Linear differential equations and model solutions
Solve first-order linear equations with integrating factors and higher-order constant-coefficient equations with complementary and particular solutions.
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Find integrating factors for first-order linear equations
- Find complementary functions and suitable particular integrals
- Use given substitutions and initial conditions
- Interpret a solution in the context of a differential-equation model
An integrating factor turns a derivative into a product
Write a first-order linear equation as y′+P(x)y=Q(x). The integrating factor is μ=e^{∫P(x)dx}; then (μy)′=μQ and one integration gives the general solution.
Use an initial condition only after finding the general solution. It selects the constant and may determine whether the model is physically meaningful over the required interval.
Constant-coefficient equations split into two parts
For a homogeneous linear equation, form the auxiliary polynomial. Distinct real roots give exponential terms; a repeated root adds a factor of x; a conjugate pair gives e^{ax}(C cos bx+D sin bx).
A general solution is complementary function plus a particular integral. Choose the particular form to match the forcing term; if it duplicates a complementary term, multiply the trial by x.
A specified substitution may transform an equation into a constant-coefficient or separable form. Follow the given substitution and transform derivatives consistently.
Pause and explain
Which habit best checks your work on linear differential equations and model solutions?
Worked example
Solve y′+2y=e⁻ˣ for the general solution.
Show the worked solution
- The integrating factor is e^{∫2dx}=e²ˣ.
- Then (e²ˣy)′=eˣ, so e²ˣy=eˣ+C.
- Divide by e²ˣ.
Answer y=e⁻ˣ+Ce⁻²ˣ.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve y′−y=2x, given y(0)=0.
Give me a hint
Use integrating factor e⁻ˣ, then integrate and apply the condition.
Compare my reasoning
- The integrating factor is e⁻ˣ, so (e⁻ˣy)′=2xe⁻ˣ.
- Integrating gives e⁻ˣy=−2(x+1)e⁻ˣ+C.
- Hence y=−2x−2+Ceˣ; y(0)=0 gives C=2.
y=2eˣ−2x−2.
Look for these in your work
- Integrating factor correct
- General solution formed
- Initial value applied
Solve y″−3y′+2y=eˣ for a particular integral and the general solution.
Give me a hint
The complementary equation has roots 1 and 2, so eˣ overlaps with the complementary function.
Compare my reasoning
- The auxiliary equation m²−3m+2=0 gives m=1,2; CF=C₁eˣ+C₂e²ˣ.
- Try yₚ=Axeˣ because eˣ is already in the CF.
- Substitution gives −Aeˣ=eˣ, so A=−1.
y=C₁eˣ+C₂e²ˣ−xeˣ.
Look for these in your work
- Complementary roots correct
- Resonant trial adjusted by x
- Particular coefficient verified
A model satisfies y′=−ky for k>0 and y(0)=y₀>0. Solve it and explain its long-term behaviour.
Give me a hint
Separate variables or recognise a first-order linear equation, then use the initial value.
Compare my reasoning
- dy/y=−k dt, so ln y=−kt+C.
- Thus y=Ae⁻ᵏᵗ; the condition y(0)=y₀ gives A=y₀.
- Because k>0, e⁻ᵏᵗ tends to zero as t grows, so the model decays toward zero.
y(t)=y₀e⁻ᵏᵗ; it decreases toward zero for t≥0.
Look for these in your work
- Differential equation solved
- Initial condition used
- Model behaviour interpreted
Common mistakes
- Using e^{∫Qdx} instead of e^{∫Pdx} as the integrating factor
- Forgetting the complementary function when finding a particular solution
- Failing to multiply the trial by x when it duplicates a complementary term
What can you explain now?
For y′+P(x)y=Q(x), what is the integrating factor?
Compare with the explanation
e^{∫P(x)dx}
It makes the left side equal to the derivative of μy.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do an integrating factor and a complementary-function/particular-integral split solve different linear equations?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.