third order runge kutta

Runge-kutta third order method:

%rk3:runge kutta of thirdorder
clc;
clear all;
close all;
% y' = y-x ode condition
f = @(x,y) y-x;
fex = @(x) exp(x)+x+1;    % exact solution
a=0;
b= 3.2;
n =16;
h=(b-a)/n;
y(1) =2;     %initial value
 
i = 0;
for x= a:h:b
    i = i+1;
    K1 = f(x,y(i));         %initializing solution
     K2 = f(x+h*0.5,y(i)+h*K1*0.5);
     K3 = f(x+h, y(i)-h*K1 +2*K2*h);
y(i+1) =y(i)+h*(1/6)*(K1 +4*K2+K3);
 
g(i) = fex(x);
xx(i) = x;
Error(i) = abs(g(i) - y(i));     %error obtain
end
%plot result
plot(xx,y(1:n+1),'k',xx,g,'y')
legend('RK3','Exact solution')
xlabel('x')
ylabel('y')
title('RK3 vs exact solution')

I am getting wrong error value.. please check my code

[KSSV]

Wrong error value????

NOTE: initialze the variable inside the loop.

[SHIVANI TIWARI]

am not getting the right value of absolute error.

http://www.ijesi.org/papers/Vol(7)i4/Version-3/D0704032531.pdf

i have taken same problem 2. but am not getting the right value of absolute error.

and plz also my program for butcher 5 for solving 1st order ode

clc;

clear all;

close all;

% y' = y-x ode condition

f = @(x,y) y-x;

fex = @(x) exp(x)+x+1; % exact solution

a=0;

b= 3.2;

n =16;

h=(b-a)/n;

y(1) =2; %initial value

i = 0;

for x= a:h:b

i = i+1;

K1 = f(x,y(i));

K2 = f(x+h*(1/4),y(i)+h*(1/4)*K1);

K3 = f(x+h*0.5, y(i)+ h*K1*(1/8)+h*K2*(1/8));

K4 = f(x+h*0.5,y(i)- h*K2*0.5+h*K3);

K5 = f(x+3*h*(1/4),y(i)+ (3/16)*h*K1+(9/16)*h*K4);

K6 = f(x+h,y(i)- 3*h*K1*(1/7)+2*h*K2*(1/7) + (12/7)*h*K3 -(12/7)*h*K4 + (8/7)*h*K5);

y(i+1) =y(i)+h*(1/90)*(7*K1 +32*K3+12*K4+ 32*K5 +7*K6);

g(i) = fex(x);

xx(i) = x;

Error(i) = abs(g(i) - y(i)); %error obtain

%plot result

plot(xx,y(1:n+1),'r',xx,g,'g','LineWidth',2)

legend('butcher5','Exact solution')

xlabel('x')

ylabel('y')

title('butcher5 vs exact solution')

[Ana Paula Cervantes Martínez]

excuse me, how do you determine those formulas for finding the values for K's?

[SHIVAM]

thank u so much