Converting poles/zeros to differential/difference equation solutions
Okay, I'm gonna attempt an answer to this while I am still convinced the question is ill-formed.  I am answering the title of the question without really referring to the text of the question, which I don't understand.

First, for discrete-time:  A general IIR (or even FIR in which all of the poles are at the origin, where $z=0$) transfer function is:

$$\begin{align}
 H(z) \triangleq \frac{Y(z)}{X(z)} &amp; = H_0 \frac{(z-q_1)(z-q_2)(z-q_3)...(z-q_M)}{(z-p_1)(z-p_2)(z-p_3)...(z-p_N)} \\
\\
 &amp; = H_0 \frac{z^M (1-q_1 z^{-1})(1-q_2 z^{-1})(1-q_3 z^{-1})...(1-q_M z^{-1})}{z^N (1-p_1 z^{-1})(1-p_2 z^{-1})(1-p_3 z^{-1})...(1-p_N z^{-1})}  \\
\\
 &amp; = z^{M-N} \ \frac{H_0 \ (1-q_1 z^{-1})(1-q_2 z^{-1})(1-q_3 z^{-1})...(1-q_M z^{-1})}{(1-p_1 z^{-1})(1-p_2 z^{-1})(1-p_3 z^{-1})...(1-p_N z^{-1})}  \\
\\
 &amp; = z^{M-N} \ \frac{b_0 + b_1 z^{-1} + b_2 z^{-2} + b_3 z^{-3} + ... + b_M z^{-M} }{ 1 + a_1 z^{-1} + a_2 z^{-2} + a_3 z^{-3} + ... +  + a_N z^{-N} }
\end{align}$$

$N$ is the number of poles, $p_i$, and $M$ is the number of zeros, $q_i$.

The difference equation comes from moving a few of the above things around.

$$\begin{align} Y(z) &amp; \left(  1 + a_1 z^{-1} + a_2 z^{-2} + ... + a_N z^{-N}  \right) \\ 
&amp; \quad = X(z) z^{M-N} \left( b_0 + b_1 z^{-1} + b_2 z^{-2} + ... + b_M z^{-M} \right) \end{align}$$

$$\begin{align} Y(z) = \ &amp; b_0 z^{M-N} X(z) + b_1 z^{M-N-1} X(z) + b_2 z^{M-N-2} X(z) + ...  \\
 &amp; \quad ... + b_M z^{-N} X(z) - a_1 z^{-1}Y(z) - a_2 z^{-2}Y(z) - ... - a_N z^{-N}Y(z) \end{align}$$

Inverting the Z Transform, this becomes 

$$\begin{align} y[n] = \ &amp; b_0 \ x[n+M-N] + b_1 \ x[n+M-N-1] + b_2 \ x[n+M-N-2] + ...  \\
 &amp; \quad ... + b_M \ x[n-N] - a_1 \ y[n-1] - a_2 \ y[n-2] - ... - a_N \ y[n-N] \end{align}$$

For a causal system or filter, we cannot look into the future for any input sample $x[n]$.  That means $M \le N$ so that there is no negative delays on the $x[n+M-N]$ term.

Often $M=N$ (the numbers of poles and zeros are the same) and you get the more familiar

$$\begin{align} y[n] = \ &amp; b_0 \ x[n] + b_1 \ x[n-1] + b_2 \ x[n-2] + ... + b_N \ x[n-N] \\
 &amp; \quad - a_1 \ y[n-1] - a_2 \ y[n-2] - ... - a_N \ y[n-N] \end{align}$$

and the more familiar transfer function

$$\begin{align}
 H(z) &amp; = H_0 \frac{(z-q_1)(z-q_2)(z-q_3)...(z-q_N)}{(z-p_1)(z-p_2)(z-p_3)...(z-p_N)} \\
\\
 &amp; = \frac{H_0 \ (1-q_1 z^{-1})(1-q_2 z^{-1})(1-q_3 z^{-1})...(1-q_N z^{-1})}{(1-p_1 z^{-1})(1-p_2 z^{-1})(1-p_3 z^{-1})...(1-p_N z^{-1})}  \\
\\
 &amp; = \frac{b_0 + b_1 z^{-1} + b_2 z^{-2} + b_3 z^{-3} + ... + b_N z^{-N} }{ 1 + a_1 z^{-1} + a_2 z^{-2} + a_3 z^{-3} + ... +  + a_N z^{-N} }
\end{align}$$

That demonstrates where the difference equation (and the feedforward, $b_i$, and feedback, $-a_i$, coefficients) come from the poles, $p_i$, and zeros, $q_i$.

The problem remaining, which is the hard problem, is that I didn't bother to multiply it out.  So you have to blast this out:

$$\begin{align}
 H_0 \ &amp; (1-q_1 z^{-1})(1-q_2 z^{-1})(1-q_3 z^{-1})...(1-q_M z^{-1})  \\
 &amp; \quad = b_0 + b_1 z^{-1} + b_2 z^{-2} + b_3 z^{-3} + ... + b_M z^{-M}
\end{align}$$

and

$$\begin{align}
  &amp; (1-p_1 z^{-1})(1-p_2 z^{-1})(1-p_3 z^{-1})...(1-p_N z^{-1})  \\
 &amp; \quad = 1 + a_1 z^{-1} + a_2 z^{-2} + a_3 z^{-3} + ... + a_N z^{-N}
\end{align}$$

You can see right away that $b_0 = H_0$ and that $a_0=1$ (I never really had an "$a_0$" term anyway), but you have to blast out the multiplication of the binomials to get the polynomials.  You can also see that $b_M = (-1)^M H_0 q_1 q_2 q_3 ... q_M$ and $a_N = (-1)^N p_1 p_2 p_3 ... p_N$   That's not a simple expression, but I hope you understand what you must do here.