# Runge-Kutta-Fehlberg Method Problem

Suppose all infected individuals remained in the population to spread the disease. A more realistic proposal is to introduce a third variable $z(t)$ to represent the number of individuals who are removed from the affected population at a given time $t$ by isolation, recovery and consequent immunity, or death. This quite naturally complicates the problem, but it can be shown that an approximate solution can be given in the form

$x(t) = x(0)e^{\frac{-k1}{k2} z(t)}$ and $y(t)=m-x(t)-z(t)$,

where $k_1$ is the infective rate, $k_2$ is the removal rate, and $z(t)$ is determined from the differential equation

$z'(t)=k_2(m-z(t)-x(0)e^{\frac{-k1}{k2} z(t)})$.

Find an approximation to $z(30)$,$y(30)$, and $x(30)$, assuming that $m=100,000$,$x(0)=99,000$, $k_1$ = $2\times 10^{-6}$, and $k_2=10^{-4}$.

Now, the question would've been easier for me if there's only one variable, but I don't know how to make it work with three variables on Maple.

So can anyone help me with this?

The differential equation you wrote for $z(t)$ only has one dependent variable $z$. Everything else is a constant. Once you have $z(30)$, you can plug that into the equations for $x(t)$ and $y(t)$.
• Yes, this is what I have so far on Maple: diff(z(t), t) = 10-(1/10000)*z(t)-(99/10)*e^(-(1/50)*z(t)), C := InitialValueProblem(deq, z(0) = 0, t = 30, method = rungekutta, submethod = rkf, numsteps = ?, output = information, digits = 8) The problem I'm having here is that I don't the numsteps and also I got a hint from my prof that $z(0)=0$. Why this is true I have no idea. I think $t$ is $30$ because the question relates to a previous question in Burden and Faires Numerical Analysis (Ninth edition). – user87274 Nov 3 '13 at 19:08