# Vector input for ODE45

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Oday Shahadh on 22 Feb 2017
Commented: Walter Roberson on 23 Feb 2017
Sirs, I have something like the below formula:
T=(A+B+C)+(D*T) dt where A, B and C are vectors with let me say (1,6000) length How Can I integrate this formula using ODE45, please help, and thanks in advance

Walter Roberson on 22 Feb 2017
x0 = ... something same length as A;
[T, X] = ode45( @(t, x) f(t, X, A, B, C, D), tspan, x0);
function dy = f(t, y, A, B, C, D)
dy = (A + B + C) + (D*t);
... except of course if that were really your formula you would construct
x0 = ... something same length as A;
ApBpC = A + B + C;
[T, X] = ode45( @(t, x) f(t, X, ApBpC, D), tspan, x0);
function dy = f(t, y, ApBpC, D)
dy = ApBpC + (D*t);

Walter Roberson on 23 Feb 2017
In your "for i" loop, you overwrite all of Tdot. The effect is going to be the same as if you had only done the last of the iterations.
You should avoid using global: global are the slowest form of variables. https://www.mathworks.com/help/matlab/math/parameterizing-functions.html
Oday Shahadh on 23 Feb 2017
how to develope the loop? to avoid overwriting a previues values?
Walter Roberson on 23 Feb 2017
i_vals = 1 : 10 : t;
num_i = length(i_vals);
Tdot = zeros(num_i, 1);
for i_idx = 1 : num_i
i = i_vals(i_idx);
Tdot(i_idx) = A(i) + B(i) + C(i) - T;
end
result = TDot;
Note that the resulting value changes length as t increases. If t represents the time parameter to the ode, then this would mean that you are trying to return a different number of derivatives as time goes on.