Deriving the Explicit Logistic Solution

From a Rate Model to an Explicit Population Formula

A logistic differential equation describes a population whose growth rate depends on both its current size and a limiting scale. Its rate is proportional to the population, but the proportional growth factor decreases as the population approaches the carrying capacity.

In this tutorial, we derive an explicit formula from that differential equation. The key steps are separation of variables, partial fractions, and careful handling of the absolute value that appears when integrating. That absolute value matters: the derivation must preserve solutions both below and above the carrying capacity.

Definition: The logistic differential equation with growth rate $r>0$ and carrying capacity $K>0$ is $$ \frac{dP}{dt}=rP\left(1-\frac{P}{K}\right), $$ where $P=P(t)$ is the population at time $t$.

The equilibrium populations are $P=0$ and $P=K$: substituting either constant value into the differential equation gives a rate of zero. We will first derive the formula for a non-equilibrium solution, so assume $P\ne 0$ and $P\ne K$ on the interval being considered. These exclusions let us divide by $P$ and $K-P$; the equilibrium solutions must be checked separately.

Separate Variables and Use Partial Fractions

Rewrite the right-hand side using $1-P/K=(K-P)/K$. For a non-equilibrium solution, the equation can then be separated as follows:

\frac{dP}{dt}=\frac{r}{K}P(K-P) \qquad\Longrightarrow\qquad \frac{K}{P(K-P)}\,dP=r\,dt.

The factor involving $P$ can be decomposed into simpler fractions. Indeed, combining the fractions on the right gives the expression on the left:

\frac{K}{P(K-P)} = \frac{1}{P}+\frac{1}{K-P}, \qquad \frac{1}{P}+\frac{1}{K-P} = \frac{K-P+P}{P(K-P)} = \frac{K}{P(K-P)}.

Now integrate both sides with respect to their variables. Because the derivative of $K-P$ with respect to $P$ is $-1$, the second logarithmic term has a minus sign:

\int\left(\frac{1}{P}+\frac{1}{K-P}\right)dP = \int r\,dt $$ $$ \ln|P|-\ln|K-P|=rt+C

Combining the logarithms gives a relation involving the absolute value of a ratio. Exponentiating yields the absolute value of that ratio, not automatically the ratio itself:

\ln\left|\frac{P}{K-P}\right|=rt+C \qquad\Longrightarrow\qquad \left|\frac{P}{K-P}\right|=e^C e^{rt}.

Since $e^C>0$, write it as a positive constant $B$. The absolute-value equation allows the ratio to be either positive or negative. We can express both possibilities at once by introducing a nonzero signed constant $A$:

\frac{P}{K-P}=A e^{rt}, \qquad A\ne 0.

This signed constant is essential. Restricting $A$ to positive values would retain the solutions with $0<P<K$ but discard solutions with $P>K$.

Rearrange to Make Population Explicit

Starting from the signed ratio relation, multiply by $K-P$, collect the terms involving $P$, and divide by the resulting factor. The formula is valid wherever its denominator is nonzero.

\frac{P}{K-P}=Ae^{rt} $$ $$ P=Ae^{rt}(K-P) =KAe^{rt}-PAe^{rt} $$ $$ P(1+Ae^{rt})=KAe^{rt} $$ $$ P=\frac{KAe^{rt}}{1+Ae^{rt}} =\frac{K}{1+A^{-1}e^{-rt}}.

It is common to rename the nonzero constant $A^{-1}$ as $C$. This gives the explicit logistic formula for non-equilibrium solutions. The formula also gives $P=K$ when $C=0$, so that equilibrium solution can be included in the same expression. The other equilibrium, $P=0$, is not represented by a finite value of $C$ and must be included separately.

Theorem: For $r>0$ and $K>0$, every non-equilibrium solution of the logistic equation has the form $$ P(t)=\frac{K}{1+Ce^{-rt}}, \qquad C\ne 0, $$ on any interval where $1+Ce^{-rt}\ne 0$. The equilibrium solutions $P(t)=0$ and $P(t)=K$ must also be considered. Allowing $C=0$ in the displayed formula gives the equilibrium $P(t)=K$.

Why the Formula Works

The separation and integration show how to obtain the formula from a non-equilibrium solution. To check the result in the other direction, differentiate the proposed formula and compare its derivative with the right-hand side of the logistic equation.

Let $D(t)=1+Ce^{-rt}$, so $P=K/D$. At any time when $D\ne 0$, differentiating and simplifying gives:

D'(t)=-rCe^{-rt} $$ $$ P'(t)=-\frac{K D'(t)}{D(t)^2} =\frac{rKCe^{-rt}}{D(t)^2}.

For the right-hand side, first express $1-P/K$ using the same denominator:

1-\frac{P}{K} = 1-\frac{1}{D} = \frac{D-1}{D} = \frac{Ce^{-rt}}{D}.

Therefore, the logistic rate calculated from the proposed population is:

rP\left(1-\frac{P}{K}\right) = r\left(\frac{K}{D}\right)\left(\frac{Ce^{-rt}}{D}\right) = \frac{rKCe^{-rt}}{D^2} = P'(t).

Thus the explicit formula satisfies the differential equation wherever it is defined. The constant solutions $P=0$ and $P=K$ also satisfy the equation by direct substitution. This check establishes that both signs of $C$ are allowed; the differential equation does not require the population to be below $K$.

Worked Example: Derive the Formula for a Specific Logistic Equation

Find the explicit non-equilibrium solutions of $\frac{dP}{dt}=\frac12P(1-\frac{P}{8})$.

Here $r=\frac12$ and $K=8$. Rewrite the equation using $8-P$, then separate variables. For $P\ne 0,8$, the steps are:

\frac{dP}{dt} = \frac{1}{2}P\left(\frac{8-P}{8}\right) = \frac{P(8-P)}{16} $$ $$ \frac{8}{P(8-P)}\,dP = \frac{1}{2}\,dt.

Decompose the fraction. Combining the two terms verifies the decomposition:

\frac{8}{P(8-P)} = \frac{1}{P}+\frac{1}{8-P}, \qquad \frac{1}{P}+\frac{1}{8-P} = \frac{8-P+P}{P(8-P)} = \frac{8}{P(8-P)}.

Integrate, keeping the sign from the derivative of $8-P$:

\ln|P|-\ln|8-P| = \frac{t}{2}+C $$ $$ \left|\frac{P}{8-P}\right| = e^C e^{t/2}.

Let $A\ne 0$ be a signed constant. Then $\frac{P}{8-P}=Ae^{t/2}$. Solving for $P$ gives:

P=Ae^{t/2}(8-P) $$ $$ P(1+Ae^{t/2})=8Ae^{t/2} $$ $$ P=\frac{8}{1+A^{-1}e^{-t/2}}.

Writing $C=A^{-1}$, the non-equilibrium family is $P=\frac{8}{1+Ce^{-t/2}}$, with $C\ne 0$ and a nonzero denominator. To check it, differentiate and compare with the equation:

P'=\frac{4Ce^{-t/2}}{(1+Ce^{-t/2})^2} $$ $$ \frac12P\left(1-\frac{P}{8}\right) = \frac12\left(\frac{8}{1+Ce^{-t/2}}\right) \left(\frac{Ce^{-t/2}}{1+Ce^{-t/2}}\right) = \frac{4Ce^{-t/2}}{(1+Ce^{-t/2})^2} = P'.

The constant solutions $P=0$ and $P=8$ are also solutions. The first is separate from this family; the second is obtained by setting $C=0$.

Worked Example: Keep Both Branches of a Logarithmic Relation

Suppose separation and integration for a logistic equation with $K=6$ and $r=3$ have produced $\ln\left|\frac{P}{6-P}\right|=3t+\ln 2$. Obtain both explicit branches.

Exponentiating gives an absolute value. Since $e^{\ln 2}=2$, the result is:

\left|\frac{P}{6-P}\right| = e^{3t+\ln 2} = 2e^{3t}.

The ratio can have either sign, so there are two cases. In the positive case, $\frac{P}{6-P}=2e^{3t}$, which corresponds to $A=2$. In the negative case, $\frac{P}{6-P}=-2e^{3t}$, which corresponds to $A=-2$. Rearranging each case gives:

A=2: \qquad P=\frac{6}{1+\frac12e^{-3t}} $$ $$ A=-2: \qquad P=\frac{6}{1-\frac12e^{-3t}}.

Both branches satisfy the original logarithmic relation: in each case, $\left|\frac{P}{6-P}\right|=2e^{3t}$. They also satisfy the differential equation wherever defined. Check at $t=0$. For $A=2$, the formula gives $P=6/(1+1/2)=4$, and:

P'(0) = \frac{3\cdot 6\cdot (1/2)}{(1+1/2)^2} = \frac{9}{9/4} = 4 $$ $$ 3P(0)\left(1-\frac{P(0)}{6}\right) = 3\cdot4\left(1-\frac{4}{6}\right) = 12\cdot\frac13 = 4.

For $A=-2$, the formula gives $P=6/(1-1/2)=12$, and:

P'(0) = \frac{3\cdot 6\cdot (-1/2)}{(1-1/2)^2} = \frac{-9}{1/4} = -36 $$ $$ 3P(0)\left(1-\frac{P(0)}{6}\right) = 3\cdot12\left(1-\frac{12}{6}\right) = 36(-1) = -36.

In particular, the negative branch gives a population above $K=6$ at $t=0$. Taking only the positive sign after exponentiating would omit this valid solution.

Worked Example: A Negative Constant and Its Domain

For $r=2$, $K=10$, and $C=-3$, write the explicit solution, identify where it is defined, and check it at $t=\ln 3$.

Substitution into the explicit formula gives:

P(t)=\frac{10}{1-3e^{-2t}}.

The denominator is zero when $1-3e^{-2t}=0$. Solving gives $e^{-2t}=1/3$, so $t=(\ln 3)/2$. The formula is undefined at that time; it defines solutions on intervals that do not include the zero of the denominator. At $t=\ln 3$, which is greater than $(\ln 3)/2$, we have $e^{-2t}=1/9$, so:

P(\ln 3) = \frac{10}{1-3/9} = \frac{10}{2/3} = 15.

Differentiate the formula and evaluate at that time. Then calculate the rate from the logistic equation using the same population value:

P'(t)=\frac{20(-3)e^{-2t}}{(1-3e^{-2t})^2} $$ $$ P'(\ln 3) = \frac{-60(1/9)}{(2/3)^2} = \frac{-20/3}{4/9} = -15 $$ $$ 2P(\ln 3)\left(1-\frac{P(\ln 3)}{10}\right) = 2\cdot15\left(1-\frac{15}{10}\right) = 30\left(-\frac12\right) = -15.

The two rates agree. This solution has $P>K$ at the chosen time, and the negative derivative is consistent with the logistic equation above the carrying capacity.

A Common Sign Error

A frequent mistake is to integrate correctly, obtain a logarithm of an absolute value, and then discard the absolute value as if the ratio were known to be positive. For example, from $\ln|P/(6-P)|=3t+\ln 2$, it is incorrect to conclude that the ratio must equal $2e^{3t}$. That conclusion silently assumes $P/(6-P)>0$.

Correction: Exponentiating the logarithmic relation first gives $\left|P/(6-P)\right|=2e^{3t}$. Therefore $P/(6-P)=2e^{3t}$ or $P/(6-P)=-2e^{3t}$. Keep both branches unless a condition on the population justifies choosing one.

For example, a stated condition $0<P<6$ makes $P/(6-P)>0$, so it selects the positive branch. Without such a condition, both signs must be retained. A separate practical check is to inspect the denominator of the explicit formula: the solution only exists on intervals where that denominator is nonzero.

What the Formula Provides

The explicit formula makes the population available directly as a function of time rather than as an implicit logarithmic relation. It also shows how the integration constant affects which solution is represented. For $C>0$, the denominator $1+Ce^{-rt}$ is greater than $1$, so the formula gives $0<P<K$. For $C<0$, the denominator can vanish, and the resulting solution must be restricted to an interval that avoids that time. In particular, negative constants are not algebraic errors; they encode branches that the absolute-value integration permits.

To derive or use the formula reliably, keep the equilibrium solutions in view, preserve the absolute value through exponentiation, and state the nonzero-denominator condition. Those checks ensure that the formula describes all the relevant solution branches rather than only the most familiar one.

Check Your Understanding

Answer these questions using the separation, integration, and checks in this tutorial.

Why must $P=0$ and $P=K$ be checked separately when separating variables?
What partial-fraction identity is used for $K/[P(K-P)]$, and how can it be verified?
What two possibilities follow from $\left|P/(K-P)\right|=Be^{rt}$, where $B>0$?
For $P=K/(1+Ce^{-rt})$, what condition must hold on the denominator?
Why can a negative value of $C$ not be discarded solely because $P$ represents a population?

Deriving the Explicit Logistic Solution

What You'll Learn

From a Rate Model to an Explicit Population Formula

Separate Variables and Use Partial Fractions

Rearrange to Make Population Explicit

Why the Formula Works

Worked Example: Derive the Formula for a Specific Logistic Equation

Worked Example: Derive the Formula for a Specific Logistic Equation

Worked Example: Keep Both Branches of a Logarithmic Relation

Worked Example: Keep Both Branches of a Logarithmic Relation

Worked Example: A Negative Constant and Its Domain

Worked Example: A Negative Constant and Its Domain

A Common Sign Error

What the Formula Provides

Check Your Understanding

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Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

Deriving the Explicit Logistic Solution

What You'll Learn

From a Rate Model to an Explicit Population Formula

Separate Variables and Use Partial Fractions

Rearrange to Make Population Explicit

Why the Formula Works

Worked Example: Derive the Formula for a Specific Logistic Equation

Worked Example: Derive the Formula for a Specific Logistic Equation

Worked Example: Keep Both Branches of a Logarithmic Relation

Worked Example: Keep Both Branches of a Logarithmic Relation

Worked Example: A Negative Constant and Its Domain

Worked Example: A Negative Constant and Its Domain

A Common Sign Error

What the Formula Provides

Check Your Understanding

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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