System of differential equations with constant (as variables) coefficients

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I have 4 differential equations that I need to solve, but the coefficients for each term in each equation are different unknown constants. I read about solving it using a matrix and came up with:
syms A(t) B(t) C(t) D(t) k1 kn1 k2 kn2 k3 kn3 k4 kn4 k5 kn5 k6 kn6
Z = [A;B;C;D];
X = [-(k1+k3+k5), kn1, kn3, kn5; k1, -(kn1+k2+k6), kn6, kn2; k3, k6, -(kn3+k4+kn6), kn4; k5, k2, k4, -(kn2+kn4+kn5)];
Y = zeros(4,1);
eqn = diff(Z) == X*Z + Y;
[ASol(t), BSol(t), CSol(t), DSol(t),] = dsolve(eqn);
however that does not seem to work. Any help would be appreciated

Risposta accettata

Torsten il 28 Ott 2016
Your system of ODEs is too complicated to be solved symbolically.
Specify the constants and use a numerical solver (e.g. ODE15s).
Best wishes
  3 Commenti
Marc Rosales
Marc Rosales il 28 Ott 2016
^Exactly my question too.
To be completely honest, this is a problem a chemistry professor gave me, and I too am at a complete loss why I need a symbolic solution hahaha. Anyway, the constants may be any non-negative value afaik.
If I get a system of ODEs of 4 equations, around how many constants(as variables) can I use to still be solvable by Matlab? or do I need to eliminate the number of equations too? :/

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Più risposte (1)

Teja Muppirala
Teja Muppirala il 28 Ott 2016
As has been mentioned, with the k's explicitly accounted for, it's very complicated.
But if you consider the matrix X simply as some constant matrix X, regardless of what's in it, then the analytical solution is
Z(t) = expm(X*t)*x0
where x0 is the initial condition.
For example
% Make some random stable X matrix
X = randn(3);
X = X*X';
X = -X;
x0 = [1;2;3]; % Some initial condition
ode45(@(t,Z)X*Z,[0 3],x0) % Using ODE45
Z = [];
tList = 0:0.01:3;
for t = tList;
Z(:,end+1) = expm(X*t)*x0; % Using matrix exponential solution
hold on;
plot(tList, Z','k','linewidth',1);
title('Same answer with ode45 and expm')
  2 Commenti
Marc Rosales
Marc Rosales il 29 Ott 2016
I've looked at the results after tweaking the initial conditions ([1,0,0]), but the graph always slopes downwards, whereas I'm expecting a downwards slope for only one graph, and upwards for the rest until they stabilize.

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