code for eulers method help

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Michelle
Michelle 2020-10-5
评论： Walter Roberson 2020-11-15
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
  2 个评论
Rik
Rik 2020-10-5
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
Walter Roberson 2020-11-15
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.

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回答（1 个）

James Tursa
James Tursa 2020-10-5
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

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