If I have a linear homogeneous recurrence relation
$$y_n=c_1y_{n-1}+\ldots+c_ky_{n-k},$$
I can get its characteristic equation, which is
$$r^k=c_1r^{k-1}+\ldots+c_k.$$
In particular for $y_n=cy_{n-1},$ I get $r=c.$ While I see that this obviously gives the right solution, something seems not right. For a differential equation $y'=cy,$ I get a similar characteristic equation: $r=c$, however, as I understand it, the the analogy between differential equations and recurrence relations is given by $y'\sim y_n-y_{n-1},$ not $y'\sim y_n.$ By this logic, I think the characteristic equation for the recurrence relation $y_n=cy_{n-1}$ should be $r=c-1,$ because the recurrence relation is equivalent to $y_n-y_{n-1}=(c-1)y_{n-1}.$ Could you help me understand why the analogy breaks down here (or why it doesn't)?
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 $y'=cy.$
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 $e^{cx}.$