Some background/context
A typical Linear Constant Coefficient Difference Equation (a.k.a. LCCDE, or recursive relation) might look like $$y[n] = \frac 34y[n-1]-\frac 18y[n-2]$$ Which is sometimes written more suggestively: $$y[n] - \frac 34y[n-1] + \frac 18y[n-2]=0$$ I've been learning how to solve these, and to my great surprise, there seems to be some kind of profound connection to differential equations. In the above example, the solution is as follows:
Assume $y[n]=\lambda^n$ and plug in: (for brevity, some steps are skipped) $$\lambda^{n-2}(\lambda^2-\frac 34\lambda+\frac 18) = 0 \\ \lambda=\frac 12, \frac 14$$ So now we say that $y[n]=C_1(\frac 12)^n+C_2(\frac 14)^n$
What I'm trying to do
I mean this is just screaming differential equations, so I'm trying to interpret an LCCDE as sampling an ODE's solution. I can get about this far: if we allow $p_1$ and $p_2$ (and in general, $p_k$) to be the roots of an ODE whose solution (sampled every $T_s$ seconds) is exactly equal to the above $y[n]$, then we can say $$e^{p_kt}|_{t=nT_s}=e^{p_kT_sn}=\lambda_k^n\ \\p_k=\frac{\ln(\lambda_k)}{T_s}$$ And in our specific example (if we just want a sampling interval of 1 second), $$ p_1=\ln(\frac 12), p_2=\ln(\frac 14)$$ Working backwards even further, we can even write down the "time-domain" differential equation which has these roots in its characteristic (homogenous) equation: $$y''(t)-\ln(\frac 18)y'(t)+\ln(\frac 14)\cdot \ln(\frac 12)\cdot y(t)=0$$
What my question is
Unfortunately, the whole point of doing all this was because I'm trying to understand how LCCDEs can be so closely connected to ODEs. Given an LCCDE, I can convert to an ODE, but not without first solving all the roots and working backwards, which isn't helping me to see what's going on. Is there another way to think about this, that might explain why difference equations are so similar to differential equations?