May you help me please to continue my syytem seconder order DE function?

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function [dF1dx,dF2dx]=funode1(x,F)
dF1dx=[F1(2);(Q/K1)*F1(1)-P/K1];
dF2dx=[F2(2);(Q/K2)*F2(1)-P/K2];
function res12=myfunbc12(F1a,F1b,F2a,F2b)
res=[F1a(1);F1b(1)-F2b(1);F2a(1);F1a(2)-F2b(2)];
The boundary conditions are as follows
F1(0)=0;
F2(l)=0;
F1(l-u)=F2(l-u);
dF1/dx(l-u)=dF2/dx(l-u))
If this is true how to continue?
  2 Comments
John D'Errico
John D'Errico on 30 Sep 2017
How to continue with what? You have code. Does it do what you need? If not, then what does it lack? How can we read your mind?

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Answers (1)

Josh Meyer
Josh Meyer on 2 Oct 2017
Since you have boundary conditions you'll want to use bvp4c or bvp5c. See Boundary Value Problems in the doc for more info.

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