2018 Practice Test 2 - IVP with Regular EM

A = [0, 1; (-19/4) (-10)];
n = [50, 75, 100, 250, 500];
x_range = 0:10/10000:10;
y_correct = ones(1, length(x_range));

%Creates a vector y_correct with the values for the analytical solution,
%using dx = 1/1000
for j = 1:length(x_range)
    y_correct(j) = (-1*(19/2)*exp(-1*x_range(j)/2)) + 0.5*exp((-19/2)*x_range(j));
end

for i = 1:length(n)
    EM(n(i))
    xlabel('Time')
    ylabel('y(t)')
    title('Solving an IVP for y(t) using varying n')
end

plot(x_range, y_correct, 'DisplayName', 'Analytical Solution')
legend Show


function EM(n)
    dt = 10/n;
    %Range of values, xi
    t_range = 0:dt:10;
    %Creates an array of zeros 2 high, and n+1 wide
    y = zeros(2, n+1);
    %Assigns the initial values to the matrix, representing the intial
    %values for y(t) and y'(t)
    y(:, 1) = [-9, 0];
    %Creating matrix A
    A = [0, 1; (-19/4) (-10)];
    %Applying the transformation A to [y; y'], which is [y'; y'']
    %When multiplied by dt, this allows us to find our next value of y(t)
    %and y'(t) respectively, when added to y(t-1) and y'(t-1) respectively
    for j = 2:length(t_range)
        y(:, j) = y(:, j-1) + dt.*(A*y(:, j-1));
    end
    plot(t_range, y(1, 1:n+1),'DisplayName',strcat('n=',num2str(n)))
    grid on
    hold on
end
%I recommend using n = 100, as it reproduces the analytical solution with a
%similar shape and nearly identical values, if not for a deviation of a few
%tenths