Euler Method IVP Coding Help

조회 수: 9 (최근 30일)
Bailey Smith
Bailey Smith 2018년 7월 30일
답변: John 2023년 8월 5일
I have an IVP problem I'm trying to solve with Euler's Method. The problem is: dy/dx = 1-x+1, y(0) = 1. My time step is 0.01 and my x final is 3. I get an "Index exceeds matrix dimensions." error when I try and run the code. Where am I going wrong? Here is my code:
clear; clc;
%inputs
xo=input('Enter x initial: ');
yo=input('Enter y initial: ');
dx=input('Enter delta x (time step): ');
xf=input('Enter x final: ');
x(1)=xo;
y(1)=yo;
n=1;
% function [out] = Euler(x,y)
% out=1-x+y;
% end
while x(n)<=xf
y(n+1)=y(n)+Euler(x(n),y(n))*(dx);
n=n+1;
end

답변 (2개)

Aquatris
Aquatris 2018년 7월 30일
Your x vector only has a single element in it. After while loop goes once, it tries to reach x(2), which does not exist. That is why you are getting index exceeds error.

John
John 2023년 8월 5일
w0=0.5;
a=0;
b=2;
hh=[0.5 0.2 0.1 0.05 0.01 0.005 0.001];
f=@(t,y) y-t^2+1;
for j=1:7
we(1,j)=w0;
wm(1,j)=w0;
wr(1,j)=w0;
h=hh(j);
n=(b-a)/h;
for i=1:n
t=(i-1)*h;
we(i+1,j)=we(i,j)+h*f(t,we(i,j));
wm(i+1,j)=wm(i,j)+h*f(t+h/2,wm(i,j)+(h/2)*f(t,wm(i,j)));
k1=h*f(t,wr(i,j));
k2=h*f(t+h/2,wr(i,j)+k1/2);
k3=h*f(t+h/2,wr(i,j)+k2/2);
k4=h*f(t+h,wr(i,j)+k3);
wr(i+1,j)=wr(i,j)+(1/6)*(k1+2*k2+2*k3+k4);
end
error_we(j)=abs(we(n+1,j)-y(2));
error_wm(j)=abs(wm(n+1,j)-y(2));
error_wr(j)=abs(wr(n+1,j)-y(2));
end
% y=@(t)....
t=a:h(3):b;
plot(t,y(t),)

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