# Thread: Formulation of seconder order difference equation

1. ## Formulation of seconder order difference equation

If $p_{n}$ denotes the population of a rapidly growing insect colony at the end of year n:
(i) write down an expression for the increase in population during year n;
(ii) write down an expression for the increase in population during year n-1;
(iii) given that the increase in population each year is three times the increase in the previous year formulate a second order differential equation for $p_{n}$. and show that the general solution is $p_{n}=A(3)^n+B$

my answer for i)
let $G_{n}$ represent the growth of the insect in year n; $P_{n+1}=P_{n}+G_{n}$
ii) $P_{n-1}=P_{n-2}+G_{n-2}$

I'm wonder how to formulate a second order differential equation for this model? thanks

2. Originally Posted by hazeleyes
If $p_{n}$ denotes the population of a rapidly growing insect colony at the end of year n:
(i) write down an expression for the increase in population during year n;
(ii) write down an expression for the increase in population during year n-1;
(iii) given that the increase in population each year is three times the increase in the previous year formulate a second order differential equation for $p_{n}$. and show that the general solution is $p_{n}=A(3)^n+B$

my answer for i)
let $G_{n}$ represent the growth of the insect in year n; $P_{n+1}=P_{n}+G_{n}$
ii) $P_{n-1}=P_{n-2}+G_{n-2}$

I'm wonder how to formulate a second order differential equation for this model? thanks
(iii)Defining...

(1)

... and considering the grow rate equation...

(2)

... You arrive at the second order difference equation...

(3)

... which has 'characteristic equation'...

(4)

... that has soltions r=3 and r=1 so that the general solution of (3) is...

(5)

The constants a and b are derived from the knowledge of and ...

Kind regards

$\chi$ $\sigma$

3. Thanks so much. If I have the information for the population for two years ago and one year ago only. to find a and b. should I substitute in (5) ? thanks

4. Like two years ago the population was 2000 and one year ago it was 3000 determine the population 3 years from now.
substitute in (5),
for two years ago: $p_{n-2}=a3^{n-2}+b$
make b the subject, $b=2000-a3^{n-2}$
and
one year ago: $p_{n-1}=a3^{n-1}+b$
plug b in this equation: $1000=a3^{n-1}-a3^{n-2}$

I am not sure is this correct?

5. Originally Posted by hazeleyes
Like two years ago the population was 2000 and one year ago it was 3000 determine the population 3 years from now.
substitute in (5)
for two years ago: $p_{n-2}=a3^{n-2}+b$
make b the subject, $b=2000-a3^{n-2}$
and
one year ago: $p_{n-1}=a3^{n-1}+b$
plug b in this equation: $1000=a3^{n-1}-a3^{n-2}$

I am not sure is this correct?
Setting and You arrive at the pair of equations...

a + b = 2000

3 a + b = 3000 (1)

... the solution of which is a=500, b=1500. Till three years the insect population will be...

Kind regards

$\chi$ $\sigma$

6. Thanks so much!!

7. [QUOTE=chisigma;646472](iii)Defining...

(1)

Hi I'm sorry. I'm wondering how did you define (1). Thanks