Solving coupled differential equations

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JaeSung Choi
JaeSung Choi on 15 Oct 2017
Answered: Josh Meyer on 16 Oct 2017
T o solve my coupled differential equations. I made the following function
======================
function x_value = reservoir(t,y);
s1 = y(1); s2 = y(2); s3 = y(3);
p1 = y(4); p2 = y(5); p3 = y(6);
x1 = 5.45*10^(-6);
x_value = [s1; s2; s3; p1; p2; p3];
h = 0.7844; b = 5*10^(-7); e = 8.8750*10^(-2); v = 7.54; g1 = 2500; g2= 2500; g3 =2500; x1 = 5.45*10^(-6); x2 = 5.45*10^(-6); x3 = 5.45*10^(-6);
% Define constants.
Dp1 = diff(p1); Dp2 = diff(p2); Dp3 = diff(p3);
Ds1 = diff(s1); Ds2 = diff(s2); Ds3 = diff(s3);
% Generate differentiations.
diffqp1 = Dp1 == h*b*s1*(g1-p3)-(e/v)*p1;
diffqp2 = Dp2 == h*b*s2*(g2-p1)-(e/v)*p2;
diffqp3 = Dp3 == h*b*s3*(g3-p2)-(e/v)*p3;
diffqs1 = Ds1 == x1/v-h*b*s1*(g1-p3)-(e/v)*s1;
diffqs2 = Ds2 == x2/v-h*b*s2*(g2-p1)-(e/v)*s2;
diffqs3 = Ds3 == x3/v-h*b*s3*(g3-p2)-(e/v)*s3;
% Generate differential equations.
end
===========================
And to derive explicit s1~p3 values, i used the code below.
===========================
[s1, s2, s3, p1, p2, p3] = ode45(@reservoir,[0 1],[0; 0; 0; 0; 0; 0]) ===========================
But it said " While using 'ode45' there wasn't proper substitute to output parameter 'varargout{4}'
My terminal goal is to derive s1~p3 values at each t=1,2,....so on, when i want!! How can I derive this?
Please help me!

Accepted Answer

Josh Meyer
Josh Meyer on 16 Oct 2017
The immediate problem is that you provide too many output arguments to ode45. The components of y will be returned as columns if you just use [t,y] = ode45(...).
Once you fix that, the next issue is that y is returned as all zeros. So you need to check your equations and/or initial condition.

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