%Question 2 - An algorithmic implementation of the Euler Method

clc; close all;

%Part 1a

k = (2/3);
h = 0.01; %time step
myeuler(1) = 10;
timesteps1 = 85/h;

for t = [2:1:timesteps1+1]
    myeuler(t) = myeuler(t-1) - h*k*myeuler(t-1);
end

timevect1 = [0:h:85];
figure;
plot(timevect1,myeuler)
title('Figure 5: h = 0.01 implementation of the Euler algorithm')

anasoln1 = myeuler(1)*exp(-(2/3)*timevect1);
mse21a = mse(anasoln1 - myeuler)

%Part 1b

k = (2/3);
h2 = 0.001; %time step
myeuler1(1) = 10;
timesteps2 = 85/h2;

for t = [2:1:timesteps2+1]
    myeuler1(t) = myeuler1(t-1) - h2*k*myeuler1(t-1);
end

timevect2 = [0 : h2 : 85];
figure;
plot(timevect2,myeuler1)
title('Figure 6: h = 0.001 implementation of the Euler algorithm')

anasoln2 = myeuler1(1)*exp(-(2/3)*timevect2);
mse21b = mse(anasoln2 - myeuler1)

%The error for the Runge-Kutta Method is in the order of 10^(-11) while the
%error for the Euler method in part b is in the order of 10^(-8), showing
%that the Runge-Kutta method is more accurate, but the Euler method is good
%for first approximations.