solve these non-homogeneous recurrence relation an+ 4an-1 − 5an-2=n+2 where a0=1 & a1=-1
A solution "b_n" to the non-homogeneous recurrence is similar to f(n).
Guess:
"b_n=cn+d"
Then:
"cn+d+4(c(n-1)+d)-5(c(n-2)+d)=n+2"
"n-4=0"
"c=d=b_n=0"
Solution of the given recurrence is
"a_n=b_n+h_n"
where "h_n" is a solution for the associated homogeneous recurrence:
"h_n+4h_{n-1}-5h_{n-2}=0"
In our case: "a_n=h_n"
Characteristic equation:
"r^2+4r-5=0"
"r_1=\\frac{-4-\\sqrt{16+20}}{2}=-5, r_2=1"
"a_n=\\alpha_1r_1^n+\\alpha_2r_2^n"
"a_n=\\alpha_1(-5)^n+\\alpha_2"
"a_0=\\alpha_1+\\alpha_2=1"
"a_1=-5\\alpha_1+\\alpha_2=-1"
"-5\\alpha_1+(1-\\alpha_1)=-1"
"\\alpha_1=1\/3,\\alpha_2=2\/3"
Solution:
"a_n=\\frac{1}{3}(-5)^n+\\frac{2}{3}"
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