Applied Maths: Difference Equations (Strand 8) — Higher Level Notes

First and second differences, recurrence relations, solving first- and second-order difference equations, and the interest and repayment problems built on them.

1. Differences tell you the shape

Given a sequence, subtract each term from the next:

ΔT_n = T_n+1 - T_n

The pattern identifies the sequence immediately:

First differences constant — the sequence is linear, T_n = an + b.
Second differences constant — quadratic. And usefully, the second difference equals 2a, so it hands you the leading coefficient directly.
Constant ratio rather than constant difference — geometric.

2. Recurrence relations

A recurrence defines each term from the ones before it. The two you know already:

Arithmetic: u_n+1 = u_n + d
Geometric: u_n+1 = ru_n

And the famous one, Fibonacci: u_n = u_n-1 + u_n-2, which is second-order because it reaches back two terms.

3. First-order difference equations

T_n+1 = aT_n + b

The standard solution splits into a general part and a particular part. The general solution of the homogeneous equation is A aⁿ; the particular solution for constant b is the fixed point, found by setting T_n+1 = T_n = k and solving. Add them, then use the initial condition to pin down A.

The fixed point has a meaning worth noticing: it is the value the sequence settles at if it converges, which is why it appears in loan and population problems as the balance that never changes.

4. Interest and repayments

These are first-order difference equations in disguise. A loan of P at interest rate i with repayment R each period gives:

A_n+1 = (1 + i)A_n - R

Set up the recurrence from the words, solve it in general, then substitute. The common exam question asks for the repayment that clears the loan after n periods, which means setting A_n = 0 and solving for R.

Exam technique — write the recurrence before anything else

Most marks in this chapter are lost in translation, not in algebra. Turn the sentence into a relation between consecutive terms first, and check it against the first two terms you can compute by hand. If the recurrence is right, the rest is mechanical.

5. Second-order homogeneous equations

pu_n+2 + qu_n+1 + ru_n = 0

Try u_n = kⁿ. That produces the characteristic quadratic pk² + qk + r = 0. Then:

Two distinct roots k₁, k₂: the solution is A k₁ⁿ + B k₂ⁿ.
One repeated root k: the solution is (A + Bn)kⁿ.

Two initial conditions fix A and B. For inhomogeneous equations, add a particular solution of the same form as the right-hand side.

Practise this chapter

QuizPerCard has Difference Equations as concept cards, a formula reference and multiple-choice practice — every question with the full derivation, and an explanation of why each wrong option is wrong. Open the practice.