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 June 17th, 2013, 06:24 AM #1 Newbie   Joined: Apr 2013 Posts: 20 Thanks: 0 Solving Recurrence equation Hey, so I have: $a_{n} -7a_{n-1} +10a_{n-2}=16n$ I also have the initial conditions, a0 = 5 and a1 = 4. I have to 'solve' it but I having issues. Answer for general solution is: $a_{n}=A2^n+B5^n+13+4n$ And it then applies initial conditions to get solution: $a_{n}=-9(2^n)+5^n+13+4n$ My attempt so far has been unsuccessful. I know we use the homogenous equation but I cannot figure out how. Any help? Thanks!
 June 17th, 2013, 08:28 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Solving Recurrence equation We first solve the associated homogenous equation: $a_{n}-7a_{n-1}+10a_{n-2}=0$ which has the auxiliary equation: $r^2-7r+10=(r-2)(r-5)=0$ Hence, a general solution to the homogeneous equation is: $h_n=c_12^n+c_25^n$ Since the inhomogeneous term is $16n$, we seek a particular solution of the form: $p_n=An+B$ where the constants $A$ and $B$ are to be determined. Substituting this expression for $p_n$ into the recursion, we find: $(An+B)-7(A(n-1)+B)+10(A(n-2)+B)=16n$ $An+B-7(An-A+B)+10(An-2A+B)=16n$ $An+B-7An+7A-7B+10An-20A+10B=16n$ $4An-13A+4B=16n+0$ Equating coefficients, we find: $4A=16\,\therefore\,A=4$ $-13A+4B=0\,\therefore\,B=13$ Thus, we have found: $p_n=4n+13$ and so: $a_n=h_n+p_n=c_12^n+c_25^n+4n+13$ Now, we may determine the parameters $c_i$ from the given initial values: $a_0=c_12^0+c_25^0+4(0)+13=c_1+c_2+13=5\,\therefore \,c_1+c_2=-8$ $a_1=c_12^1+c_25^1+4(1)+13=2c_1+5c_2+17=4\,\therefo re\,2c_1+5c_2=-13$ Solving this linear 2X2 system, we find: $c_1=-9,\,c_2=1$ and thus we finally have: $a_n=-9\cdot2^n+5^n+4n+13$
 June 17th, 2013, 01:20 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Solving Recurrence equation Nicely done!

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