Each year the population1 grows by 4% and population2 by 2%. Also each year 3% of population1 leaves it and go to population2 and 1% vice versa, leaves population2 and goes to population1.

Kamden Larson

Kamden Larson

Answered question

2022-10-28

differential equation of a population growth and change
I want to formulate a system of equations and initial conditions of the following data: Each year the population1 grows by 4% and population2 by 2%. Also each year 3% of population1 leaves it and go to population2 and 1% vice versa, leaves population2 and goes to population1.
The initial populations are: p o p u l a t i o n 1 = 20000000 , p o p u l a t i o n 2 = 5000000.
I thought of the next formalism of this:
x ( t + 1 ) = 1.04 x ( t ) 0.03 x ( t ) + 0.01 y ( t )
y ( t + 1 ) = 1.02 y ( t ) 0.01 y ( t ) + 0.03 x ( t )
where x(t) is the size of population1 and y(t) is the size of population2. Is this is correct in those term or maybe I should reformulate it in another way ?
I think that I need some how to get something like this (ignore the concrete number and function):
X ¯ = ( 1 1 2 2 ) X ¯ + ( e t 2 e t )
with an initial conditions like that:
X ¯ ( 0 ) = ( 1 0 )
but I don't know how to do it, so will be grateful for some help.
An idea I had is:
x ( t ) = x ( t + 1 ) x ( t ) ( t + 1 t ) = ( 1.04 0.03 1 ) x ( t ) + 0.01 y ( t )
is it correct ?

Answer & Explanation

Alannah Yang

Alannah Yang

Beginner2022-10-29Added 22 answers

Step 1
For population denoted by x, you get
x = .04 x .03 x + .01 y
and for the second population denoted by y you get
y = .02 y + .03 x .01 y
Step 2
That simplifies to
x = .01 x + .01 y
and
y = .03 x + .01 y
Now you write it in matrix form.
bergvolk0k

bergvolk0k

Beginner2022-10-30Added 4 answers

Step 1
For a year you're modeling the overall change in the first population, x ( t + 1 ) x ( t ), as being the result of:
- Some sort of inherent growth, adding 0.04 × x ( t ).
- A loss to population 2, removing 0.03 × x ( t ).
- New additions coming from population 2, adding 0.01 × y ( t ).
That gives you
x ( t + 1 ) x ( t ) = ( 0.04 0.03 ) x ( t ) + 0.01 y ( t ) .
Same sort of thing for population 2:
y ( t + 1 ) y ( t ) = 0.03 x ( t ) + ( 0.02 0.01 ) y ( t ) .
Maybe easier to write
x ( t + Δ t ) x ( t ) ) = ( A Δ t ) x ( t ) + ( B Δ t ) y ( t ) y ( t + Δ t ) y ( t ) ) = ( C Δ t ) x ( t ) + ( D Δ t ) y ( t )
where A, B, C, D are rates since it looks like you might have an eye on shifting to a continuous model ( Δ t 0). In that case you get the system
x ( t ) = A x ( t ) + B y ( t ) y ( t ) = C x ( t ) + D y ( t ) ,
and typically you'd have initial conditions.
Step 2
But, if that's not what you're after, the original problem would end up looking like
[ x ( t + 1 ) y ( t + 1 ) ] = [ 0.01 0.01 0.03 0.01 ] [ x ( t ) y ( t ) ] ,
and if you're working from initial conditions you end up with
[ x ( n ) y ( n ) ] = [ 0.01 0.01 0.03 0.01 ] n [ x ( 0 ) y ( 0 ) ] .

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