You have the recurrence $a_{n+1}=ba_n+cn$, presumably with some initial value $a_0$. Multiply both sides by $x^n$ and sum over $n\ge 0$: $$\sum_{n\ge 0}a_{n+1}x^n=b\sum_{n\ge 0}a_nx^n+c\sum_{n\ge 0}nx^n\;.\tag{1}$$ For convenience let $$A(x)=\sum_{n\ge 0}a_nx^n\;.$$ Then for starters we can rewrite $(1)$ as $$\sum_{n\ge 0}a_{n+1}x^n=bA(x)+c\sum_{n\ge 0}nx^n\;.$$ Now look more closely at the sum on the lefthand side of $(1)$: it’s $$a_1+a_2x+a_3x^2+\dots\;,\tag{2}$$ compared with $$A(x)=a_0+a_1x+a_2x^2+\dots\;.$$ If you multiply $(2)$ by $x$ and add $a_0$, you get exactly $A(x)$, so we can further rewrite $(2)$ as $$\frac{A(x)-a_0}x=bA(x)+c\sum_{n\ge 0}nx^n\;,$$ or $$A(x)-a_0=bxA(x)+c\sum_{n\ge 0}nx^{n+1}\;,\tag{3}$$ which can be solved for $A(x)$ as soon as we evaluate the summation in the last term. From what you’ve written, I suspect that you may already be okay up to here. Now
$$\begin{align*}
\sum_{n\ge 0}nx^{n+1}&=x^2+2x^3+3x^4+\dots\\
&=x^2(1+2x+3x^2+\dots)\\\\
&=x^2\sum_{n\ge 0}(n+1)x^n\;,
\end{align*}$$
and $$\begin{align*}\sum_{n\ge 0}(n+1)x^n&=\frac{d}{dx}\left(\sum_{n\ge 0}x^{n+1}\right)\\
&=\frac{d}{dx}\left(\sum_{n\ge 0}x^n\right)\tag{4}\\
&=\frac{d}{dx}\left(\frac1{1-x}\right)\\
&=\frac1{(1-x)^2}\;.
\end{align*}$$
The step at $(4)$ is justified because $\sum_{n\ge 0}x^{n+1}$ and $\sum_{n\ge 0}x^n$ differ only by the constant $1$, so they have the same derivative. $(3)$ now becomes $$A(x)-a_0=bxA(x)+\frac{c}{(1-x)^2}\;,$$ so $$\begin{align*}A(x)&=\frac1{1-bx}\left(a_0+\frac{c}{(1-x)^2}\right)\\
&=\frac{a_0}{1-bx}+\frac{c}{(1-bx)(1-x)^2}\;.
\end{align*}$$
This is the desired generating function, and if you really want a closed form solution to the recurrence, you can decompose this into partial fractions, convert them to power series, and read off the coefficient of $x^n$.