Question

The starting population of a certain kind of migratory birds in a forest is \(10000\).

\(3\%\) of the population die each year.

Reproduction rate is \(4\%\) per year and \(500\) birds immigrate to the forest each year.

Estimate the population of birds at the end of \(30\) th year

Answer 1

let \(P(n)\) denote the population after \(n\) years. we're given that \(P(0)=10000\) and the recurrence relation $$P(n)=0.04P(n-1)-0.03P(n-1)+500\\P(n)=0.01P(n-1)+500$$

Answer 2

so $$P(n)-0.01P(n-1)=500$$ignoring our initial condition, a constant particular solution \(P(n)=k\) gives \(k-0.01k=500\implies 0.99k=500\implies k=500/0.99=\frac{50000}{99}\)

Answer 3

Ahh that looks much better than what im doing..
i ended up with a geometric series after writing out the first few terms

Answer 4

now consider the homogeneous recurrence relation $$P(n)-0.01P(n-1)=0$$suppose that we had a geometric sequence \(P(n)=r^n\) which gives $$r^n-0.01r^{n-1}=0\\r^{n-1}\left(r-0.01\right)=0\implies r=0.01$$ so our homogeneous solution looks like \(P(n)=C(0.01)^n\) and thus the general solution to our original recurrence is \(P(n)=C(0.01)^n+\frac{50000}{99}\)

Answer 5

now we want to solve for \(C\) using our initial condition: $$P(0)=10000\\C(0.01)^0+\frac{50000}{99}=10000\\C=10000(1-5/99)\approx 9494.95$$

Answer 6

oops, I messed up the recurrence -- it should've actually read

Answer 7

\(P(n)=C(1.01)^n+\frac{50000}{99}\)

?

Answer 8

$$P(n)=P(n-1)+0.04P(n-1)-0.03P(n-1)+500\\P(n)=1.01P(n-1)+500$$ in which case we'd get \(r=1.01\) and our particular solution would be different as well

Answer 9

Ohk gotcha!

Answer 10

$$k=1.01k+500\\-0.01k=500\\C=-50000$$ so the general solution is \(P(n)=C(1.01)^n-50000\)