Please help us to find the general solution of this recurrence:
$$x_{n+1}=5^{n-1} x_n+3^n.$$
We found the solution for the associated homogeneous recurrence $x_{n+1}=5^{n-1} x_n$ which is $$x_n = 5^{(n-1)(n-2)/2}x_1,$$ and we tried to use it to find a particular solution of the nonhomogeneous recurrence, but unfortunately we were not able to find it.
On
This is the route I usually take but....
$$\begin{align} x_{n+1}&=5^{n-1} x_n+3^n \\ &=5^{n-1}(5^{n-2} x_{n-1}+3^{n-1})+3^n \\ &\quad = 5^{2n-(1+2)}x_{n-1}+(3 \cdot 5)^{n-1} +3^n \\ &=5^{2n-(1+2)}(5^{n-3}x_{n-2}+3^{n-2})+(3\cdot5)^{n-1}+3^n \\ &\quad =5^{3n-(1+2+3)}x_{n-2}+5^{2n-(1+2)}3^{n-2}+5^{n-1}3^{n-1}+3^n \\ &\vdots \\ &\vdots \quad \text{seeing a pattern...after i steps} \\ &\vdots \\ &=5^{(i+1)n+\frac{(i+1)(i+2)}{2}}x_{n-i}+\sum_{k=0} ^{i} 5^{kn-\frac{k(k+1)}{2}}3^{n-k} \\ &\vdots \\ &\vdots \quad \text{supposing that you do it n times, and set i=n...} \\ \\ &\vdots \\ &=5^{\frac{(n+1)(3n+2)}{2}}x_0+\sum_{k=0} ^{n} 5^{kn-\frac{k(k+1)}{2}}3^{n-k} \\ \end{align}$$
I'm not sure if this is any help to you but I couldn't resist taking a stab.
(Nothing original here.)
If $x_{n+1} =u_{n+1} x_n+v_{n+1} $, let $U_n =\prod_{k=1}^n u_k $ so $\dfrac{U_n}{u_n} =\prod_{k=1}^{n-1} u_k =U_{n-1} $.
Then $\dfrac{x_{n+1}}{U_{n+1}} =\dfrac{u_{n+1}}{U_{n+1}} x_n+\dfrac{v_{n+1}}{U_{n+1}} =\dfrac{x_{n}}{U_{n}}+\dfrac{v_{n+1}}{U_{n+1}} $.
Letting $a_n =\dfrac{x_{n}}{U_{n}} $ and $b_n =\dfrac{v_{n}}{U_{n}} $, this becomes $a_{n+1} =a_n+b_{n+1} $.
Therefore $a_{n+1}-a_n =b_{n+1} $.
Summing $a_n-a_1 =\sum_{k=1}^{n-1}(a_{k+1}-a_k) =\sum_{k=1}^{n-1}b_{k+1} $ so $a_n =a_1+\sum_{k=1}^{n-1}b_{k+1} =a_1+\sum_{k=2}^{n}b_{k} $.
Replacing these by their definitions, $\dfrac{x_{n}}{U_{n}} =\dfrac{x_{1}}{U_{1}}+\sum_{k=2}^{n}\dfrac{v_{k}}{U_{k}} $ so $x_{n} =\dfrac{x_{1}U_n}{U_{1}}+\sum_{k=2}^{n}\dfrac{U_nv_{k}}{U_{k}} =x_{1}\prod_{k=1}^n u_k+\sum_{k=1}^{n-1}v_k\prod_{j=k+1}^n u_j $.
Put in $u_n = 5^{n-2}, v_n=3^{n-1} $ and see what you get.
You probably ought to check my math, also.