# How to include a negative number in initial conditions for Eulers method code?

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Alex on 13 Dec 2023
Edited: Torsten on 13 Dec 2023
clear
close all
h=0.1; % step size
x=-1:h:2; % x interval define here
y=zeros(size(x));
y(-1)=8; % intial condition
n=numel(y);
for i = 1:n-1
dydx=((2.*x.*y.^2+4)/(2.*(3-y.*x.^2)))
y(i+1) = y(i)+dydx*h;
end
plot(x,y,'ro');
hold on
f1=exp(x);
plot(x,f1,'b');
grid on;
I'm writing some code for Eulers method and my initial condition is y(-1)=8. However when I run this it comes up with the error message saying array indicies must be positive integers or logical values. Any advice on how i can include my intial condition without these error messages?
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Alex on 13 Dec 2023
I don't fully understand what you're saying
Torsten on 13 Dec 2023
Edited: Torsten on 13 Dec 2023
y(i) means the value of the function y at x(i).
It does not mean the value of the function y at x = i.
Since
x=-1:h:2; % x interval define here
you thought you could use y(-1) for y (x=-1).
But you must use y(1) for y @ x(1) = -1.

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### Answers (1)

Les Beckham on 13 Dec 2023
It is always better to post example code as text rather than a screenshot. Nevertheless...
Since the first element of x is -1 and you want to specify the value of y when x is -1, just define the first element of y, y(1), with your initial condition
y(1) = 8;
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Alex on 13 Dec 2023
I've tried this but it still doesn't produce the answer im after. Is there anything else incorrect with the code?
Torsten on 13 Dec 2023
dydx=((2.*x(i).*y(i).^2+4)/(2.*(3-y(i).*x(i).^2)))
instead of
dydx=((2.*x.*y.^2+4)/(2.*(3-y.*x.^2)))
And your denominator becomes 0 in the course of the integration - thus your solution has a singularity.

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