

$$\boxed{v'(t)=g-\frac{\gamma}{m}v(t)}.$$

This is a first-order linear equation. Put it in standard form:

$$v'+\frac{\gamma}{m}v=g.$$

Its integrating factor is

$$\mu(t)=e^{\gamma t/m}.$$

Multiplying the equation by $\mu$ creates a product derivative:

$$\frac{d}{dt}\left(e^{\gamma t/m}v(t)\right) =ge^{\gamma t/m}.$$

Integrating gives

$$e^{\gamma t/m}v(t) =\frac{mg}{\gamma}e^{\gamma t/m}+C.$$

If $v(0)=v_0$, then $C=v_0-mg/\gamma$, so

$$\boxed{ v(t)=\frac{mg}{\gamma} +\left(v_0-\frac{mg}{\gamma}\right)e^{-\gamma t/m} }.$$

Velocity under linear drag

Since $m>0$ and $\gamma>0$,

$$e^{-\gamma t/m}\longrightarrow0 \qquad\text{as }t\to\infty.$$

Therefore,

$$\boxed{ \lim_{t\to\infty}v(t)=\frac{mg}{\gamma} }.$$

The value $mg/\gamma$ is the terminal velocity. It is also a stable equilibrium:

• an initial velocity above $mg/\gamma$ decreases toward it;
• an initial velocity below $mg/\gamma$, including a negative one, increases toward it;
• an initial velocity equal to $mg/\gamma$ remains constant.