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

조회 수: 1 (최근 30일)
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)
  댓글 수: 4
Jeffrey Clark
Jeffrey Clark 2022년 6월 16일
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);

댓글을 달려면 로그인하십시오.

답변 (0개)

카테고리

Help CenterFile Exchange에서 Resizing and Reshaping Matrices에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by