How to solve this infinite system of equations? I'm trying to solve an infinite set of coupled equations.  j is a real integer index where $-\infty<j<\infty$. {a,b,c,d,f} are variables, and K's are constants.  I need to find closed form solutions for $\{a_{j},b_{j},c_{j}\}$ in terms of only the K's. I believe the respective solutions for $\{a_{j},b_{j},c_{j}\}$ should have the same form for all j.  I don't need explicit forms for the d's or f's.
$$a_{j}+b_{j}+c_{j}=K_{j}$$
$$3a_{j}+2b_{j}+c_{j}=c_{j+1}$$
$$6a_{j}+3b_{j}+d_{j}+3f_{j}=0$$
$$3a_{j}+2b_{j}+d_{j}+2f_{j}=0$$
$$d_{j}+f_{j}-f_{j-1}=0$$
Can this be done?

Given
$$
\left\{ \begin{array}{l}
 a_j  + b_j  + c_j  = K_j  \\ 
 3a_j  + 2b_j  + c_j  = c_{j + 1}  \\ 
 6a_j  + 3b_j  + d_j  + 3f_j  = 0 \\ 
 3a_j  + 2b_j  + d_j  + 2f_j  = 0 \\ 
 d_j  + f_j  - f_{j - 1}  = 0 \\ 
 \end{array} \right.
$$
we can use 3rd and 4th rows to enucleate $d$ anf $f$ as a combination of  $a$ and $b$
$$
\left\{ \begin{array}{l}
 f_j  =  - 3a_j  - b_j  \\ 
 d_j  = 3a_j  \\ 
 \end{array} \right.
$$
Thereafter we can enucleate $a$ as
$$
\begin{array}{l}
 \left\{ \begin{array}{l}
 a_j  + b_j  + c_j  = K_j  \\ 
 3a_j  + 2b_j  + c_j  = c_{j + 1}  \\ 
 3a_{j - 1}  - b_j  + b_{j - 1}  = 0 \\ 
 \end{array} \right.\;\; \Rightarrow \;\left\{ \begin{array}{l}
 a_j  + b_j  + c_j  = K_j  \\ 
 3a_j  + 2b_j  + c_j  = c_{j + 1}  \\ 
 3a_j  + b_j  = b_{j + 1}  \\ 
 \end{array} \right.\;\; \Rightarrow  \\ 
  \Rightarrow \;\left\{ \begin{array}{l}
 a_j  = K_j  - b_j  - c_j  \\ 
 \left\{ \begin{array}{l}
 3K_j  - b_j  - 2c_j  = c_{j + 1}  \\ 
 3K_j  - 2b_j  - 3c_j  = b_{j + 1}  \\ 
 \end{array} \right. \\ 
 \end{array} \right. \\ 
 \end{array}
$$
where in the last row we have increased the index: we shall take that in due account, after solving the system, by accomodating the initial conditions.
So  we are left with
$$
\left\{ \begin{array}{l}
 3K_j  - b_j  - 2c_j  = c_{j + 1}  \\ 
 3K_j  - 2b_j  - 3c_j  = b_{j + 1}  \\ 
 \end{array} \right.\;\; \Rightarrow \left( {\begin{array}{*{20}c}
   {b_{j + 1} }  \\   {c_{j + 1} }  \\
\end{array}} \right) =  - \left( {\begin{array}{*{20}c}
   2 & 3  \\   1 & 2  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   {b_j }  \\   {c_j }  \\
\end{array}} \right) + 3K_j \left( {\begin{array}{*{20}c}
   1  \\   1  \\
\end{array}} \right)
$$
The matrix diagonalizes as
$$
\left( {\begin{array}{*{20}c}
   2 & 3  \\
   1 & 2  \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
   1 & {\sqrt 3 }  \\
   1 & { - \sqrt 3 }  \\
\end{array}} \right)^{ - 1} \left( {\begin{array}{*{20}c}
   {2 + \sqrt 3 } & 0  \\
   0 & {2 - \sqrt 3 }  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   1 & {\sqrt 3 }  \\
   1 & { - \sqrt 3 }  \\
\end{array}} \right)
$$
and thus finally we get the decoupled difference equations
$$
\left( {\begin{array}{*{20}c}
   {u_{j + 1} }  \\   {v_{j + 1} }  \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
   1 & {\sqrt 3 }  \\   1 & { - \sqrt 3 }  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   {b_{j + 1} }  \\   {c_{j + 1} }  \\
\end{array}} \right) =  - \left( {\begin{array}{*{20}c}
   {2 + \sqrt 3 } & 0  \\   0 & {2 - \sqrt 3 }  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   {u_j }  \\   {v_j }  \\
\end{array}} \right) + 3K_j \left( {\begin{array}{*{20}c}
   1  \\   1  \\
\end{array}} \right)
$$
which can be easily solved .
