third order runge kutta
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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
4 个评论
Ana Paula Cervantes Martínez
2021-4-30
excuse me, how do you determine those formulas for finding the values for K's?
回答(0 个)
另请参阅
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