MATLAB Script:
close all
clear
clc
% Given
f = @(t,y) t^2 * y; % Given ODE
t0 = 0; tf = 1; % Intervals of x
y0 = 1; % Initial condition
fprintf('Part (a)\n----------------------------------\n')
h1 = 0.1; % Step Size 1
x1 = t0:h1:tf;
y1 = my_euler(t0, y0, tf, h1, f);
fprintf('For h = 0.1, y(1) = %.6f\n', y1(end))
fprintf('\nPart
(b)\n----------------------------------\n')
h2 = 0.05; % Step Size 2
x2 = t0:h2:tf;
y2 = my_euler(t0, y0, tf, h2, f);
fprintf('For h = 0.05, y(1) = %.6f\n', y2(end))
fprintf('\nPart
(c)\n----------------------------------\n')
% Exact Solution
syms y(t)
ODE = diff(y,t) == t^2 * y; % Given ODE
cond = y(0) == 1; % Initial condition
y_sol = dsolve(ODE, cond); % Solver
fprintf('For h = 0.1, Error = %.6f\n', abs(subs(y_sol, 1) -
y1(end)))
fprintf('For h = 0.05, Error = %.6f\n', abs(subs(y_sol, 1) -
y2(end)))
disp('Lower step-size gives better results.')
function y = my_euler(t0, y0, tf, h, f)
y(1) = y0;
t = t0:h:tf;
for i = 1:length(t)-1
y(i+1) = y(i) + h*f(t(i), y(i)); % Euler Update
end
end
Output:
Part (a)
----------------------------------
For h = 0.1, y(1) = 1.320016
Part (b)
----------------------------------
For h = 0.05, y(1) = 1.355880
Part (c)
----------------------------------
For h = 0.1, Error = 0.075597
For h = 0.05, Error = 0.039732
Lower step-size gives better results.
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