The solution is:[tex]a_n=\alpha_1 \cdot3^n+\alpha _2\cdot2^n+{(\frac{49}{20})}\cdot7^n[/tex]
Given:- [tex]a_n =5a_{n-1}-6a_{n-2} +7^n[/tex]
To find:- all solutions of the recurrence relation
The solutions of the associated homogeneous recurrence are
[tex]a^{(h)}_n = \alpha _13^n + \alpha _22^n[/tex]
Because F(n) =[tex]7^n[/tex] a reasonable trial solution is [tex]a^{(p)}_n[/tex][tex]n = C\cdot7^n[/tex]. Making the substitution and solving for C gives C = [tex]\frac{49}{ 20}[/tex], and hence [tex]a^{(h)}_n ={(\frac{49}{20})}\cdot7^n[/tex]
is a particular solution. Hence, a complete parametrization of solutions is:
[tex]a_n=\alpha_1 \cdot3^n+\alpha _2\cdot2^n+{(\frac{49}{20})}\cdot7^n[/tex]
#SPJ3
