Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 6A (Differential Equations)

Table of contents
- a 6A
  - Solution a
- b 6A
  - Solution b

a

Solution

The homogeneous characteristic equation is

r^{2} + r - 6 = (r - 2) (r + 3) = 0.

A linear particular solution

u_{n} = an + b

gives

- 4 an + 11 a - 4 b = n,

so

a = - 1/4

and

b = - 11/16

. Thus

u_{n} = A 2^{n} + B (- 3)^{n} - \frac{n}{4} - \frac{11}{16} .

The initial data yield

A = 1

and

B = - 5/16

, hence

u_{n} = 2^{n} - \frac{5}{16} (- 3)^{n} - \frac{n}{4} - \frac{11}{16} .

For the ordinary generating function of a recurrence, multiply by

x^{n}

and sum for

n \geq 2

. Using

u_{0} = 0, u_{1} = 2

gives

(1 + x - 6 x^{2}) G (x) = x + \frac{x}{( 1 - x ) ^{2}} .

Equivalently,

G (x) = \frac{1}{1 - 2 x} - \frac{5}{16 ( 1 + 3 x )} - \frac{x}{4 ( 1 - x ) ^{2}} - \frac{11}{16 ( 1 - x )} .

Expanding each geometric series gives exactly the displayed formula for

u_{n}

, verifying consistency.

b

Solution

The floor-half generating function is

H (x) = \sum_{n \geq 1} ⌊ \frac{n}{2} ⌋ x^{n} = \frac{x ^{2}}{( 1 - x ) ^{2} ( 1 + x )} .

Summing

u_{n} - 2 u_{n - 1} = ⌊ n /2 ⌋

for

n \geq 1

gives

G (x) - 1 - 2 x G (x) = H (x),

so

G (x) = \frac{1 + \frac{x ^{2}}{( 1 - x ) ^{2} ( 1 + x )}}{1 - 2 x} .

Since

⌊ \frac{n}{2} ⌋ = \frac{n}{2} - \frac{1}{4} + \frac{1}{4} (- 1)^{n},

a particular solution is

- \frac{n}{2} - \frac{3}{4} + \frac{1}{12} (- 1)^{n} .

Adding the homogeneous term

C 2^{n}

and imposing

u_{0} = 1

gives

C = 5/3

. Therefore

u_{n} = \frac{5}{3} 2^{n} - \frac{n}{2} - \frac{3}{4} + \frac{1}{12} (- 1)^{n} .

 Ancestors (9)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version