Why is my plot not showing for improved Euler method?

I am currently learning about the improved Euler formula and so I am testing it with the DE shown below. However, I am trying to obtain the graph for it but it does not show up when I try to plot it. I am new to MATLAB so I would appreciate if someone could explain why this is happening and how I could fix it. This is what I'm working with:

dy=@(t,y) -(0.1/10^(1/2))^2*(2*9.8*y)^(1/2); %ODE

t=0; %initial t
y=2*sqrt(10); %Initial y

h=0.05; %Half step
a=1136; %approximate value

fprintf('\t(Euler Improved)\n');
fprintf('t \t\t y\n');

while a-t >= -10^(-5) %loop until given approximate value
    
    dty = dy(t,y);

    tn = t+h;  %new t value
    
    yp = y + h*dty;  %placeholder y value
    
    dtnyp = dy(tn,yp);

    yn = y + (h/2)*(dty + dtnyp);    %new y value

    
    fprintf('%0.2f\t %f\n',t ,y)
    
    t = tn;  %update t value
    y = yn;  %update y value

end

plot(t, y, 'b')
grid 

Answer 1

[Sarvesh Kale]

Hi Anthony Aquino,

You are updating the time t and y values, they are of size 1x1 so the final plot will only show you a point, you have to append to the time vector and y vector and then use it to plot, here is your own modified code, I have defined two empty arrays t1 and y1 and then appending to them inside the while loop.

dy=@(t,y) -(0.1/10^(1/2))^2*(2*9.8*y)^(1/2); %ODE
t=0; %initial t
y=2*sqrt(10); %Initial y
h=0.05; %Half step
a=1136; %approximate value

t1=[];y1=[]; %%%% added line %%%%%%%


while a-t >= -10^(-5) %loop until given approximate value
    dty = dy(t,y);
    tn = t+h;  %new t value
    yp = y + h*dty;  %placeholder y value
    dtnyp = dy(tn,yp);
    yn = y + (h/2)*(dty + dtnyp);    %new y value
    t = tn;  %update t value
    y = yn;  %update y value

    t1=[t1,t]; %%% append to time vector
    y1=[y1,y]; %%% append to ouptput y vector

end

figure ; 
grid on ;
title ('Solution using Euler Imrpoved method')
plot(t1, y1, 'b','lineWidth',1.5) %% the vectors accumulated are used in plotting 

I hope this answers your query, please accept the answer if it does

Thank you

[Anthony Aquino]

I see, that makes sense when I think about it now. Thank you so much.

Answer 2

[Alan Stevens]

You need to keep track of each step. Your original simply overwrites the values of t and y at each step.

dy=@(t,y) -(0.1/10^(1/2))^2*(2*9.8*y)^(1/2); %ODE

t(1)=0; %initial t
y(1)=2*sqrt(10); %Initial y

h=0.05; %Half step
a=1136; %approximate value    ... of time???

% fprintf('\t(Euler Improved)\n');
% fprintf('t \t\t y\n');
step = 0;
err = 1;
while err >= 10^(-5) %loop until given approximate value
    step = step+1;
    dty = dy(t(step),y(step));
     
    tn = t(step)+h;  %new t value
    
    yp = y(step) + h*dty;  %placeholder y value
    
    dtnyp = dy(tn,yp);

    yn = y(step) + (h/2)*(dty + dtnyp);    %new y value

    
%     fprintf('%0.2f\t %f\n',t ,y)
    
    t(step+1) = tn;  %update t value
    y(step+1) = yn;  %update y value
    err = abs(a-t(step+1));
end

plot(t, y, 'b')
grid 
xlabel('t'), ylabel('y')

Answer 3

[FERNANDO CALVO RODRIGUEZ]

Good morning Anthony,

From what I have seen in your code, the result you finally get for t and y is a single value because t and y are variables and not vectors.

Try designating them as zeros vectors with the necessary length:

t = zeros(N,1) % Create a 0's vector with N rows and 1 cols.
y = ones(N,1) % Create a 1's vector with N rows and 1 cols.

To multiply or divide a number between each position of the vector you must use ".*", "./", ".^".

I have given some numerical methods and it is more advisable to use a for loop followed by the while loop as it facilitates the indexing of the vectors.

If you want to plot each iteration, I still recommend you to use the for loop and save each iteration in a vector to plot it later.

Sorry for not being able to help you more.