# Why is my plot not showing for improved Euler method?

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Anthony Aquino on 6 Mar 2023
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

Sarvesh Kale on 6 Mar 2023
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
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
Thank you
Anthony Aquino on 6 Mar 2023
I see, that makes sense when I think about it now. Thank you so much.

Alan Stevens on 6 Mar 2023
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')

FERNANDO CALVO RODRIGUEZ on 6 Mar 2023
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.

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