The idea behind using the characteristic equation to solve the recurrence relation is to postulate exponential solutions, $y_n=r^n.$ The roots of the characteristic equation, assuming none are multiple roots, then give a set of basis functions, and the general solution is a linear combination of these basis functions,
$$y_n=\alpha_1r_1^n+\alpha_2r_2^n+\ldots+\alpha_kr_k^n.$$
The idea behind using the characteristic equation to solve the differential equation is similar. We postulate an exponential solution $y=e^{rx}.$ The roots of the characteristic equation, again assuming no multiple roots, give a set of basis functions, and the general solution is a linear combination of these basis functions,
$$y=\alpha_1e^{r_1x}+\alpha_2e^{r_2x}+\ldots+\alpha_ke^{r_kx}.$$
Let's see how the two methods are related in your examples, $y_n=cy_{n-1}$ and $y'=cy.$
- For the recurrence we get $r=c,$ and therefore the general solution $y_n=\alpha c^n$.
- For the differential equation we also get $r=c.$ This gives the general solution $y=\alpha e^{cx}.$
Let's now solve a discrete approximation of the differential equation using the recurrence equation method. Approximating the derivative, we have
$$\frac{y(x+\Delta x)-y(x)}{\Delta x}\approx cy(x).$$
If we discretize the $x$-axis, letting $x=n\Delta x,$ we get
$$\frac{y((n+1)\Delta x)-y(n\Delta x)}{\Delta x}\approx cy(n\Delta x).$$
If we write $y_n$ for $y(n\Delta x),$ we get
$$y_{n+1}-y_n\approx cy_n\Delta x$$
or
$$y_{n+1}\approx(1+c\Delta x)y_n.$$
The quantity in parentheses plays the role of $c$ in the recurrence relation solved above. Adapting that solution gives $r=1+c\Delta x$ and therefore
$$y_n\approx\alpha(1+c\Delta x)^n.$$
Since $\Delta x=x/n,$ this is the same as
$$y_n\approx\alpha(1+cx/n)^n.$$
This is consistent with the exact solution to the differential equation since, for large $n,$ $(1+cx/n)^n$ is approximately $e^{cx}.$