# How to add variable to ode45 method?

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I G on 7 May 2018
Answered: Torsten on 7 May 2018
I am solving system of two ODEs with ode45 method, and code is working for constant cross section, that means constant radius R:
function f=fun(z,p)
R=2; sig=1; beta=1;
f(1)=-32*beta/(R^4*p(1));
f(2)=(-(2-sig)*8*f(1)/(sig*R)-f(1)*p(2))/p(1);
[zv,pv]=ode45('fun',[1 0],[1; 0])
But I need to implement solution where R=2-z, z is longitudinal coordinate, it is there already in first row of function. I wrote z=0:0.001:1; and R=2-z, like this:
function f=fun(z,p)
z=0:0.001:1;
R=2-z; sig=1; beta=1;
f(1)=-32*beta/(R^4*p(1));
f(2)=(-(2-sig)*8*f(1)/(sig*R)-f(1)*p(2))/p(1);
[zv,pv]=ode45('fun',[1 0],[1; 0])
But I got errors matrix dimensions must agree, what should I do to correct this? On the other side z need to be equal to zv, do I need to call z=0:0.001:1; different?

Torsten on 7 May 2018
The actual longitudinal coordinate z is transferred to your function "fun" by ODE45:
function f=fun( *z*,p)
You cannot reset this value by
z=0:0.001:1;
and I don't see any reason to do so. Thus just delete the line
z=0:0.001:1;
Best wishes
Torsten.

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