How to solve difference equation in MATLAB

2020 10 7
3 个回答
回答已采纳
更新时间：2023 8 29
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0 个投票

How to solve the difference equation for
yn+1 =5/2 yn +yn-1 ,y0 =y1 =1 in terms of the roots of its characteristic equation in MATLAB ?

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 采纳的回答

Ameer Hamza
Ameer Hamza 2020-10-7
编辑：Ameer Hamza 2020-10-7

0 个投票

An alternative is to use filter()
a = [1 -5/2 -1];
b = 0;
ic = [1 1];
n = 50; % 50 terms
y1 = [ic(1) filter(b, a, ones(1, n-1), ic)];

3 个评论
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Betty Johnson
Betty Johnson 2020-10-7
Thanks for your response.It gives the following error "Unrecognized function or variable 'Y'."
Ameer Hamza
Ameer Hamza 2020-10-7
Sorry! There was a typo. Try again.
Betty Johnson
Betty Johnson 2020-10-7
Thank you for your help.After finding y1 according to your code,I am trying to check the behavior of the sequence {yn} as n → infinity based on following code but it doesn't look right.
clear all
close all
%function for fixed point iteration
yn=@(y1,y0) (5/2)*y1+y0;
%all initial guess
yy0=1; yy1=1;
y_val(1)=yy0; y_val(2)=yy1;
cnt=2; tt(1)=0;tt(2)=1;
%loop for all y
for i=1:10
cnt=cnt+1;
y_val(cnt)=yn(yy1,yy0);
yy0=yy1; yy1=y_val(cnt);
tt(cnt)=cnt-1;
end
%plotting the result
plot(tt,y_val)
xlabel('iterations')
ylabel('y_n')
title('y_n vs. iteration plot')
fprintf('The solution is diverging and tends to infinity at t tends to infinite.\n')

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更多回答（2 个）

KSSV
KSSV 2020-10-7

1 个投票

n = 50 ;
y = zeros(1,n) ;
y(1:2) = 1 ;
for i = 2:n-1
y(i+1) =5/2*y(i) +y(i-1) ;
end

3 个评论
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Betty Johnson
Betty Johnson 2020-10-7
Thank you for your response.I am trying to check the behavior of the sequence {yn} as n → infinity based on your code.It gives me error.
Walter Roberson
Walter Roberson 2020-10-7
Ran in:
Ran in:
No error in the KSSV posted.
n = 50 ;
y = zeros(1,n) ;
y(1:2) = 1 ;
for i = 2:n-1
y(i+1) =5/2*y(i) +y(i-1) ;
end
disp(y(end-4:end))
1.0e+21 * 0.1255 0.3577 1.0198 2.9073 8.2881
You cannot, of course, run this out to infinity.
Walter Roberson
Walter Roberson 2023-8-29
There are no negative coefficients, and no coefficients with absolute value less than one, and the initial values are positive. Each value is at least 5/2 times the previous one, so a lower bound would be (5/2)^(n-1) and therefore the bound to infinity is infinite

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mohammed hussain
mohammed hussain 2023-8-29

0 个投票

a = [1 -5/2 -1];
b = 0;
ic = [1 1];
n = 50; % 50 terms
y1 = [ic(1) filter(b, a, ones(1, n-1), ic)];

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