Use the substitution $A_n=r^n$ to solve each of the recurrence relation:
(a)$2A_n=7A_{n-1}-3A_{n-2}; n \ge 0;A_0=A_1=1$
(b)$2S_n=7S_{n-1}-3S_{n-2}; n \ge 0;S_0=S_1=1$
This is a problem in my study guide and (a) and (b) seems the same problem to me. We can solve (a) by letting $A=r^n$ and we can solve (b) by letting $S=t^n$ or something. Both (a) & (b) should have the same solution. I am a bit perplexed why the professor will put the same problem twice. Am I missing something?
HINT: with your Substitution we get $$2q^n=7q^{n-1}-3q^{n-2}$$ this gives the solution $$a_n=\frac{1}{5}2^{-n}(4+6^n)$$ which is the same as your result.