Forward Euler method error

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ISABELLA CORONADO
ISABELLA CORONADO 2022-4-28
编辑： Torsten 2022-5-3
I am trying to create a loop to solve a system of differential equations, where x' = -y and y' = x. The initial conditions given are x(0) = r and y(0) = 0, but r can be set as any arbitrary number and it won't change the solution too much. I am getting this error that says "Array indices must be positive integers or logical values", but I am confused because I am nonly trying to set initial conditions. ANY help is greatly appreciated!!!

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Torsten
Torsten 2022-4-28
编辑：Torsten 2022-4-28
Array indices in MATLAB start at 1, not at 0.
Thus you will have to replace
x(0)=1
y(0)=0
by
x(1)=1
y(1)=0
And you solve the wrong equations. What you solve is
x' = -t
y' = y
Look at how you call dxdt and dydt.
In both cases, the call must be
x(n+1) = x(n) + h*dxdt(x(n),y(n))
y(n+1) = y(n) + h*dydt(x(n),y(n))
Further, by the line
t(n+1) = a+n*h
you overwrite the t-vector you've already created by
t=a:h:b
So you should delete this line.
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ISABELLA CORONADO
ISABELLA CORONADO 2022-5-3
I see. I was confused by your answer involving t, as manually solving the differential equation does not involve t. I ended up using the code attached, and it worked perfectly. I am not familiar enough with matlab to understand if this is equivalent to your suggestion, so I am not sure if what you suggested would have provided the correct results, but I appreciate your help in my process.
Torsten
Torsten 2022-5-3
编辑：Torsten 2022-5-3
As I wrote in my first response, the call to dydt must be
x(n+1) = x(n) + h*dxdt(x(n),y(n))
y(n+1) = y(n) + h*dydt(x(n),y(n))
instead of
x(n+1) = x(n) + h*dxdt(x(n),t(n))
y(n+1) = y(n) + h*dydt(y(n),t(n))
as you first did, and that's what you finally found out by yourself. That's optimal.

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更多回答（1 个）

KSSV
KSSV 2022-4-28
The loop should be taken as:
for i = 1:round(N)
In your present case, it is a fraction and you are getting non integers as the indices and error is pooped.
% Demo
N = 10.2;
for i = 1:N
i
A(i) = rand ; % error, index cannot be fraction
end

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