MathLab with Mr. Ahmed

MathLab with Mr. Ahmed is a math learning website by Ahmed Amasha, a mathematics teacher teaching the American curriculum. Explore Algebra 2, Precalculus, Calculus, and SAT and PSAT Math Prep through lessons, focused practice and printable resources.

MathLabwith Mr. Ahmed
Differential Equations

6.3 Separation of Variables and the Logistic Equation

Focus on separation of variables, logistic equilibria, fastest logistic growth.

Free preview Explore the lesson before creating an account.

Play games for this section

Before you solve

Ideas in this lesson

01Separation of variablesWorked steps · five questions02Logistic equilibriaWorked steps · five questions03Fastest logistic growthWorked steps · five questions
Idea 1 of 3

Separation of variables

Move yy-dependent factors to dydy and xx-dependent factors to dxdx, integrate, and use the initial condition. Check for lost equilibrium solutions.

Worked example 1

Solve y′=2xy,y(0)=1y' = 2xy, y(0) = 1. Find y(1)y(1).
  1. dyy=2xdx\dfrac{dy}{y} = 2x dx
  2. ln⁡y=2x22+C\ln y = \dfrac{2x^{2}}{2} + C; y(0)=1  ⟹  C=0y(0) = 1 \implies C = 0
  3. y=e2x22y = e^{\dfrac{2x^{2}}{2}}
  4. y(1)=e≈2.72y(1) = e \approx 2.72

Answer: e≈2.72e \approx 2.72

Worked example 2

Solve y′=3xy,y(0)=1y' = 3xy, y(0) = 1. Find ln⁡y(2)\ln y(2).
  1. dyy=3xdx\dfrac{dy}{y} = 3x dx
  2. ln⁡y=3x22+C\ln y = \dfrac{3x^{2}}{2} + C; y(0)=1  ⟹  C=0y(0) = 1 \implies C = 0
  3. y=e3x22y = e^{\dfrac{3x^{2}}{2}}
  4. ln⁡y(2)=3(2)22=6\ln y(2) = \dfrac{3(2)^{2}}{2} = 6

Answer: 66

Idea 2 of 3

Logistic equilibria

In P′=P' = rP (1−PK)(1 - \frac{P}{K}), zero and KK are equilibria; KK is the carrying capacity for the standard growth model.

Worked example 1

A population obeys P′=0.1P(1−P200)P' = 0.1P(1 - \dfrac{P}{200}). Find its positive equilibrium.
  1. P′=0.1P(1−P200)=0  ⟹  P=0,200P' = 0.1P(1 - \dfrac{P}{200}) = 0 \implies P = 0, 200
  2. dP′dP=0.1(1−2P200)=0  ⟹  P=100\dfrac{dP'}{dP} = 0.1(1 - \dfrac{2P}{200}) = 0 \implies P = 100
  3. Positive equilibrium =200= 200

Answer: 200200

Worked example 2

A population obeys P′=0.2P(1−P300)P' = 0.2P(1 - \dfrac{P}{300}). Find the population where growth is fastest.
  1. P′=0.2P(1−P300)=0  ⟹  P=0,300P' = 0.2P(1 - \dfrac{P}{300}) = 0 \implies P = 0, 300
  2. dP′dP=0.2(1−2P300)=0  ⟹  P=150\dfrac{dP'}{dP} = 0.2(1 - \dfrac{2P}{300}) = 0 \implies P = 150
  3. Population for fastest growth =150= 150

Answer: 150150

Idea 3 of 3

Fastest logistic growth

The rate is a concave-down quadratic in PP. Its maximum occurs at half the carrying capacity.

Worked example 1

A population obeys P′=0.1P(1−P200)P' = 0.1P(1 - \dfrac{P}{200}). Find its positive equilibrium.
  1. P′=0.1P(1−P200)=0  ⟹  P=0,200P' = 0.1P(1 - \dfrac{P}{200}) = 0 \implies P = 0, 200
  2. dP′dP=0.1(1−2P200)=0  ⟹  P=100\dfrac{dP'}{dP} = 0.1(1 - \dfrac{2P}{200}) = 0 \implies P = 100
  3. Positive equilibrium =200= 200

Answer: 200200

Worked example 2

A population obeys P′=0.2P(1−P300)P' = 0.2P(1 - \dfrac{P}{300}). Find the population where growth is fastest.
  1. P′=0.2P(1−P300)=0  ⟹  P=0,300P' = 0.2P(1 - \dfrac{P}{300}) = 0 \implies P = 0, 300
  2. dP′dP=0.2(1−2P300)=0  ⟹  P=150\dfrac{dP'}{dP} = 0.2(1 - \dfrac{2P}{300}) = 0 \implies P = 150
  3. Population for fastest growth =150= 150

Answer: 150150

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §6.3,p6.3, p. 415415. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.