code for eulers method help

The derivative equation is y'=r(1-(y/L))*y-((p*y^2)/(q+y^2)) where L can be 5.4,8.1,16.3

Heres the code I have so far but it is not giving me a value.

p=1.2;
q=1;
r=0.65;
L1=5.4;
L2=8.1;
L3=16.3;

h=0.01; %step size
t=0:h:60; %0 days to 60 days
y=zeros(size(t));
y(1)=100; %inital amount of 100
n=numel(y);
for i=1:n-1
    f=@(y)r(1-(y/L))*y-((p*y^2)/(q+y^2));
    y(i+1)=y(i)+h*f(y)
end

[Rik]

This code returns an error, because L is missing. After that it errors because r is indexed with 1-(y/L), which is unlikely to only return 1 (which is the only value it is allowed to have, because r is a scalar. Once you get past that, there are many products and divisions with y, and none of them are element-wise operations, which is probably not how it should be. After that I stopped looking for issues with your code.

[Walter Roberson]

Michelle, if you feel that the question is unclear, then as you are the person who posted the question, you should be the one who clarifies it to make it more clear.

Answer 1

[James Tursa]

This needs to be outside of and prior to your loop, and r needs a multiply operator:

f=@(y)r*(1-(y/L))*y-((p*y^2)/(q+y^2));

This needs to have y indexed when passed into f:

y(i+1)=y(i)+h*f(y(i));

Finally, you should define the L you are going to use prior to creating the f function handle. E.g.,

L = L1;
f=@(y)r*(1-(y/L))*y-((p*y^2)/(q+y^2));

You can change L manually and run your code three times, or perhaps you can put all of the L selection stuff in a loop around your other code. E.g.,

for L=[L1,L2,L3]
    f = @(y)r*(1-(y/L))*y-((p*y^2)/(q+y^2));
    % your other code
end