Runge Kutta 2nd order 1st order ODE equations

4 views (last 30 days)
Maria Ruiz
Maria Ruiz on 26 Nov 2021
Commented: Jan on 28 Nov 2021
Please help, those are the EDO´S i need to put in the graph, the graph should look like this:
I've been trying to fix it, but runge-kutta is a difficult method for me, thank you!
%RK de 2nd order
% dx/dt = MU*xI - Kd*xI
%ds/dt = (-a/Yps)*(mu*xI);
%dp/dt = a*mu*xI;
%Initial indep. values
Kd = 0.0032;
a = 0.6212;
Yps = 0.5820;
Ki = 243.95
MU = 0.5557;
Ks = 0.0191;
p = 30.7600
xI =1.25;
sI =86.63;
pI =0;
%F_xt = @(x,t) mu*xI - Kd*xI;
%F_st = @(s,t) -a/Yps*mu*xI;
%F_pt = @(p,t) a*mu*xI;
for i=1:length(t)-1
mu(i) = MU*sI(i)/(Ks + sI(i)+ (sI(i)^2/Ki))*((1-pI(i)/p)^1);
k1f1(i)= h*(mu(i)*xI(i) - Kd*xI(i));
k2f1(i)= h*(mu(i)*(xI(i)+k1f1(i)) - Kd*(xI(i)+k1f1(i)));
xI(i+1) =xI(i) + 0.5*(k1f1(i) + k2f1(i));
k1f2(i) = h*((-a/Yps)*mu(i)*xI(i));
k2f2(i) = h*((-a/Yps)*mu(i)*(xI(i)));
sI(i+1) =sI(i) + 0.5*(k1f2(i) + k2f2(i));
k1f3(i) = h*(a*mu(i)*xI(i));
k2f3(i) = h*(a*mu(i)*(xI(i)));
pI(i+1) = pI(i) + 0.5*(k1f3(i) + k2f3(i));
  1 Comment
Jan on 28 Nov 2021
The code gets much easier, if you split the function to be integrated and the code to integrate. Do not run 3 different integrations, but combine the 3 components to a vector y =[x,s,p].
A problem is, that your ODE does not depend on the variables at all:
dx/dt = MU*xI - Kd*xI
ds/dt = (-a/Yps)*(mu*xI);
dp/dt = a*mu*xI;
These are all constants, which do not depend on x,s,p or t. What is mu in the the code?

Sign in to comment.

Answers (0)


Find more on Numerical Integration and Differential Equations in Help Center and File Exchange




Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by