Question

1. Consider the following differential equation. ag = ty, y(0)=1. dt (a) Use Euler's Method with At = .1 to approximate y(1). (b) Use Euler's Method with At = .05 to approximate y(1). (c) Find the exact solution to the problem. Use this solution to compare the error for the different values of At. What does this say about the method? Note: On the course page there are notes describing an implementation of Euler's method on a spread sheet.

Answer 1

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.