Can someone explain to me in a little bit more detail how you can get to this point. I know its explained here but i'm trying to apply the way he did this problem to this one
\begin{equation*} u_n-7u_{n-1}=3\times 7^n \end{equation*} But I get it wrong.
I'm sorry if this question has been asked before but I can't seem to find a linear recurrence problem the same as this one for me to understand how to do a problem like this.
Going into more detail about the solved example in the OP may not help with the recurrence $u_n-7u_{n-1}=3\cdot 7^n$. So we concentrate on solving that recurrence.
The general solution of the homogeneous recurrence $u^{(c)}_n-7u^{(c)}_n-1=0$ is $7^n A$. I expect you found that easily.
Now we need to find a particular solution of the inhomogeneous recurrence $u_n^-7u_{n-1}=3\cdot 7^n$. We can look for a solution of shape $7^nC$. However, when we substitute, we get the equation $7^n C-7\cdot 7^{n-1}C=3\times 7^n$. The left side is $0$, so cannot be equal to the right side. We conclude that there is no particular solution of the desired shape. That is probably what stopped you in your attempt to imitate the solved example.
What to do? You may remember a trick that is used when the characteristic equat equation has multiple roots. We look for a particular solution of the shape $n7^nC$. When you plug in, you will find that this works if we put $C=3$.
So the general solution of our recurrence is $u_n=7^nA+3n7^n$. Now if you have been given an initial condition (not mentioned in the post) you can calculate $A$.