Hi Aleily,
I understand that you intend to modify your existing code to implement the "midpoint" Euler method, also known as the "modified" Euler method.
To plot the solution along with the Explicit Euler function and exact solutions, I recommend adjusting the "ode_euler" function. The midpoint method involves taking an Euler step to estimate the midpoint, then using the derivative at the midpoint to take another whole step.
For your reference, I am attaching the modified code that supports the Midpoint method and plots the solution:
% EXPLICIT EULER AND MIDPOINT METHOD
clear;
% INPUT VARIABLES
a = 2; % FIRST VALUE OF X
b = 6; % SECOND VALUE OF X
N = 4; % INTERVAL SIZE
y_initial = -6.5; % INITIAL VALUE OF Y
% FUNCTION TO SOLVE ODE USING EULER METHOD
[x_euler,y_euler] = ode_euler(@Fun_ODE, a, b, N, y_initial);
% FUNCTION TO SOLVE ODE USING MIDPOINT METHOD
[x_midpoint,y_midpoint] = ode_midpoint(@Fun_ODE, a, b, N, y_initial);
% PLOTTING
xp = a:0.1:b;
yp = 3./xp - 2*xp.^2; % Exact solution
figure
hold on
plot(x_euler, y_euler, '--r', 'LineWidth', 1.5)
plot(x_midpoint, y_midpoint, '--b', 'LineWidth', 1.5)
plot(xp, yp, 'k', 'LineWidth', 1)
xlabel('x'); ylabel('y');
legend("Euler solution", "Midpoint solution", "Exact solution");
title('Comparison of Euler, Midpoint, and Exact Solutions');
grid on
% Euler method function
function [x,y] = ode_euler(ODE, a, b, N, y_initial)
x(1) = a; y(1) = y_initial;
h = (b-a)/N;
for i = 1:N
x(i+1) = x(i) + h;
y(i+1) = y(i) + ODE(x(i),y(i))*h;
end
end
% Midpoint method function
function [x,y] = ode_midpoint(ODE, a, b, N, y_initial)
x(1) = a; y(1) = y_initial;
h = (b-a)/N;
for i = 1:N
x(i+1) = x(i) + h;
% First, take a half-step
y_half = y(i) + ODE(x(i), y(i)) * (h/2);
% Then, use the derivative at the midpoint to take a full step
y(i+1) = y(i) + ODE(x(i) + h/2, y_half) * h;
end
end
% ODE function
function dydx = Fun_ODE(x, y)
dydx = -(y/x) - 6*x;
end
To summarize changes in the new code, a new function "ode_midpoint" is introduced. This function implements the Midpoint method. The main differences from the initial approach are as under:
- A half-step forward is taken using the Euler method to estimate the midpoint.
- The derivative at this midpoint ("y_half") is then used to take a full step from the original point.
Additionally, the plotting section is updated to include the solution from the Midpoint method. This allows for a direct comparison between the Explicit Euler method, the Midpoint method, and the exact solution.
Hope this helps.
Regards,
Nipun
