First-Order Linear Equations and Applications: Learn It 5

Solving the Logistic Differential Equation

The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution.

Step 1: Find the equilibrium solutions

Setting the right-hand side equal to zero gives us [latex]P = 0[/latex] and [latex]P = K[/latex] as constant solutions.

  • When [latex]P = 0[/latex]: If no organisms are present, the population will never grow
  • When [latex]P = K[/latex]: If the population starts at carrying capacity, it remains constant

Step 2: Separate the variables

Rewrite the differential equation as:

[latex]\frac{dP}{dt} = \frac{rP(K-P)}{K}[/latex]

Multiply both sides by [latex]dt[/latex] and divide by [latex]P(K-P)[/latex]:

[latex]\frac{dP}{P(K-P)} = \frac{r}{K}dt[/latex]

Step 3: Prepare for integration

Multiply both sides by [latex]K[/latex]:

[latex]\int \frac{K}{P(K-P)}dP = \int r dt[/latex]

Step 4: Use partial fraction decomposition

The left side can be decomposed using partial fractions:

[latex]\frac{K}{P(K-P)} = \frac{1}{P} + \frac{1}{K-P}[/latex]

You can verify this decomposition by finding a common denominator on the right side and checking that it equals the left side.

Step 5: Integrate both sides

[latex]\int \left(\frac{1}{P} + \frac{1}{K-P}\right)dP = \int r dt[/latex]

[latex]\ln|P| - \ln|K-P| = rt + C[/latex]

[latex]\ln\left|\frac{P}{K-P}\right| = rt + C[/latex]

Step 6: Solve for [latex]P[/latex]

Exponentiate both sides to eliminate the natural logarithm:

[latex]e^{\ln\left|\frac{P}{K-P}\right|} = e^{rt+C}[/latex]

[latex]\left|\frac{P}{K-P}\right| = e^C \cdot e^{rt}[/latex]

Defining [latex]C_1 = e^C[/latex], we get: [latex]\frac{P}{K-P} = C_1 e^{rt}[/latex]

Step 7: Solve for [latex]P(t)[/latex]

Starting from [latex]\frac{P}{K-P} = C_1 e^{rt}[/latex], multiply both sides by [latex]K-P[/latex]:

[latex]P = C_1 e^{rt}(K-P)[/latex]

[latex]P = C_1 K e^{rt} - C_1 P e^{rt}[/latex]

Collect all terms containing [latex]P[/latex] on the left side:

[latex]P + C_1 P e^{rt} = C_1 K e^{rt}[/latex]

Factor out [latex]P[/latex] and solve:

[latex]P(1 + C_1 e^{rt}) = C_1 K e^{rt}[/latex]

[latex]P(t) = \frac{C_1 K e^{rt}}{1 + C_1 e^{rt}}[/latex]

Step 8: Find the constant [latex]C_1[/latex]

To determine [latex]C_1[/latex], use the initial condition. When [latex]t = 0[/latex], [latex]P(0) = P_0[/latex].

Substitute into [latex]\frac{P}{K-P} = C_1 e^{rt}[/latex]:

[latex]\frac{P_0}{K-P_0} = C_1 e^{r(0)} = C_1[/latex]

Therefore: [latex]C_1 = \frac{P_0}{K-P_0}[/latex]

Step 9: Complete the solution

Substitute [latex]C_1 = \frac{P_0}{K-P_0}[/latex] into our equation:

[latex]P(t) = \frac{C_1 K e^{rt}}{1 + C_1 e^{rt}} = \frac{\frac{P_0}{K-P_0} \cdot K e^{rt}}{1 + \frac{P_0}{K-P_0} e^{rt}}[/latex]

Multiply numerator and denominator by [latex]\left(K-{P}_{0}\right)[/latex] to simplify:

[latex]\begin{array}{cc}\hfill P\left(t\right)& =\frac{\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\hfill \\ & =\frac{\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\cdot \frac{K-{P}_{0}}{K-{P}_{0}}\hfill \\ & =\frac{{P}_{0}K{e}^{rt}}{\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}}.\hfill \end{array}[/latex]

We state this result as a theorem.

logistic growth solution

Consider the logistic differential equation subject to an initial population of [latex]{P}_{0}[/latex] with carrying capacity [latex]K[/latex] and growth rate [latex]r[/latex]. The solution to the corresponding initial-value problem is:

[latex]P\left(t\right)=\dfrac{{P}_{0}K{e}^{rt}}{\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}}[/latex].

Finding the Point of Inflection

The logistic growth curve has a point of inflection where the growth rate changes from increasing to decreasing. To find this point, we set the second derivative equal to zero.

Starting with our solution, we can find the derivatives:

[latex]\begin{array}{ccc}\hfill P\left(t\right)& =\hfill & \frac{{P}_{0}K{e}^{rt}}{\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}}\hfill \\ \hfill {P}^{\prime }\left(t\right)& =\hfill & \frac{r{P}_{0}K\left(K-{P}_{0}\right){e}^{rt}}{{\left(\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}\right)}^{2}}\hfill \\ \hfill P^{\prime\prime}\left(t\right)& =\hfill & \frac{{r}^{2}{P}_{0}K{\left(K-{P}_{0}\right)}^{2}{e}^{rt}-{r}^{2}{P}_{0}{}^{2}K\left(K-{P}_{0}\right){e}^{2rt}}{{\left(\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}\right)}^{3}}\hfill \\ & =\hfill & \frac{{r}^{2}{P}_{0}K\left(K-{P}_{0}\right){e}^{rt}\left(\left(K-{P}_{0}\right)-{P}_{0}{e}^{rt}\right)}{{\left(\left(K-{P}_{0}\right)+{P}_{0}{e}^{rt}\right)}^{3}}.\hfill \end{array}[/latex]

Setting the numerator equal to zero,

[latex]{r}^{2}{P}_{0}K\left(K-{P}_{0}\right){e}^{rt}\left(\left(K-{P}_{0}\right)-{P}_{0}{e}^{rt}\right)=0[/latex].

 

As long as [latex]{P}_{0}\ne K[/latex], the entire quantity before and including [latex]{e}^{rt}[/latex] is nonzero, so we can divide it out:

[latex]\left(K-{P}_{0}\right)-{P}_{0}{e}^{rt}=0[/latex].

Solving for [latex]t[/latex],

[latex]\begin{array}{ccc}\hfill {P}_{0}{e}^{rt}& =\hfill & K-{P}_{0}\hfill \\ \hfill {e}^{rt}& =\hfill & \frac{K-{P}_{0}}{{P}_{0}}\hfill \\ \hfill \text{ln}{e}^{rt}& =\hfill & \text{ln}\frac{K-{P}_{0}}{{P}_{0}}\hfill \\ \hfill rt& =\hfill & \text{ln}\frac{K-{P}_{0}}{{P}_{0}}\hfill \\ \hfill t& =\hfill & \frac{1}{r}\text{ln}\frac{K-{P}_{0}}{{P}_{0}}.\hfill \end{array}[/latex]
If [latex]P_0 > K[/latex] (initial population exceeds carrying capacity), this expression becomes undefined because we’d be taking the natural logarithm of a negative number. In this case, the population decreases monotonically toward [latex]K[/latex] with no inflection point.

The inflection point marks where the population growth changes from accelerating to decelerating. This is where the “leveling off” begins as the population approaches the carrying capacity and growth becomes limited by available resources.

A population of rabbits in a meadow is observed to be [latex]200[/latex] rabbits at time [latex]t=0[/latex]. After a month, the rabbit population is observed to have increased by [latex]4\text{%}[/latex]. Using an initial population of [latex]200[/latex] and a growth rate of [latex]0.04[/latex], with a carrying capacity of [latex]750[/latex] rabbits,

  1. Write the logistic differential equation and initial condition for this model.
  2. Draw a slope field for this logistic differential equation, and sketch the solution corresponding to an initial population of [latex]200[/latex] rabbits.
  3. Solve the initial-value problem for [latex]P\left(t\right)[/latex].
  4. Use the solution to predict the population after [latex]1[/latex] year.