unable to perform assignment becuase the left and right sides have a different amount of elements HELPPP!!

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f1=@(x,y1,y2) (-6*y2-5*y1)/.5
f2=@(x,y1,y2) diff(y1)
% Condiciones iniciales -----------------------------------------------
%y1(0)=1;
%y2(0)=0;
x=0
xn=5
y1=1
y2=0
h=0.5
% Método de RK4Orden ---------------------------------------------------
while x(end)<=xn
k11= f1(x(end),y1(end),y2(end));
k12= f2(x(end),y2(end),y2(end));
k21= f1(x(end)+.5*h,y1(end)+.5*k11*h,y2(end)+.5*k12*h);
k22= f2(x(end)+.5*h,y1(end)+.5*h*k11,y2(end)+.5*h*k12);
k31= f1(x(end)+.5*h,y1(end)+.5*k21*h,y2(end)+.5*k22*h);
k32= f2(x(end)+.5*h,y2(end)+.5*k22*h,y2(end)+.5*k22*h);
k41= f1(x(end)+h,y1(end)+k31*h,y2(end)+k32*h);
k42= f2(x(end)+h,y2(end)+k32*h,y2(end)+k32*h);
x(end+1)=x(end)+h;
y1(end+1)=y1(end)+1/6*(k11+2*k21+2*k31+k41)*h;
y2(end+1)=y2(end)+1/6*(k12+2*k22+2*k32+k42)*h;
end
plot(x,y1)
  4 comentarios
Jeffrey Clark
Jeffrey Clark el 16 de Jun. de 2022
Two more lines like k12 need fixing:
k32= f2(x(end)+.5*h,y2(end)+.5*k22*h,y2(end)+.5*k22*h);
k42= f2(x(end)+h,y2(end)+k32*h,y2(end)+k32*h);

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