Euler's Method/Improved Euler's Method

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Having trouble working out the bugs in my Improved Euler's Method code. I previously had trouble with the normal Euler's method code, but I figured it out.
Euler's Method (working code):
syms t y
h=0.01;
N=200;
y(1)=1;
t(1)=0;
for n=1:N
k1=1-t(n)+4*y(n);
y(n+1)=y(n)+h*k1;
t(n+1)=t(n)+h;
end
plot(t,y)
And here is my attempt at Improved Euler's Method:
h=0.01;
N=200;
y(1)=1;
t(1)=0;
for n=1:N
k1=1-t(n)+4*y(n);
k2=1-t(n+1)+4*(y(n)+h*k1);
y(n+1)=y(n)+(h/2)*(k1+k2);
t(n+1)=t(n)+h;
end
plot(t,y)
The error message that pops up is "Index exceeds the number of array elements (1)." I'm rather new at MATLAB, and don't know what this means, can someone help me rework this? Thank you!
  2 Comments
Lucas Howarth
Lucas Howarth on 9 Oct 2020
Here is the initial value problem: y'=1-t+4*y with y(0)=1 on the interval [0, 2] using a step size of h = 0.01
Mike Asmanis
Mike Asmanis on 18 Jun 2021
Hey , how would i be able to solve this : y'(t)=cos(t + y) y(0)=0 t[0,3] exact solution y(t)=-t + 2arctan(t)
using your code?

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Accepted Answer

Sudhakar Shinde
Sudhakar Shinde on 9 Oct 2020
May be position of t(n+1)=t(n)+h; coulb be at the starting of loop.
  6 Comments
Lucas Howarth
Lucas Howarth on 9 Oct 2020
For the Runge-Kutta Method for approximation, k2 and k3 are done with the "t" value halfway between the current step and the next step. I'm not sure how to do this in MATLAB and still keeping integer values.

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More Answers (1)

J. Alex Lee
J. Alex Lee on 9 Oct 2020
The error is telling you that at the first step of your loop (n=1), you are trying to access the n=2nd element of t and y, but at the stage, t and y are only scalars (arrays with only 1 element) variables. You are trying to access an element of the "arrays" that doesn't exist.
if you are trying to implement implicit Euler, your problem is math, not coding.
  2 Comments
J. Alex Lee
J. Alex Lee on 9 Oct 2020
my bad, i didn't look very closely. i guess you are doing a 2 step RK, and it is probably right according to Sudhakar's answer.

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