Question

LAB. Implement the classical procedure (8.82), and apply it to equation (8.66). Solve it with stepsizes of h = 0.25 and 0.125. (8.82) h h h Runge-Kutta methods of higher order can also be developed. A popular classical method is the following fourth-order procedure: 01 -f(x,y) V2 = 1 (**+ 2 Yu + 2" h lg = f(x+ Yu + (8.82) V4 = f(x+h, yn + hvy) h Yn+1 = y +7(+202 +223 +24] The truncation error in this method is (h). When the differential equation is simply Y'(x) = f(x) (8.66) To keep the initial explanations as intuitive as possible, we will develop a Taylor method for the problem Y'(x) = -Y(x) + 2 cos(x). Y(0)=1 (8.66) (tips) Elementary Numerical Analysis Help Please show steps! (you can write it or code it using Matlab or Python program)

Answer 1

clc

clear all

close all

format long

f=@(x,y) -y+2*cos(x);

disp('Solution for h=0.25')

[X1,Y1]=runge4(f,[0,1],1,0.25)

[X2,Y2]=runge4(f,[0,1],1,0.125)

plot(X1,Y1,X2,Y2);

legend('h=0.25','h=0.125')

function [x,y]=runge4(f,tspan,y0,h)

x = tspan(1):h:tspan(2); % Calculates upto y(3)

y = zeros(length(x),1);

y(1,:) = y0; % initial condition

for i=1:(length(x)-1) % calculation loop

k_1 = f(x(i),y(i,:));

k_2 = f(x(i)+0.5*h,y(i,:)+0.5*h*k_1);

k_3 = f((x(i)+0.5*h),(y(i,:)+0.5*h*k_2));

k_4 = f((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; % main equation

end

end