Differential equations: 6 exam questions solved step by step

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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

6 questions, one method

  • 1. Separate: y terms left, x terms right.
  • 2. Integrate both sides, one + c.
  • 3. Find the constant from the condition.
  • 4. Logs to find k; exact until the end.
Your turn

Question 1 · Achieved

  • dy/dx = 6x − 2/x, for x > 0,
  • and y = 4 when x = 1. Find y.
Worked example

Q1: dy/dx = 6x − 2/x

  • Integrate, keep the + c.
  • The point finds c.
  1. y = ∫ (6x − 2/x) dxintegrate
  2. = 3x² − 2 ln x + ckeep the + c
  3. 4 = 3(1)² − 2 ln 1 + cx = 1, y = 4
  4. 4 = 3 − 0 + c, so c = 1ln 1 = 0
  5. y = 3x² − 2 ln x + 1
  6. dy/dx = 6x − 2/xdifferentiate back
Your turn

Question 2 · Achieved

  • d²y/dx² = 12x² − 6. The gradient is 3
  • at x = 1, and the curve passes
  • through (0, 2). Find y.
Worked example

Q2: d²y/dx² = 12x² − 6

  • Integrate twice.
  • Two constants, two facts.
  1. dy/dx = ∫ (12x² − 6) dxintegrate once
  2. = 4x³ − 6x + c₁
  3. 3 = 4 − 6 + c₁, so c₁ = 5x = 1, dy/dx = 3
  4. y = ∫ (4x³ − 6x + 5) dxintegrate again
  5. = x⁴ − 3x² + 5x + c₂a new constant
  6. 2 = 0 + c₂, so c₂ = 2point (0, 2)
  7. y = x⁴ − 3x² + 5x + 2
Your turn

Question 3 · Merit

  • dy/dx = 2x/y, and y = 3 when x = 0.
  • Find y in terms of x.
Worked example

Q3: dy/dx = 2x/y

  • Separate the variables.
  • One + c, one side only.
  1. ∫ y dy = ∫ 2x dxseparate
  2. y²/2 = x² + cone constant
  3. 9/2 = 0 + c, so c = 9/2x = 0, y = 3
  4. y²/2 = x² + 9/2
  5. y² = 2x² + 9× 2, every term
  6. y = ±√(2x² + 9)
  7. y = −√(2x² + 9)gives y = −3 at x = 0
  8. y = √(2x² + 9)
Your turn

Question 4 · Merit, in context

  • Kiwi on a predator-free island grow at
  • a rate proportional to the number, N.
  • N = 120 now, and 180 after 5 years.
  • When will there be 400 kiwi?
Worked example

Q4: the model, and k

  • Rate ∝ N: dN/dt = kN.
  • Two facts: A, then k.
  1. dN/dt = kNrate ∝ number
  2. ∫ (1/N) dN = ∫ k dtseparate
  3. ln N = kt + cN > 0
  4. N = Ae^(kt)A = e^c
  5. 120 = Ae⁰, so A = 120t = 0
  6. 180 = 120e^(5k)t = 5
  7. e^(5k) = 1.5÷ 120
  8. k = (ln 1.5)/5 ≈ 0.081093ln both sides
Worked example

Q4: when N = 400

  • Keep k exact.
  • Answer in years.
  1. N = 120e^(kt), k = (ln 1.5)/5
  2. 400 = 120e^(kt)N = 400
  3. e^(kt) = 10/3÷ 120
  4. kt = ln(10/3)ln both sides
  5. t = 5 ln(10/3) / ln 1.5k exact
  6. ≈ 14.847
  7. k ≈ 0.08 gives t ≈ 15.05k rounded too early
  8. after about 14.8 years
Your turn

Question 5 · Excellence, in context

  • A pie at 90 °C cools in a room at 18 °C,
  • by Newton's law: dT/dt = −k(T − 18).
  • After 5 minutes it is 60 °C.
  • When will it reach 30 °C?
Worked example

Q5: solve the equation

  • Separate, integrate.
  • T(0) = 90 finds A.
  1. dT/dt = −k(T − 18)Newton's law
  2. ∫ dT/(T − 18) = ∫ −k dtseparate
  3. ln(T − 18) = −kt + cT > 18
  4. T − 18 = Ae^(−kt)A = e^c
  5. 90 − 18 = Ae⁰, so A = 72t = 0
  6. T = 18 + 72e^(−kt)
Worked example

Q5: find k

  • Isolate the e first.
  • Then ln both sides.
  1. T = 18 + 72e^(−kt)
  2. 60 = 18 + 72e^(−5k)t = 5, T = 60
  3. 42 = 72e^(−5k)subtract 18
  4. e^(−5k) = 7/12÷ 72
  5. −5k = ln(7/12)ln both sides
  6. k = ln(12/7)/5 ≈ 0.10780÷ −5, flip the fraction
Worked example

Q5: when T = 30

  • Same steps again.
  • Answer in minutes.
  1. T = 18 + 72e^(−kt), k ≈ 0.10780
  2. 30 = 18 + 72e^(−kt)T = 30
  3. 12 = 72e^(−kt)subtract 18
  4. e^(−kt) = 1/6÷ 72
  5. kt = ln 6ln both sides
  6. t = 5 ln 6 / ln(12/7)k exact
  7. ≈ 16.62
  8. after about 16.6 minutes
Your turn

Question 6 · Excellence, in context

  • A drink at 4 °C is left in a room at 22 °C.
  • dT/dt = −k(T − 22). After 10 minutes
  • it is 10 °C. Find T in terms of t,
  • and when it reaches 16 °C.
Worked example

Q6: warming, not cooling

  • T starts below the room.
  • So A is negative.
  1. dT/dt = −k(T − 22)Newton's law
  2. T < 22, so T − 22 < 0the drink warms
  3. ln|T − 22| = −kt + cseparate, integrate
  4. T − 22 = Ae^(−kt)A = ±e^c
  5. 4 − 22 = Ae⁰, so A = −18t = 0
  6. T = 22 − 18e^(−kt)
  7. T = 22 + 18e^(−kt)T(0) would be 40
Worked example

Q6: k, then T = 16

  • Isolate the e each time.
  • k exact, round at the end.
  1. T = 22 − 18e^(−kt)
  2. 10 = 22 − 18e^(−10k)t = 10, T = 10
  3. e^(−10k) = 12/18 = 2/3−22, then ÷ −18
  4. k = −ln(2/3)/10 = ln 1.5/10ln both sides
  5. 16 = 22 − 18e^(−kt)T = 16
  6. e^(−kt) = 6/18 = 1/3−22, then ÷ −18
  7. t = 10 ln 3 / ln 1.5 ≈ 27.10
  8. after about 27.1 minutes
Summary

The method that solved all six

  • Separate, integrate, find the constant, then k.
StepHow
y′ = f(x)integrate, then find c
y″ = f(x)twice: two constants
Separatey terms left, x terms right
Constantone c, found from the start
Find kisolate e, then take ln
Answerexact k, round at the end

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