Problem Calling my own Function in my Code

I am trying to call a function in my code and I keep getting an error message that reads ...

"Error using odearguments (line 95)@(T,Q)NONLINEAR2(W,Q,T) returns a vector of length 1, but the length of initial conditions vector is 2.

The vector returned by @(T,Q)NONLINEAR2(W,Q,T) and the initial conditions vector must have the same

number of elements."

I am confused on what this means and how I go about fixing it. If you could help me fix my function call in my code that would be amazing! I will post my code below as well as the function.

Main Code:

tspan = [0 80]; % time

qinit = [0 ; 1]; % q initial

for w = [0, 0.5, 1, 2, 4, 8, 16] % w (calculates the IVP solution as w changes)

[t,q] = ode45(@(t,q) nonlinear2(w,q,t),tspan,qinit); % function call; % function call

figure(1)

plot(t,q(:,1),'-') % plots IVP Solution vs time

grid on % adds a grid

xlabel("Time") % x-axis label (time)

ylabel("IVP Solutiion") % y-axis label (IVP Solution)

title("IVP Solution vs Time (Question 3)") % Adds a title to the plot

hold on

end

Function :

function qp = nonlinear2(w,q,t) % defines function and variables needed

qp = [-4* q(1) + 5 * q(2) + 10 * cos(w * t)];

end

Answer 1

[Walter Roberson]

You have two boundary conditions, such as for q and q'. You need to return one entry for each input boundary condition.

If your equation is of the form

q''(t) = function in q'(t) and q(t)

then you would (typically) use the q(t) and q'(t) boundary conditions to calculate the current q''(t) and return it as the second entry in the column vector, and you would also in that case set the first entry as dp(1) = q(2) implying that q'(t) is the integral of q''(t). But remember, column vector!