NCEA Level 3Differential equationsAS 91579

Differential equations: growth, decay and Newton's cooling

What this lesson covers

  • A differential equation gives a rate; solving it gives the quantity itself.
  • Separating variables and integrating both sides leaves one constant, found from the starting condition.
Key idea

A rate, and the quantity behind it

  • A differential equation gives a rate.
  • Solving it gives the quantity.
  • Every integration brings a constant.
Worked example

dy/dx = 6x² − 4, y = 5 at x = 1

  • Integrate, with + c.
  • Put in the point to find c.
  1. dy/dx = 6x² − 4
  2. y = ∫ (6x² − 4) dxintegrate
  3. = 2x³ − 4x + ckeep the + c
  4. 5 = 2(1)³ − 4(1) + cx = 1, y = 5
  5. 5 = −2 + c, so c = 7
  6. y = 2x³ − 4x + 7
Worked example

A second derivative: a = 4e²ᵗ

  • Integrate twice.
  • Two constants, in turn.
  1. a = 4e²ᵗat rest at the origin
  2. v = ∫ 4e²ᵗ dt = 2e²ᵗ + c₁integrate
  3. 0 = 2e⁰ + c₁ = 2 + c₁v = 0 at t = 0
  4. c₁ = −2
  5. v = 2e²ᵗ − 2
  6. s = ∫ (2e²ᵗ − 2) dtintegrate again
  7. = e²ᵗ − 2t + c₂a new constant
Worked example

s = 0 at t = 0 does not mean c = 0

  • e⁰ = 1, not 0.
  • Always substitute.
  1. s = e²ᵗ − 2t + c₂
  2. c₂ = 0assumed, not found
  3. 0 = e⁰ − 2(0) + c₂s = 0 at t = 0
  4. 0 = 1 + c₂, so c₂ = −1e⁰ = 1
  5. s = e²ᵗ − 2t − 1
  6. s(0) = 1 − 0 − 1 = 0
  7. c₂ = 0 gives s(0) = 1not 0
Worked example

Separating variables: dy/dx = ky

  • Separate y and x.
  • The c goes into the power.
  1. dy/dx = ky
  2. ∫ (1/y) dy = ∫ k dxseparate
  3. ln y = kx + cy > 0, no modulus
  4. y = e^(kx + c)undo the log
  5. y = Ae^(kx)A = e^c
Worked example

One constant, all the way: a 2025 error

  • The c goes into the power.
  • Call it A and keep it.
  1. y = e^(kx) + cc moved
  2. y = e^(kx + c)
  3. = e^(kx) × e^cindex law
  4. = Ae^(kx)one constant, A
Worked example

Growth: N = Ae^(kt)

  • t = 0 gives A.
  • A second reading gives k.
  1. N = Ae^(kt)
  2. 200 = Ae⁰ = At = 0
  3. 800 = 200e^(2k)t = 2
  4. e^(2k) = 4divide by 200
  5. 2k = ln 4ln both sides
  6. k = (ln 4)/2 = ln 2 ≈ 0.6931
Worked example

When is N = 6400?

  • Isolate the exponential.
  • Answer in context.
  1. 6400 = 200e^(kt)
  2. e^(kt) = 32divide by 200
  3. kt = ln 32ln both sides
  4. t = ln 32 / ln 2k = ln 2
  5. = 5 ln 2 / ln 2 = 532 = 2⁵
  6. after 5 hours
Worked example

Newton's law of cooling

  • Separate and integrate.
  • T(0) = 100 gives A.
  1. dT/dt = −k(T − 20)
  2. ∫ dT/(T − 20) = ∫ −k dtseparate
  3. ln(T − 20) = −kt + cT > 20
  4. T − 20 = Ae^(−kt)A = e^c
  5. 100 − 20 = Ae⁰, so A = 80t = 0
  6. T = 20 + 80e^(−kt)
Worked example

Cooling towards the room

  • It falls towards 20 °C, never below.
  • T(10) = 60 gives k.
  • Then find when T = 40.
Cooling towards the room
Worked example

Finding k from T(10) = 60

  • Isolate the exponential.
  • Then ln both sides.
  1. T = 20 + 80e^(−kt)
  2. 60 = 20 + 80e^(−10k)t = 10, T = 60
  3. 40 = 80e^(−10k)subtract 20
  4. e^(−10k) = 1/2divide by 80
  5. −10k = ln(1/2) = −ln 2ln both sides
  6. k = ln 2 / 10 ≈ 0.0693divide by −10
Worked example

When is the water at 40 °C?

  • Isolate, then ln.
  • Keep k exact.
  1. T = 20 + 80e^(−kt)
  2. k = ln 2 / 10
  3. 40 = 20 + 80e^(−kt)T = 40
  4. 20 = 80e^(−kt)subtract 20
  5. e^(−kt) = 1/4divide by 80
  6. −kt = ln(1/4) = −ln 4ln both sides
  7. t = ln 4 / k = 10 ln 4 / ln 2
  8. = 10(2 ln 2)/ln 2 = 20ln 4 = 2 ln 2
  9. after 20 minutes
Worked example

Keep k exact

  • Isolate, then ln both sides.
  • Round only at the end.
  1. k = ln 2 / 10exact
  2. t = ln 4 / k = 20
  3. k ≈ 0.07rounded too early
  4. t = ln 4 / 0.07 ≈ 19.80drifts
Key idea

Where the marks are

  • Achieved: integrate, find the constant.
  • Merit: separate, use starting values.
  • Excellence: models in context, logs.
Summary

What to take away

  • Integrate, keep one constant, and find it by substituting.
y′ = f(x)integrate, find c
y″ = f(x)twice, two constants
y′ = kyy = Ae^(kx)
CoolingT − room = Ae^(−kt)
Find kisolate, then ln
cnever assume zero
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Question 2 has two constants.
  • Question 4: a room at 15 °C.
GivenFind
1dy/dx = e²ˣ, y(0) = 3y
2y″ = 12x, y′(0) = 1y, if y(1) = 5
3dN/dt = kN, N(0) = 50N(8), if N(4) = 150
4room 15 °C, T(0) = 75T(30), if T(10) = 45
Show answer

1. y = (1/2)e²ˣ + c; 3 = 1/2 + c, so y = (1/2)e²ˣ + 5/2. 2. y′ = 6x² + 1, y = 2x³ + x + c₂; y(1) = 5 gives c₂ = 2, so y = 2x³ + x + 2. 3. N = 50e^(kt); e^(4k) = 3, so N(8) = 50 × 3² = 450. 4. T − 15 = 60e^(−kt); e^(−10k) = 1/2, so T(30) = 15 + 60 × 1/8 = 22.5 °C.

Worked example

The answers: questions 1 and 2

  1. 1. y = ∫ e²ˣ dx = (1/2)e²ˣ + creverse chain rule
  2. 3 = (1/2)e⁰ + c, so c = 5/2x = 0, y = 3
  3. y = (1/2)e²ˣ + 5/2
  4. 2. y′ = ∫ 12x dx = 6x² + c₁integrate
  5. 1 = 6(0)² + c₁, so c₁ = 1y′(0) = 1
  6. y = ∫ (6x² + 1) dx = 2x³ + x + c₂
  7. 5 = 2 + 1 + c₂, so c₂ = 2y(1) = 5
  8. y = 2x³ + x + 2
Worked example

The answers: questions 3 and 4

  1. 3. N = 50e^(kt)N(0) = 50
  2. 150 = 50e^(4k), so e^(4k) = 3N(4) = 150
  3. N(8) = 50e^(8k) = 50(e^(4k))²index law
  4. = 50 × 3² = 450
  5. 4. T − 15 = 60e^(−kt)T(0) = 75, so A = 60
  6. 30 = 60e^(−10k), so e^(−10k) = 1/2
  7. T(30) = 15 + 60e^(−30k)
  8. = 15 + 60(1/2)³ = 15 + 7.5e^(−30k) = (1/2)³
  9. = 22.5 °C

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