Hard A-Level Differential equations Questions

Challenging, exam-style A-Level Differential equations questions with worked solutions. Stretch yourself on the hardest separation of variables, particular solution, boundary condition, general solution problems.

separation of variablesparticular solutionboundary conditiongeneral solutionexponentialintegration
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A liquid cools according to Newton's law of cooling. The object starts at 100∘C100^{\circ}\text{C} in surroundings at 20∘C20^{\circ}\text{C}, and after 55 it is 60∘C60^{\circ}\text{C}. Find θ\theta as a function of tt.
Show worked solution

Worked solution

  1. Set up the differential equation

    dθdt=−k(θ−20)\frac{d\theta}{dt} = -k(\theta - 20)

    Newton's law: the cooling rate is proportional to the excess temperature.

  2. Separate the variables

    1θ−20 dθ=−k dt\frac{1}{\theta - 20}\,d\theta = -k\,dt

    Divide by the temperature difference.

  3. Integrate both sides

    ∫1θ−20 dθ=∫−k dt\int \frac{1}{\theta - 20}\,d\theta = \int -k\,dt

    The left side integrates to a logarithm.

  4. Carry out the integration

    ln⁡(θ−20)=−kt+c\ln(\theta - 20) = -kt + c

    Include a constant of integration.

  5. Exponentiate to get the general model

    θ=20+Ae−kt\theta = 20 + A e^{-kt}

    The temperature tends to the ambient value as tt grows.

  6. Apply the condition at t=0t=0

    100=20+A⇒A=80100 = 20 + A \Rightarrow A = 80

    At t=0t=0 the exponential is 1.

  7. Write the model so far

    θ=20+80e−kt\theta = 20 + 80 e^{-kt}

    The initial temperature is now included.

  8. Use the temperature at t=5t=5

    60=20+80e−5k60 = 20 + 80 e^{-5k}

    Substitute the second reading to find kk.

  9. Solve for kk

    e−5k=12⇒k=ln⁡(2)5e^{-5k} = \frac{1}{2} \Rightarrow k = \frac{\ln{\left(2 \right)}}{5}

    Rearrange and take logarithms.

  10. Write the complete model

    θ=20+80⋅2−t5\theta = 20 + 80 \cdot 2^{- \frac{t}{5}}

    Using e−kt=12t/5e^{-kt}=\frac{1}{2}^{t/5} gives a clean power form.

  11. Check the initial value

    t=0: θ=100t=0:\ \theta = 100

    The model reproduces the starting amount.

  12. Check the second data point

    t=5: θ=60t=5:\ \theta = 60

    The model reproduces the second reading.

  13. Interpret the model

    exponential change with constant k\text{exponential change with constant } k

    The behaviour is exponential, as expected for a proportional rate.

  14. Note the long-term behaviour

    approaches the limiting value\text{approaches the limiting value}

    This describes what happens for large times.

  15. State the model

    θ=20+80⋅2−t5\theta = 20 + 80 \cdot 2^{- \frac{t}{5}}

    This particular solution fits both temperature readings.

Answer
θ=20+80⋅2−t5\theta = 20 + 80 \cdot 2^{- \frac{t}{5}}
Question 2
8 markschallenging
A population grows at a rate proportional to its size. The amount starts at 500 units500\ \text{units} and is 1000 units1000\ \text{units} after 55. Find PP as a function of tt.
Show worked solution

Worked solution

  1. Set up the differential equation

    dPdt=kP\frac{dP}{dt} = kP

    The rate of change is proportional to the amount present.

  2. Separate the variables

    1P dP=k dt\frac{1}{P}\,dP = k\,dt

    Divide by P and multiply by dtdt.

  3. Integrate both sides

    ∫1P dP=∫k dt\int \frac{1}{P}\,dP = \int k\,dt

    Both integrals are standard.

  4. Carry out the integration

    ln⁡P=kt+c\ln P = kt + c

    Integrating 1/P1/P gives a natural logarithm.

  5. Exponentiate to get the general model

    P=AektP = A e^{kt}

    Write A=ecA=e^{c} for the arbitrary constant.

  6. Apply the condition at t=0t=0

    500=A500 = A

    At t=0t=0 the exponential equals 1, so AA is the initial amount.

  7. Write the model so far

    P=500ektP = 500 e^{kt}

    The initial value is now built in.

  8. Use the value at t=5t=5

    1000=500e5k1000 = 500 e^{5k}

    Substitute the second data point to find kk.

  9. Solve for kk

    k=15ln⁡2=ln⁡(2)5k = \frac{1}{5}\ln 2 = \frac{\ln{\left(2 \right)}}{5}

    Rearranging gives the growth/decay constant.

  10. Write the complete model

    P=500⋅2t5P = 500 \cdot 2^{\frac{t}{5}}

    Using ekt=2t/5e^{kt}=2^{t/5} gives a clean power form.

  11. Check the initial value

    t=0: P=500t=0:\ P = 500

    The model reproduces the starting amount.

  12. Check the second data point

    t=5: P=1000t=5:\ P = 1000

    The model reproduces the second reading.

  13. Interpret the model

    exponential change with constant k\text{exponential change with constant } k

    The behaviour is exponential, as expected for a proportional rate.

  14. Note the long-term behaviour

    as t→∞\text{as } t\to\infty

    This describes what happens for large times.

  15. State the model

    P=500⋅2t5P = 500 \cdot 2^{\frac{t}{5}}

    This particular solution fits both data points.

Answer
P=500⋅2t5P = 500 \cdot 2^{\frac{t}{5}}
Question 3
8 markschallenging
A radioactive material decays proportionally to the mass present. The amount starts at 100 g100\ \text{g} and is 50 g50\ \text{g} after 1010. Find the exact value of the constant kk in dNdt=kN\frac{dN}{dt}=kN.
Show worked solution

Worked solution

  1. Set up the differential equation

    dNdt=kN\frac{dN}{dt} = kN

    The rate of change is proportional to the amount present.

  2. Separate the variables

    1N dN=k dt\frac{1}{N}\,dN = k\,dt

    Divide by N and multiply by dtdt.

  3. Integrate both sides

    ∫1N dN=∫k dt\int \frac{1}{N}\,dN = \int k\,dt

    Both integrals are standard.

  4. Carry out the integration

    ln⁡N=kt+c\ln N = kt + c

    Integrating 1/N1/N gives a natural logarithm.

  5. Exponentiate to get the general model

    N=AektN = A e^{kt}

    Write A=ecA=e^{c} for the arbitrary constant.

  6. Apply the condition at t=0t=0

    100=A100 = A

    At t=0t=0 the exponential equals 1, so AA is the initial amount.

  7. Write the model so far

    N=100ektN = 100 e^{kt}

    The initial value is now built in.

  8. Use the value at t=10t=10

    50=100e10k50 = 100 e^{10k}

    Substitute the second data point to find kk.

  9. Solve for kk

    k=110ln⁡12=−ln⁡(2)10k = \frac{1}{10}\ln \frac{1}{2} = - \frac{\ln{\left(2 \right)}}{10}

    Rearranging gives the growth/decay constant.

  10. Write the complete model

    N=100⋅2−t10N = 100 \cdot 2^{- \frac{t}{10}}

    Using ekt=12t/10e^{kt}=\frac{1}{2}^{t/10} gives a clean power form.

  11. Check the initial value

    t=0: N=100t=0:\ N = 100

    The model reproduces the starting amount.

  12. Check the second data point

    t=10: N=50t=10:\ N = 50

    The model reproduces the second reading.

  13. Interpret the model

    exponential change with constant k\text{exponential change with constant } k

    The behaviour is exponential, as expected for a proportional rate.

  14. Note the long-term behaviour

    approaches the limiting value\text{approaches the limiting value}

    This describes what happens for large times.

  15. State the constant

    k=−ln⁡(2)10k = - \frac{\ln{\left(2 \right)}}{10}

    This constant controls the rate of growth/decay.

Answer
−ln⁡(2)10- \frac{\ln{\left(2 \right)}}{10}
Question 4
8 markschallenging
A quantity grows at a rate proportional to its size. The amount starts at 500 units500\ \text{units} and is 1000 units1000\ \text{units} after 55. Find the exact value of the constant kk in dPdt=kP\frac{dP}{dt}=kP.
Show worked solution

Worked solution

  1. Set up the differential equation

    dPdt=kP\frac{dP}{dt} = kP

    The rate of change is proportional to the amount present.

  2. Separate the variables

    1P dP=k dt\frac{1}{P}\,dP = k\,dt

    Divide by P and multiply by dtdt.

  3. Integrate both sides

    ∫1P dP=∫k dt\int \frac{1}{P}\,dP = \int k\,dt

    Both integrals are standard.

  4. Carry out the integration

    ln⁡P=kt+c\ln P = kt + c

    Integrating 1/P1/P gives a natural logarithm.

  5. Exponentiate to get the general model

    P=AektP = A e^{kt}

    Write A=ecA=e^{c} for the arbitrary constant.

  6. Apply the condition at t=0t=0

    500=A500 = A

    At t=0t=0 the exponential equals 1, so AA is the initial amount.

  7. Write the model so far

    P=500ektP = 500 e^{kt}

    The initial value is now built in.

  8. Use the value at t=5t=5

    1000=500e5k1000 = 500 e^{5k}

    Substitute the second data point to find kk.

  9. Solve for kk

    k=15ln⁡2=ln⁡(2)5k = \frac{1}{5}\ln 2 = \frac{\ln{\left(2 \right)}}{5}

    Rearranging gives the growth/decay constant.

  10. Write the complete model

    P=500⋅2t5P = 500 \cdot 2^{\frac{t}{5}}

    Using ekt=2t/5e^{kt}=2^{t/5} gives a clean power form.

  11. Check the initial value

    t=0: P=500t=0:\ P = 500

    The model reproduces the starting amount.

  12. Check the second data point

    t=5: P=1000t=5:\ P = 1000

    The model reproduces the second reading.

  13. Interpret the model

    exponential change with constant k\text{exponential change with constant } k

    The behaviour is exponential, as expected for a proportional rate.

  14. Note the long-term behaviour

    as t→∞\text{as } t\to\infty

    This describes what happens for large times.

  15. State the constant

    k=ln⁡(2)5k = \frac{\ln{\left(2 \right)}}{5}

    This constant controls the rate of growth/decay.

Answer
ln⁡(2)5\frac{\ln{\left(2 \right)}}{5}
Question 5
8 markschallenging
A culture increases at a rate proportional to its size. The initial amount is 1000 cells1000\ \text{cells} and after 22 it is 3000 cells3000\ \text{cells}. Find the time taken to reach 27000 cells27000\ \text{cells}.
Show worked solution

Worked solution

  1. Set up the differential equation

    dPdt=kP\frac{dP}{dt} = kP

    The rate of change is proportional to the amount present.

  2. Separate the variables

    1P dP=k dt\frac{1}{P}\,dP = k\,dt

    Divide by P and multiply by dtdt.

  3. Integrate both sides

    ∫1P dP=∫k dt\int \frac{1}{P}\,dP = \int k\,dt

    Both integrals are standard.

  4. Carry out the integration

    ln⁡P=kt+c\ln P = kt + c

    Integrating 1/P1/P gives a natural logarithm.

  5. Exponentiate to get the general model

    P=AektP = A e^{kt}

    Write A=ecA=e^{c} for the arbitrary constant.

  6. Apply the condition at t=0t=0

    1000=A1000 = A

    At t=0t=0 the exponential equals 1, so AA is the initial amount.

  7. Write the model so far

    P=1000ektP = 1000 e^{kt}

    The initial value is now built in.

  8. Use the value at t=2t=2

    3000=1000e2k3000 = 1000 e^{2k}

    Substitute the second data point to find kk.

  9. Solve for kk

    k=12ln⁡3=ln⁡(3)2k = \frac{1}{2}\ln 3 = \frac{\ln{\left(3 \right)}}{2}

    Rearranging gives the growth/decay constant.

  10. Write the complete model

    P=1000⋅3t2P = 1000 \cdot 3^{\frac{t}{2}}

    Using ekt=3t/2e^{kt}=3^{t/2} gives a clean power form.

  11. Set the model equal to 2700027000

    27000=1000 3t/227000 = 1000\,3^{t/2}

    We need the time when the amount reaches this value.

  12. Solve for tt

    t=6t = 6

    Take logarithms (or compare powers) to find the time.

  13. Check the initial value

    t=0: P=1000t=0:\ P = 1000

    The model reproduces the starting amount.

  14. Check the second data point

    t=2: P=3000t=2:\ P = 3000

    The model reproduces the second reading.

  15. State the time

    t=6t = 6

    This is when the quantity reaches the target value.

Answer
66

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