MATLAB: Not enough input arguments in function. Cannot get any output at all

argumentode

function yp = equationb(t, y)
A=2.69; B=168; C=6270; D=59900; %coefficients
V = 10:50;
E= 15000000000; h = 0.048; I = (1/12)*h^4; l=4; rho=1600;
m = 0.177; c = 0.77; k = (48*E*I)/l^3; R = (0.5*rho*h);
yp (1) = y(2);
yp (2,1) = (R/m)*(A*V*y(2).^2 - (B*y(2).^4)/V + (C*y(2).^6)/V^3 - (D*y(2).^8)/V^5) - (c/m)*y(2) - (k/m)*y(1);
end
And my main function is
   close all; clear all; clc;
% Initial Conditions
h = 0.048;              %section width (m)
x0 = [h,10*h];          %initial displacements
v0 = zeros(length(x0)); %inital velocity
V = 10:50;              %flow velocity (m/s)
timespan = [0 30];
for j=1:length(x0)
      for k=1:length(V)
          a = [x0(1,j), v0(1,1), V(k)];
            [t,y] = ode15s(@equationb, timespan, a);      
          R = find(t>25); %returns a vector containing the linear indices of each nonzero element in array.
          LCO = y(R,1);
          maximum_amplitude(j,k) = max(LCO);  
      end
end
figure(1)
hold on; grid on; box on;
ylabel('Structural Displacement Amplitutde (m)')
xlabel('Air Flow Velocity (m/s)')
plot(V,maximum_amplitude(1,:),'go')
plot(V,maximum_amplitude(2,:),'r+')
axis([V(1) V(end) 0 0.7])

Best Answer

This is one possible approach:
function yp = equationb(t, y, V)
Eliminate the vector definition of ‘V’ within ‘equationb’.
Change the call to ‘equationb’ to:
a = [x0(1,j), v0(1,1)];
[t,y] = ode15s(@(t,y)equationb(t,y,V(k)), timespan, a);
and see if that does what you want. (Having 3 initial conditions with a function that returns two derivatives will likely throw an error anyway. This should at least avoid that problem.)
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