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

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Nadim Mhanna
Nadim Mhanna 2017-9-30
回答： Josh Meyer 2017-10-2
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?

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回答（1 个）

Josh Meyer
Josh Meyer 2017-10-2
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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