Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 1C (Differential Equations)

Table of contents
- a 1C
  - Solution a
- b 1C
  - Solution b

a

Solution

Subtract successive terms and use a telescoping series:

z_{n + 1} - z_{1} = \sum_{j = 1}^{n} (z_{j + 1} - z_{j}) = \sum_{j = 1}^{n} c_{j} .

Thus the linear recurrence relation has solution

z_{n + 1} = z_{1} + j = 1 \sum n c_{j} .

b

Solution

Put

U_{0} = 1

, so the same definition

z_{n} = x_{n} / U_{n - 1}

includes

n = 1

. Since

U_{n} = a_{n} U_{n - 1}

and every

a_{n}

is nonzero,

z_{n + 1} = \frac{a _{n} x _{n} + b _{n}}{U _{n}} = z_{n} + \frac{b _{n}}{U _{n}} .

The factor

U_{n}^{- 1}

is a discrete integrating factor for this variable-coefficient affine recurrence. Sum the increments as a telescoping series and multiply back by

U_{n}

:

z_{n + 1} - z_{n} = \frac{b _{n}}{U _{n}}, x_{n + 1} = U_{n} (x_{1} + j = 1 \sum n \frac{b _{j}}{U _{j}}) .

The identity requires no assumption that the coefficients have the same sign.

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