RK4 Function Code

The issue occurs when running the function at the 'k_1 =' part of the code, the error message says "symbolic function expected 1 input but recieved 2"
function [x,y] = rk4(dydx,xo,xf,yo,h)
x = xo:h:xf ;
y = zeros(1,length(x));
y(1)= yo ;
for i = 1:(length(x)-1)
k_1 = dydx(x(i),y(i));
k_2 = dydx(x(i)+0.5*h,y(i)+0.5*h*k_1);
k_3 = dydx((x(i)+0.5*h),(y(i)+0.5*h*k_2));
k_4 = dydx((x(i)+h),(y(i)+k_3*h));
y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;
end
dydx = @(x,y) 3.*exp(-x)-0.4*y;
%[x,y] = rk4(dydx,0,100,-0.5,0.5); %plot(x,y,'o-');
end

2 Comments

Torsten
Torsten on 5 Dec 2022
Most probably the reason is that you didn't call "rk4" correctly.
It shouldn’t have a th when called. When testing it didn’t. I forgot to delete that before posting here

Sign in to comment.

Answers (2)

dydx = @(x,y) 3.*exp(-x)-0.4*y;
[x,y] = rk4(dydx,0,100,-0.5,0.5);
plot(x,y,'o-');
function [x,y] = rk4(dydx,xo,xf,yo,h)
x = xo:h:xf ;
y = zeros(1,length(x));
y(1)= yo ;
for i = 1:(length(x)-1)
k_1 = dydx(x(i),y(i));
k_2 = dydx(x(i)+0.5*h,y(i)+0.5*h*k_1);
k_3 = dydx((x(i)+0.5*h),(y(i)+0.5*h*k_2));
k_4 = dydx((x(i)+h),(y(i)+k_3*h));
y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h;
end
end

2 Comments

For this code i get the error "Error using indexing" and " error in line 6
The above code runs without failure as you can see above. Did you save the function and run the 3-line script in the command line?

Sign in to comment.

clc
a=0;b=2;N=10;alph=0.5;
h=(b-a)/N;t(1)=a;w(1)=alph;
f=@(t,y) y-t^2+1;
for i=2:N+1
k1= h*f(t(i-1),w(i-1));
k2= h*f(t(i-1)+(h/2),w(i-1)+(k1/2));
k3= h*f(t(i-1)+(h/2),w(i-1)+(k2/2));
k4= h*f(t(i-1)+h,w(i-1)+k3);
w(i)=w(i-1)+(k1+2*k2+2*k3+k4)/6;
t(i)=a+(i-1)*h;
end
x=[t;w];
disp("t w")
t w
fprintf("%5.7f %5.7f\n",x)
0.0000000 0.5000000 0.2000000 0.8292933 0.4000000 1.2140762 0.6000000 1.6489220 0.8000000 2.1272027 1.0000000 2.6408227 1.2000000 3.1798942 1.4000000 3.7323401 1.6000000 4.2834095 1.8000000 4.8150857 2.0000000 5.3053630

Categories

Find more on Programming in Help Center and File Exchange

Tags

Asked:

on 5 Dec 2022

Edited:

on 5 Aug 2023

Community Treasure Hunt

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

Start Hunting!