I keep getting errors when trying to use the ode 45 function to integrate an acceleration equation to determine vO vs. t, transferring all constants to the function using global variables.

This is my main code:

clc 
clear

f=3500;
cd=.6;
cf=3;
m0=600;
c=2;
d=0;
dN=0;

for t=0:0.1:5
    vE=0;
    m=m0-(c*t);
    n=0;
    v=0;
    a=(f/m)-((cd/m)*v);
    vE=vE+(a*.1);
    fprintf('VEuler is %1.2f when t is %1.1f \n',vE,t)
    vO=v+vE;
    dN=((5+t)*((vO+vE)/2))+d;
end
fprintf('After initiating acceleration, the dragster has moved %1.2f meters\n',dN)

y0 = 0;
tspan=0:0.1:5;
[t,y]= ode45('fun',tspan,y0);
fprintf('\n   t        y\n');
tvalues=transpose(t);
fxvalues=transpose(fx);

With the function:

function fx=fun(t,y0)
global f cd cf m0 c;
fx=(((cf/m0-c.*t))*y0)-((cd/m0-c.*t).*y0.^2)+(f/m0-c.*t);
end

Matlab is telling me that it's returning a vector length of zero, an ode arguements error (odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options,varargin);)

and that there's an error in the ode function itself. I've explored the internet and tried several online solutions but none have worked.

[madhan ravi]

please don't close the question that have an answer , thank you for understanding , see https://www.mathworks.com/help/matlab/math/parameterizing-functions.html , what's unclear?

Answer 1

[madhan ravi]

clc 
clear
f=3500;
cd=.6;
cf=3;
m0=600;
c=2;
d=0;
dN=0;
for t=0:0.1:5
    vE=0;
    m=m0-(c*t);
    n=0;
    v=0;
    a=(f/m)-((cd/m)*v);
    vE=vE+(a*.1);
    fprintf('VEuler is %1.2f when t is %1.1f \n',vE,t)
    vO=v+vE;
    dN=((5+t)*((vO+vE)/2))+d;
end
fprintf('After initiating acceleration, the dragster has moved %1.2f meters\n',dN)
y0 = 0;
tspan=0:0.1:5;
[t,y]= ode45(@(t,y0)fun(t,y0,f, cd ,cf, m0, c),tspan,y0);
fprintf('\n   t=%f        y=%f\n',t,y);
tvalues=transpose(t);
fxvalues=transpose(y); % what's fx? should be y because it's the solution which is returned from ode45
 
function fx=fun(t,y0,f, cd ,cf, m0, c)
fx=(((cf/m0-c.*t))*y0)-((cd/m0-c.*t).*y0.^2)+(f/m0-c.*t);
end