Gilbert Strang’s Calculus: Differential Equations of Growth

Calculus

Differential Equations

Growth Models

Logistic Equation

Predator-Prey

Gilbert Strang

From exponential growth to logistic saturation and predator-prey interaction: first-order differential equations as models of growth, competition, and oscillation.

Author

Chao Ma

Published

March 28, 2026

This lecture starts from the simplest growth law and then adds more realistic effects:

pure exponential growth
constant source terms
logistic saturation
predator-prey interaction

Basic Exponential Growth

The simplest growth model is

\[ \frac{dy}{dt} = cy \]

The growth rate is proportional to the current amount $y$.

With initial value

\[ y(0) = y_0 \]

the solution is

\[ y(t) = y_0 e^{ct} \]

because

\[ \frac{d}{dt}(y_0 e^{ct}) = c y_0 e^{ct} = c y(t) \]

So:

if $c>0$, the quantity grows exponentially
if $c<0$, the quantity decays exponentially

Adding a Constant Source

Now include a source term:

\[ \frac{dy}{dt} = cy + s \]

This is a linear differential equation. Its solution can be written as

\[ y(t) = y_{\text{particular}}(t) + y_{\text{homogeneous}}(t) \]

Particular Solution

Try a constant solution. Then $dy/dt = 0$, so

\[ 0 = cy + s \qquad \Rightarrow \qquad y = -\frac{s}{c} \]

So one particular solution is

\[ y_{\text{particular}} = -\frac{s}{c} \]

Homogeneous Solution

The associated homogeneous equation is

\[ \frac{dy}{dt} = cy \]

with solution

\[ Ae^{ct} \]

Therefore the full solution is

\[ y(t) = -\frac{s}{c} + Ae^{ct} \]

Using $y(0)=y_0$ gives

\[ A = y_0 + \frac{s}{c} \]

\[ y(t) + \frac{s}{c} = \left(y_0 + \frac{s}{c}\right)e^{ct} \]

Logistic Population Growth

Pure exponential growth cannot continue forever. A common correction is the logistic equation:

\[ \frac{dP}{dt} = cP - sP^2 \]

where

$c$ is the net growth rate
$s$ is the slowdown factor from competition

This can also be written as

\[ \frac{dP}{dt} = P(c-sP) \]

Early Stage

When $P$ is small, the term $sP^2$ is negligible, so the model behaves almost like

\[ \frac{dP}{dt} \approx cP \]

which is exponential growth.

Carrying Capacity

At equilibrium,

\[ \frac{dP}{dt}=0 \]

\[ cP - sP^2 = 0 \qquad \Rightarrow \qquad P = 0 \quad \text{or} \quad P = \frac{c}{s} \]

The positive steady state

\[ P = \frac{c}{s} \]

is the carrying capacity.

Inflection Point

The logistic curve grows fastest halfway to the carrying capacity:

\[ P = \frac{c}{2s} \]

That is where the graph changes from bending upward to bending downward.

Solving the Logistic Equation by Letting $y = 1/P$

Let

\[ y = \frac{1}{P} \]

Then by the chain rule,

\[ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{1}{P}\right) = -\frac{1}{P^2}\frac{dP}{dt} \]

Substitute the logistic equation:

\[ \frac{dy}{dt} = -\frac{1}{P^2}(cP - sP^2) = s - c\frac{1}{P} = s - cy \]

So $y$ satisfies the linear equation

\[ \frac{dy}{dt} = s - cy \]

whose solution is

\[ y(t) - \frac{s}{c} = \left(y(0) - \frac{s}{c}\right)e^{-ct} \]

Since $y=1/P$, this becomes

\[ \frac{1}{P(t)} - \frac{s}{c} = \left(\frac{1}{P(0)} - \frac{s}{c}\right)e^{-ct} \]

This is the logistic solution in reciprocal form.

Predator-Prey Model

The lecture ends by moving from one population to two interacting populations.

Let

$u$ = predator population
$v$ = prey population

One standard model is

\[ \frac{du}{dt} = -cu + kuv \]

\[ \frac{dv}{dt} = Cv - suv \]

Interpretation:

predators die out without prey because of the term $-cu$
predators grow when predator-prey encounters happen, through $kuv$
prey grows naturally through $Cv$
prey decreases through predation, modeled by $-suv$

Unlike the logistic equation, which approaches a steady ceiling, the predator-prey system often produces oscillation:

prey increases first
predators then increase because food is abundant
prey falls as predation rises
predators then fall because food becomes scarce

and the cycle can repeat.

Takeaways

$\frac{dy}{dt}=cy$ gives pure exponential growth or decay.
Adding a source term still leaves a linear ODE with a particular-plus-homogeneous structure.
The logistic equation adds self-limiting competition and produces an S-curve.
The substitution $y=1/P$ turns the logistic equation into a linear equation.
Coupling two growth equations leads naturally to predator-prey oscillations.

--- title: "Gilbert Strang's Calculus: Differential Equations of Growth" author: "Chao Ma" date: "2026-03-28" categories: [Calculus, Differential Equations, Growth Models, Logistic Equation, Predator-Prey, Gilbert Strang] description: "From exponential growth to logistic saturation and predator-prey interaction: first-order differential equations as models of growth, competition, and oscillation." toc: true --- <iframe width="100%" height="480" src="https://www.youtube.com/embed/IDo4uPyqQbQ" title="Differential Equations of Growth" frameborder="0" allowfullscreen></iframe> This lecture starts from the simplest growth law and then adds more realistic effects: - pure exponential growth - constant source terms - logistic saturation - predator-prey interaction ## Basic Exponential Growth The simplest growth model is $$ \frac{dy}{dt} = cy $$ The growth rate is proportional to the current amount $y$. With initial value $$ y(0) = y_0 $$ the solution is $$ y(t) = y_0 e^{ct} $$ because $$ \frac{d}{dt}(y_0 e^{ct}) = c y_0 e^{ct} = c y(t) $$ So: - if $c>0$, the quantity grows exponentially - if $c<0$, the quantity decays exponentially ## Adding a Constant Source Now include a source term: $$ \frac{dy}{dt} = cy + s $$ This is a linear differential equation. Its solution can be written as $$ y(t) = y_{\text{particular}}(t) + y_{\text{homogeneous}}(t) $$ ### Particular Solution Try a constant solution. Then $dy/dt = 0$, so $$ 0 = cy + s \qquad \Rightarrow \qquad y = -\frac{s}{c} $$ So one particular solution is $$ y_{\text{particular}} = -\frac{s}{c} $$ ### Homogeneous Solution The associated homogeneous equation is $$ \frac{dy}{dt} = cy $$ with solution $$ Ae^{ct} $$ Therefore the full solution is $$ y(t) = -\frac{s}{c} + Ae^{ct} $$ Using $y(0)=y_0$ gives $$ A = y_0 + \frac{s}{c} $$ so $$ y(t) + \frac{s}{c} = \left(y_0 + \frac{s}{c}\right)e^{ct} $$ ## Logistic Population Growth Pure exponential growth cannot continue forever. A common correction is the logistic equation: $$ \frac{dP}{dt} = cP - sP^2 $$ where - $c$ is the net growth rate - $s$ is the slowdown factor from competition This can also be written as $$ \frac{dP}{dt} = P(c-sP) $$ ### Early Stage When $P$ is small, the term $sP^2$ is negligible, so the model behaves almost like $$ \frac{dP}{dt} \approx cP $$ which is exponential growth. ### Carrying Capacity At equilibrium, $$ \frac{dP}{dt}=0 $$ so $$ cP - sP^2 = 0 \qquad \Rightarrow \qquad P = 0 \quad \text{or} \quad P = \frac{c}{s} $$ The positive steady state $$ P = \frac{c}{s} $$ is the carrying capacity. ### Inflection Point The logistic curve grows fastest halfway to the carrying capacity: $$ P = \frac{c}{2s} $$ That is where the graph changes from bending upward to bending downward. ![](differential-equations-of-growth-1.png) ## Solving the Logistic Equation by Letting $y = 1/P$ Let $$ y = \frac{1}{P} $$ Then by the chain rule, $$ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{1}{P}\right) = -\frac{1}{P^2}\frac{dP}{dt} $$ Substitute the logistic equation: $$ \frac{dy}{dt} = -\frac{1}{P^2}(cP - sP^2) = s - c\frac{1}{P} = s - cy $$ So $y$ satisfies the linear equation $$ \frac{dy}{dt} = s - cy $$ whose solution is $$ y(t) - \frac{s}{c} = \left(y(0) - \frac{s}{c}\right)e^{-ct} $$ Since $y=1/P$, this becomes $$ \frac{1}{P(t)} - \frac{s}{c} = \left(\frac{1}{P(0)} - \frac{s}{c}\right)e^{-ct} $$ This is the logistic solution in reciprocal form. ## Predator-Prey Model The lecture ends by moving from one population to two interacting populations. Let - $u$ = predator population - $v$ = prey population One standard model is $$ \frac{du}{dt} = -cu + kuv $$ $$ \frac{dv}{dt} = Cv - suv $$ Interpretation: - predators die out without prey because of the term $-cu$ - predators grow when predator-prey encounters happen, through $kuv$ - prey grows naturally through $Cv$ - prey decreases through predation, modeled by $-suv$ Unlike the logistic equation, which approaches a steady ceiling, the predator-prey system often produces oscillation: - prey increases first - predators then increase because food is abundant - prey falls as predation rises - predators then fall because food becomes scarce and the cycle can repeat. ## Takeaways - $\frac{dy}{dt}=cy$ gives pure exponential growth or decay. - Adding a source term still leaves a linear ODE with a particular-plus-homogeneous structure. - The logistic equation adds self-limiting competition and produces an S-curve. - The substitution $y=1/P$ turns the logistic equation into a linear equation. - Coupling two growth equations leads naturally to predator-prey oscillations.