How to solve difference equation in MATLAB

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 ?

Answer 1

[Ameer Hamza]

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)];

[Betty Johnson]

Thanks for your response.It gives the following error "Unrecognized function or variable 'Y'."

[Ameer Hamza]

Sorry! There was a typo. Try again.

[Betty Johnson]

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')

Answer 2

[KSSV]

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

[Betty Johnson]

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]

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]

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

Answer 3

[mohammed hussain]

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)];

[Walter Roberson]

how does this differ from the accepted answer?