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

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?

[Torsten]

-1 is not an array index, but a function argument. You got confused by these two things.

Or do you want y(i) to be interpreted as y at x = i instead of y at x(i) ?

[Alex]

I don't fully understand what you're saying

[Torsten]

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.

Answer 1

[Les Beckham]

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;

[Alex]

I've tried this but it still doesn't produce the answer im after. Is there anything else incorrect with the code?

[Torsten]

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.