why do I get this answer

this is my code:

clc; clear all; close all;
% Euler's Method
a=160;
Q=400;
A=1200; 
t = 0:2.5:10;
delt= t(2)-t(1);
y = zeros(size(t));
for i=1:numel(t)-1
     y(i+1) = (3*Q/A*sin(t).^2 - a*1+y.^1.5/A)*delt + y(i)
end

I get this answer:

Unable to perform assignment because the left and right sides have a different number of elements.

Error in lab4 (line 10)

y(i+1) = (3*Q/A*sin(t).^2 - a*1+y.^1.5/A)*delt + y(i)

Answer 1

[Alan Stevens]

You can overcome that problem as follows

% Euler's Method
a=160;
Q=400;
A=1200; 
delt = 2.5;
t = 0:delt:10;

y = zeros(size(t));
for i=1:numel(t)-1
     y(i+1) = (3*Q/A*sin(t(i)).^2 - a*1+y(i).^1.5/A)*delt + y(i);
end

However, y quickly goes negative with the values of the constants you have. This means y(i).^1.5 results in complex numbers. Is that what you intended?

[Josue Sosa]

thanks I use ;

a*(1+y(i)).^1.5 so it does not to negative

how do I do to show the sum(y) of each interation

[Alan Stevens]

Do you mean something like this (where I've used a much smller timestep):

% Euler's Method
a=160;
Q=400;
A=1200; 
delt = 0.1;
t = 0:delt:10;

y = zeros(size(t));
for i=1:numel(t)-1
     y(i+1) = (3*Q/A*sin(t(i)).^2 - a*(1+y(i)).^1.5/A)*delt + y(i);
end

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

Answer 2

[Josue Sosa]

I meant this is the answer:

how can i get the answer of each iteration

1 iteration y= -0.333

2 iteration y= -0.333 + 0.3806

3 iteration y= -0.333+0.3806+2.1387

.... and so on

y =

         0   -0.3333         0         0         0 


y =

         0   -0.3333    0.3806         0         0


y =

         0   -0.3333    0.3806    2.1387         0


y =

         0   -0.3333    0.3806    2.1387    2.4848

Answer 3

[Alan Stevens]

This

% Euler's Method
a=160;
Q=400;
A=1200; 
delt = 2.5;
t = 0:delt:10;

y = zeros(size(t));
for i=1:numel(t)-1
     y(i+1) = (3*Q/A*sin(t(i)).^2 - a*(1+y(i)).^1.5/A)*delt + y(i);
     disp(y)
end

produces

0 -0.3333 0 0 0

0 -0.3333 0.3806 0 0

0 -0.3333 0.3806 2.1387 0

0 -0.3333 0.3806 2.1387 2.4848