Solution to Solve the following set of recurrence relations and initial conditions3an-1+n^2-3,n>1,a0=1 - Sikademy
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Solve the following set of recurrence relations and initial conditions3an-1+n^2-3,n>1,a0=1

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Let us find the solution of the recurrence relation a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3 with

a_0 = 1 and a_1 = 4.

The characteristic equation k^2-4k+3=0 of the homogeous recurrence relation a_n -4a_{n−1} + 3a_{n−2}=0 is equivalent to (k-1)(k-3)=0, and hence has the solutions k_1=1

and k_2=3. Therefore, the general solution of the relation a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3 is of the form a_n=c_1+c_23^n+b_n^p, where b_n^p=a2^n+n(bn+c)=a2^n+bn^2+cn.

It follows that

a2^n+bn^2+cn

=4(a2^{n-1}+b(n-1)^2+c(n-1))-3(a2^{n-2}+b(n-2)^2+c(n-2))+2^n+n+3

=4a2^{n-1}+4b(n^2-2n+1)+4c(n-1)-3a2^{n-2}-3b(n^2-4n+4)-3c(n-2)+2^n+n+3

=4a2^{n-1}-3a2^{n-2}+2^n+bn^2+(4b+c+1)n+(-8b+2c+3)

=(5a+4)2^{n-2}+bn^2+(4b+c+1)n+(-8b+2c+3).

It follows that 4a=5a+4,\ c=4b+c+1 and -8b+2c+3=0.

Therefore, a=-4,\ b=-\frac{1}4,\ c=\frac{1}2(8b-3)=\frac{1}2(-2-3)=-\frac{5}2.

We conclude that the general solution of the relation a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3 is of the form a_n=c_1+c_23^n-4\cdot2^n-\frac{1}4n^2-\frac{5}2n.

Since a_0 = 1 and a_1 = 4, we get

1=a_0=c_1+c_2-4 and 4=a_1=c_1+3c_2-8-\frac{1}4-\frac{5}2=c_1+3c_2-\frac{43}4.

Therefore, c_1+c_2=5 and c_1+3c_2=\frac{59}4. It follows that c_2=\frac{39}8 and c_1=\frac{1}8.


Consequently, the general solution of the relation a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3 is the following:

a_n=\frac{39}8+\frac{1}83^n-2^{n+2}-\frac{1}4n^2-\frac{5}2n.

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Question ID: mtid-5-stid-8-sqid-328-qpid-215