Solving Recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n\geq 1, a_0=2$? How can I solve recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n \geq1 $ and $a_0 = 2$?
Hint: $$a_{n+1} - 3a_n = 5 \cdot 3^{n+1} = 3 (a_n-3a_{n-1})$$
Can you start from here?

 $$a_{n+1} - 6a_n + 9a_{n-1} = 0 \implies \frac{a_{n+1}}{3^{n+1}} - 2\frac{a_n}{3^n} + \frac{a_{n-1}}{3^{n-1}}=0$$ Let $a_n = 3^n b_n$ then $$b_{n+1} - 2b_n  + b_{n-1} = 0 \implies b_{n+1} - b_{n} = b_n-b_{n-1} = \cdots = b_1 - b_0 = 7 - 2 =5$$ Therefore $b_n=b_0+5n=5n+2, a_n = 3^n (5n+2). \blacksquare$
general sequence is $a_n= (3^n)a_0 +n \times 5\times 3^n$
substitute $a(n-1)$ for $a(n-2)$ and check the pattern you can easily figure it out.