Solution to solve these non-homogeneous recurrence relation an+ 4an-1 − 5an-2=n+2 where a0=1 & a1=-1 - Sikademy
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Archangel Macsika

solve these non-homogeneous recurrence relation an+ 4an-1 − 5an-2=n+2 where a0=1 & a1=-1

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Here's the Solution to this Question

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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