help with my 4th order runge kutta code

I received this error: "Unable to perform assignment because the size of the left side is 1-by-4 and the size of the right side is 4-by-4." when trying to run my 4th order runge kutta code to solve the 4 non linear odes of an elastic pendulum. Any help on how to fix this would be appreciated. The line of code where the error is:

Y(i+1,:) = Y(i,:) + phi*h;

edit: here is my code

tspan1=[0,2*pi];

h=0.01;

r1=11;

rdot1=0;

theta1=0;

thetadot1=0;

init1= [r1;theta1;rdot1;thetadot1];

[T,Y]=rk4_fun(@spring1,tspan1,init1,h);

function [T,Y]=rk4_fun(f,tspan,y0,h)

x0 = tspan(1);xf = tspan(2);

T = (x0:h:xf)';

n = length(T);

n_eqn = length(y0);

Y = ones(n,n_eqn);

Y(1,:) = y0;

for i = 1:n-1

k1 = f(T(i),Y(i,:))';

k2 = f(T(i)+h/2,Y(i,:) + k1*h/2)';

k3 = f(T(i)+h/2,Y(i,:) + k2*h/2)';

k4 = f(T(i)+h,Y(i,:) + k3*h)';

phi = (k1+2*k2+2*k3+k4)/6;

Y(i+1,:) = Y(i,:) + phi*h;

end

end

function ydt = spring1(t, y)

g = 1000;

km = 100;

lo=10;

[rows,cols] = size(y);

ydt = zeros(rows,cols);

ydt(1) = y(3);

ydt(2) = y(4);

ydt(3) = y(1)*y(4)*y(4) + g*cos(y(2)) - km*(y(1)-lo);

ydt(4) = -(g*sin(y(2))+2*y(3)*y(4))/y(1);

end

[James Tursa]

In the future, it is best to post the entire code that produces the error so that we can get the proper context. Without seeing the rest of your code we have to make guesses.

[John D'Errico]

Without seeing the rest of the code, it is impossble to help you.

Most importantly, we need to know the size of both the phi and h variables. Not what you THINK them to be. But what they are. EXACTLY.

Answer 1

[James Tursa]

My guess without seeing the rest of your code is that phi is a column vector, so when you add phi*h to the row vector Y(i,:) it creates a 4x4 matrix. Try changing phi to a row vector, or transpose it in the equation itself. E.g.,

Y(i+1,:) = Y(i,:) + phi'*h;

Answer 2

[Vezzaz]

i posted the entire code for this part.