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 January 28th, 2012, 10:24 PM #1 Newbie   Joined: Jan 2012 Posts: 5 Thanks: 0 changing difference equation I have to change this systems of difference equation to a single higher order equation. $\ x_{n+1}=3x_{n}+2y_{n}\$ $\ y_{n+1}=3x_{n}+4y_{n}\$ The answer is $\ A(5)^n+B(2)^n\$ What I did so far was set it equal to zero and got this: $\ 0=3x_{n}+2y_{n}-x_{n+1} = 3(y_{n+1}-4y_{n})-2y_{n}-x_{n+1}=3y_{n+1}-10y_{n}-x_{n+1}\$ I am not sure what to do after. I think I am supposed to find the characteristic equation but my equation doesn't look like a polynomial. Can someone help me figure this out please? Thank you for your time!
 January 28th, 2012, 10:27 PM #2 Newbie   Joined: Jan 2012 Posts: 5 Thanks: 0 Re: changing difference equation Whoops. I made a mistake on the second equation. It is $\ y_{n+1}=x_{n}+4y_{n}\$
 January 28th, 2012, 11:01 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: changing difference equation We are given: (1) $x_{n+1}=3x_{n}+2y_{n}$ (2) $y_{n+1}=x_{n}+4y_{n}$ Method 1: Taking (2) and replacing n with n + 1 gives: $y_{n+2}=x_{n+1}+4y_{n+1}$ Using (1), substitute for $x_{n+1}$: $y_{n+2}=3x_{n}+2y_{n}+4y_{n+1}$ Solving (2) for $x_{n}$ and substituting, we then get: $y_{n+2}=3$$y_{n+1}-4y_{n}$$+2y_{n}+4y_{n+1}$ $y_{n+2}=3y_{n+1}-12y_{n}+2y_{n}+4y_{n+1}$ $y_{n+2}=7y_{n+1}-10y_{n}$ $y_{n+2}-7y_{n+1}+10y_{n}=0$ Now we have a linear homogeneous recurrence, whose associated characteristic equation is: $r^2-7r+10=0$ $(r-5)(r-2)=0$ Thus, we find the closed-form for $y_n$ is: $y_n=A\cdot5^n+B\cdot2^n$ where A and B are arbitrary constants. Method 2: Taking (1) and replacing n with n + 1 gives: $x_{n+2}=3x_{n+1}+2y_{n+1}$ Using (2), substitute for $y_{n+1}$: $x_{n+2}=3x_{n+1}+2$$x_{n}+4y_{n}$$$ $x_{n+2}=3x_{n+1}+2x_{n}+8y_{n}$ Solving (1) for $8y_n$ and substituting, we then get: $x_{n+2}=3x_{n+1}+2x_{n}+4x_{n+1}-12x_{n}$ $x_{n+2}=7x_{n+1}-10x_{n}$ $x_{n+2}-7x_{n+1}+10x_{n}=0$ Now we have a linear homogeneous recurrence, whose associated characteristic equation is: $r^2-7r+10=0$ $(r-5)(r-2)=0$ Thus, we find the closed-form for $x_n$ is: $x_n=A\cdot5^n+B\cdot2^n$ where A and B are arbitrary constants.

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