A-Level Differential equations Practice Questions

Free A-Level Differential equations practice questions with full step-by-step worked solutions. Covers separation of variables, general solution, exponential, modelling. Practise exam-style problems and check your method.

separation of variablesgeneral solutionexponentialmodellingdecaycooling
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Find the general solution of the differential equation dydx=2y\frac{dy}{dx} = 2 y.
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Worked solution

  1. Separate the variables

    1y dy=2 dx\frac{1}{y}\,dy = 2\,dx

    Collect all the y terms on one side and the x terms on the other.

  2. Integrate both sides

    ln⁡y=2x+c\ln y = 2 x + c

    Integrate each side with respect to its own variable and add one constant of integration.

  3. State the general solution

    y=Ae2xy = A e^{2 x}

    This is the general solution; the arbitrary constant absorbs ece^{c}.

Answer
y=Ae2xy = A e^{2 x}
Question 2
2 markseasy
Which of the following differential equations is separable?
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Worked solution

  1. Recall what separable means

    dydx=f(x) g(y)\frac{dy}{dx} = f(x)\,g(y)

    The right side must factor into a function of xx times a function of yy.

  2. Test xyxy

    xy=(x)(y)xy = (x)(y)

    This factorises, so it is separable.

  3. Select the separable equation

    dydx=xy\frac{dy}{dx} = xy

    This is the only one that factorises.

Answer
dydx=xy\frac{dy}{dx} = xy
Question 3
3 marksintermediate
In the model dydt=ky\frac{dy}{dt}=ky describing decay, what is the sign of kk?
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Worked solution

  1. Write the model

    dydt=ky\frac{dy}{dt} = ky

    The rate is proportional to the amount.

  2. Recall the sign of the derivative for decay

    dydt<0\frac{dy}{dt} < 0

    Decay means the amount is decreasing.

  3. Since y>0y>0, deduce the sign of kk

    k<0k < 0

    Divide the inequality by the positive quantity yy.

  4. State the conclusion

    k<0 (negative)k<0\ \text{(negative)}

    This fixes the sign of the constant.

  5. Contrast with the opposite case

    the reverse sign gives the other behaviour\text{the reverse sign gives the other behaviour}

    Flipping the sign flips growth and decay.

  6. Select the correct sign of kk

    k<0 (negative)k<0\ \text{(negative)}

    This is required for decay.

Answer
k<0 (negative)k<0\ \text{(negative)}
Question 4
5 markshard
Which differential equation represents Newton's law of cooling toward ambient temperature θ0\theta_0?
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Worked solution

  1. Recall Newton's law of cooling

    rate∝excess temperature\text{rate} \propto \text{excess temperature}

    An object cools faster when it is much hotter than its surroundings.

  2. Identify the excess temperature

    θ−θ0\theta - \theta_0

    θ0\theta_0 is the ambient temperature.

  3. The rate is proportional to this difference

    dθdt∝(θ−θ0)\frac{d\theta}{dt} \propto (\theta - \theta_0)

    Proportionality gives the structure of the equation.

  4. Choose the sign for cooling

    dθdt=−k(θ−θ0)\frac{d\theta}{dt} = -k(\theta - \theta_0)

    A hotter object cools, so the rate is negative.

  5. Note the limiting behaviour

    θ→θ0\theta \to \theta_0

    The object approaches the surrounding temperature.

  6. Reject models missing θ0\theta_0

    must depend on θ−θ0\text{must depend on } \theta - \theta_0

    Cooling depends on the difference, not the temperature alone.

  7. Compare with the given options

    examine each option in turn\text{examine each option in turn}

    Only one option has the correct structure.

  8. Eliminate the options that do not fit

    reject the mismatched forms\text{reject the mismatched forms}

    Discard options with the wrong sign, power or constant.

  9. Verify the remaining option is consistent

    the chosen form reverses correctly\text{the chosen form reverses correctly}

    Reversing the working confirms the selected statement.

  10. Identify Newton's law of cooling

    dθdt=−k(θ−θ0)\frac{d\theta}{dt} = -k(\theta - \theta_0)

    This is the standard cooling model.

Answer
dθdt=−k(θ−θ0)\frac{d\theta}{dt} = -k(\theta - \theta_0)
Question 5
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.
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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}}

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