$\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
\newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
\newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
\newcommand{\dd}{{\rm d}}%
\newcommand{\ds}[1]{\displaystyle{#1}}%
\newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
\newcommand{\fermi}{\,{\rm f}}%
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
\newcommand{\half}{{1 \over 2}}%
\newcommand{\ic}{{\rm i}}%
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}%
\newcommand{\isdiv}{\,\left.\right\vert\,}%
\newcommand{\ket}[1]{\left\vert #1\right\rangle}%
\newcommand{\ol}[1]{\overline{#1}}%
\newcommand{\pars}[1]{\left( #1 \right)}%
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}%
\newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
\newcommand{\sech}{\,{\rm sech}}%
\newcommand{\sgn}{\,{\rm sgn}}%
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}%
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{ay_{t + 2} + by_{t + 1} + cy_{t} = \fermi\pars{t}\,,\qquad t = 0, 1,2,\ldots}$
Let's introduce $\ds{{\cal Y}\pars{z} \equiv \sum_{t = 0}^{\infty}y_{t}\,z^{t}}$ and
$\ds{{\cal F}\pars{z} \equiv \sum_{t = 0}^{\infty}\fermi\pars{t}\,z^{t}}$.
Then,
$$
\sum_{t = 0}^{\infty}\pars{ay_{t + 2} + by_{t + 1} + cy_{t}}z^{t}
= \sum_{t = 0}^{\infty}\fermi\pars{t}z^{t} \equiv {\cal F}\pars{z}
$$
which yields
\begin{align}
&{\cal F}\pars{z}
=a\sum_{t = 2}^{\infty}y_{t}z^{t - 2} + b\sum_{t = 1}^{\infty}y_{t}z^{t - 1}
+ c\sum_{t = 0}^{\infty}y_{t}z^{t}
=
{a \over z^{2}}\bracks{{\cal Y}\pars{z} - y_{0} - y_{1}z}
+
{b \over z}\bracks{{\cal Y}\pars{z} - y_{0}} + c{\cal Y}\pars{z}
\\[3mm]&=
\pars{{a \over z^{2}} + {b \over z} + c}{\cal Y}\pars{z} - {ay_{0} \over z^{2}}
- {by_{0} + ay_{1} \over z}
\\[3mm]
&z^{2}{\cal F}\pars{z}=
\pars{cz^{2} + bz + a}{\cal Y}\pars{z} - ay_{0} - \pars{by_{0} + ay_{1}}z
\end{align}
$$
{\cal Y}\pars{z}
=
{{\cal G}\pars{z}\over cz^{2} + bz + a}\quad\mbox{where}\quad
{\cal G}\pars{z} \equiv z^{2}{\cal F}\pars{z} + ay_{0} + \pars{by_{0} + ay_{1}}z
$$
Now, we can expand both members of $\ds{{\cal Y}\pars{z}
=
{{\cal G}\pars{z}\over cz^{2} + bz + a}}$ in powers of $z$. By comparing 'term a term' we can find $y_{t}\,,\ \forall\ t = 0,1,2,\ldots$. Also
$$
{\cal Y}\pars{z}
=
{1 \over c}\,{{\cal G}\pars{z} \over \pars{z - z_{-}}\pars{z - z_{+}}}
\quad\mbox{where}\quad z_{\pm} = {-b \pm \root{b^{2} - 4ca} \over 2c}
$$
$$
{\cal Y}\pars{z}
=
{{\cal G}\pars{z} \over c}\,{1 \over z_{+} - z_{-}}
\pars{{1 \over z - z_{+}} - {1 \over z - z_{-}}}
=
{1 \over \root{b^{2} - 4ca}}\bracks{%
{1 \over z_{-}}\,{{\cal G}\pars{z} \over 1 - z/z_{-}}
-
{1 \over z_{+}}\,{{\cal G}\pars{z} \over 1 - z/z_{+}}}
$$
In order to proceed further, in general terms, we need additional information of the parameters $\braces{a,b,c}$.
In order to illustrate the general procedure, we'll consider a particular case:
$$
a = c = 1\,,\quad b = -2\,,\quad y_{0} = 0\,,\quad y_{1} = 1\,,\quad
\fermi\pars{t} \equiv -6t
$$
Then $\pars{~\mbox{with}\ \verts{z} < 1~}$,
\begin{align}
&{\cal F}\pars{z} = \sum_{t = 0}^{\infty}\pars{-6t}z^{t}
=-6z\,\totald{}{z}\sum_{t = 0}^{\infty}z^{t}
= -6z\,\totald{}{z}\pars{1 \over 1 - z} = -\,{6z \over \pars{1 - z}^{2}}
\\[3mm]&
\mbox{such that}\
{\cal G}\pars{z} = -\,{6z^{3} \over \pars{1 - z}^{2}} + z
=
{-6z^{3} + \pars{z - 1}^{2}z \over \pars{1 - z}^{2}}
=
{-5z^{3} -2z^{2} + z \over \pars{1 - z}^{2}}
\\[3mm]&
\mbox{and}\ {\cal Y}\pars{z} = {-5z^{3} - 2z^{2} + z \over \pars{1 - z}^{4}}
=\pars{-5z^{3} - 2z^{2} + z}\,\sum_{t = 0}d_{t}z^{t}
\\[3mm]&\mbox{where}\
d_{t} \equiv {1 \over 6}\pars{t + 1}\pars{t + 2}\pars{t + 3}
\end{align}
\begin{align}
{\cal Y}\pars{t}
&=
-5\sum_{t = 3}^{\infty}d_{t - 3}z^{t} -2\sum_{t = 2}^{\infty}d_{t - 2}z^{t}
+
\sum_{t = 1}^{\infty}d_{t - 1}z^{t}
\\[3mm]&=
d_{0}z + \pars{d_{1} - 2d_{0}}z^{2}
+
\sum_{t = 3}^{\infty}\pars{d_{t - 1} -2d_{t - 2} - 5d_{t - 3}}z^{t}
\\[3mm]&=
z + 2z^{2} +
\sum_{t = 3}^{\infty}\pars{d_{t - 1} -2d_{t - 2} - 5d_{t - 3}}z^{t}
\end{align}
The solution becomes:
\begin{align}
&y_{0} = 0\,,\quad y_{1} = 1\,,\quad y_{2} = 2\,;\qquad\qquad
y_{t} = d_{t - 1} -2d_{t - 2} - 5d_{t - 3}\,,\quad t \geq 3
\\[3mm]&\mbox{where}\
d_{t} \equiv {1 \over 6}\pars{t + 1}\pars{t + 2}\pars{t + 3}
\end{align}
