Solving Recurrence Relation

Question

a_n+2 = 5_an+1 - 6_an + 4n - 4, a_0 = 1, a_1 =4

Answer 1

Dividing by rn, we get the characteristic equation:

6r2=5r−6

Now, we can solve this quadratic equation:

6=0r2−5r+6=0

Factoring the quadratic, we get:

r−2)(r−3)=0

So, the characteristic roots are 1=2r1​=2 and 2=3r2​=3.

The general form of the solution for the homogeneous part is:

an(h)​=A⋅r1n​+B⋅r2n​

Now, let's find the particular solution for the non-homogeneous part44n−4. Since the non-homogeneous part is a linear function, we assume a particular solution of the form an(p)​=Cn+D. Substituting this into the original recurrence relation:

4C(n+2)+D=5(C(n+1)+D)−6(Cn+D)+4n−4

Simplifying, we get:

4Cn+2C+D=5Cn+5D−6Cn−6D+4n−4

Grouping terms:

4Cn+2C+D=−Cn−D+4n−4

Equating coefficients:

42C+D=4

SubstitutingC=−1 into the second equation, we get 6D=6.

So, the particular solution is6an(p)​=−n+6.

The general solution for the entire recurrence relation is the sum of the homogeneous and particular solutions:

)an​=an(h)​+an(p)​

6an​=A⋅2n+B⋅3n−n+6

Now, we use the initial conditions 0=1a0​=1 and 1=4a1​=4 to find the values of A and B:0+6=1a0​=A+B−0+6=1 1+6=4a1​=2A+3B−1+6=4

Solving these simultaneous equations, we g2A=−2 and3B=−3.

to the given recurrence relation with initial conditions 0=1a0​=1 and 1=4a1​=4 is:

6an​=−2⋅2n−3⋅3n−n+6