Clear Filters
Clear Filters

Euler method: ODE with different initial conditions

4 views (last 30 days)
I want to solve an ODE by Euler method with different initial conditions.
I have used for y(1)=0.5:0.05:1.5; as the different initial conditions. But it given an error
for y(1)=0.5:0.05:1.5;
Error: Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
How to modify the code so that I can able to run a loop for different values of y(1).
NB: Tried with
for v(1) = [0.05 0.01 0.15 0.2]
end
but got the error.
h=0.5;
x=0:h:4;
y=zeros(size(x));
for y(1)=0.5:0.05:1.5;
n=numel(y);
for i = 1:n-1
dydx= -2*x(i).^3 +12*x(i).^2 -20*x(i)+8.5 ;
y(i+1) = y(i)+dydx*h ;
fprintf('="Y"\n\t %0.01f',y(i));
end
figure;
plot(x,y);
end
  2 Comments
Torsten
Torsten on 21 Jan 2023
Make the code a function and call the function in a loop for different initial values y(1).

Sign in to comment.

Accepted Answer

Torsten
Torsten on 21 Jan 2023
Edited: Torsten on 21 Jan 2023
Y0 = 0.5:0.05:1.5;
hold on
for i = 1:numel(Y0)
y0 = Y0(i);
[x,y] = euler(y0);
plot(x,y)
end
hold off
function [x,y] = euler(y0)
h=0.5;
x=0:h:4;
n = numel(x);
y=zeros(1,n);
y(1) = y0;
for i = 1:n-1
dydx= -2*x(i).^3 +12*x(i).^2 -20*x(i)+8.5 ;
y(i+1) = y(i)+dydx*h ;
end
end

More Answers (0)

Categories

Find more on Programming in Help Center and File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!