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Differential Equations

6.2 Differential Equations: Growth and Decay

Focus on exponential growth and decay, half-life, doubling and rate interpretation.

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Ideas in this lesson

01Exponential growth and decayWorked steps · five questions02Half-lifeWorked steps · five questions03Doubling and rate interpretationWorked steps · five questions
Idea 1 of 3

Exponential growth and decay

A constant relative rate y′=y' = ky gives y=y = Ce kt^{kt}. The sign of kk distinguishes growth from decay.

Worked example 1

The function y=y = Ce 2x^{2x} solves y′=2yy' = 2y. If y(0)=1y(0) = 1, find CC.
  1. y′=2y' = 2 Ce 2x=2y^{2x} = 2y
  2. y(0)=C=1y(0) = C = 1
  3. C=1C = 1

Answer: 11

Worked example 2

The function y=y = Ce 3x^{3x} solves y′=3yy' = 3y. If y(0)=2y(0) = 2, find y(1)y(1).
  1. y′=3y' = 3 Ce 3x=3y^{3x} = 3y
  2. y(0)=C=2y(0) = C = 2
  3. y(1)=2e3=2e3≈40.17y(1) = 2e^{3} = 2e^{3} \approx 40.17

Answer: 2e3≈40.172e^{3} \approx 40.17

Idea 2 of 3

Half-life

A decay process loses the same fraction over each half-life, not the same fixed amount.

Worked example 1

A substance starts with 200200 grams and has half-life 22 hours. Find the mass after one half-life.
  1. y(t)=2002−t2y(t) = 2002^{-\dfrac{t}{2}}
  2. k=ln⁡(12)2k = \dfrac{\ln (\dfrac{1}{2})}{2}
  3. y(2)=2002=100y(2) = \dfrac{200}{2} = 100

Answer: 100100

Worked example 2

A substance starts with 300300 grams and has half-life 44 hours. Find the mass after three half-lives.
  1. y(t)=3002−t4y(t) = 3002^{-\dfrac{t}{4}}
  2. k=ln⁡(12)4k = \dfrac{\ln (\dfrac{1}{2})}{4}
  3. y(12)=3008=752≈37.5y(12) = \dfrac{300}{8} = \dfrac{75}{2} \approx 37.5

Answer: 752≈37.5\dfrac{75}{2} \approx 37.5

Idea 3 of 3

Doubling and rate interpretation

Solve the exponential model for time using logarithms. Keep time and rate units consistent.

Worked example 1

A quantity follows Q(t)=2e0.1tQ(t) = 2e^{0.1t}. Find Q(2)Q(2).
  1. Q(t)=2e0.1tQ(t) = 2e^{0.1t}; Q′(t)=0.1Q(t)Q'(t) = 0.1Q(t)
  2. Q(2)=2e0.2=2e0.2≈2.44Q(2) = 2e^{0.2} = 2e^{0.2} \approx 2.44

Answer: 2e0.2≈2.442e^{0.2} \approx 2.44

Worked example 2

A quantity follows Q(t)=3e0.2tQ(t) = 3e^{0.2t}. Find its doubling time.
  1. Q(t)=3e0.2tQ(t) = 3e^{0.2t}; Q′(t)=0.2Q(t)Q'(t) = 0.2Q(t)
  2. Doubling: e0.2t=2  ⟹  t=ln⁡20.2=ln⁡20.2≈3.47e^{0.2t} = 2 \implies t = \dfrac{\ln 2}{0.2} = \dfrac{\ln 2}{0.2} \approx 3.47

Answer: ln⁡20.2≈3.47\dfrac{\ln 2}{0.2} \approx 3.47

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §6.2,p6.2, p. 407407. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.