Finding an analogous differential equation to a given recursive relation (clarifying examples included)

Question

Example 1. The recursive relation

$$x_{n+1} = rx_n$$

with initial condition $x_0$ admits the general solution

$$x_n = x_0r^n.$$

The corresponding differential equation whose solution coincides with the above recursion over the restricted domain $\mathbb{N}$ is

$$\dfrac{dx}{dn} = (\ln r)\cdot x.$$

Example 2. The recursive relation

$$x_{n+1} = rx_n + c$$

with initial condition $x_0$ admits the general solution

$$x_n = \Big(x_0 + \dfrac{c}{r - 1}\Big)r^n - \dfrac{c}{r - 1}.$$

The analogous differential equation is thus

$$\dfrac{dx}{dn} = \ln r\Big(x + \dfrac{r}{c - 1}\Big).$$

Question. What is the differential equation corresponding to the recursive relation $x_{n+1} = \dfrac{x_n}{n}$ with initial condition $x_1$?

Looking for a hint or general direction. Thanks.

Answer 1

The solution to the recursion $$x_{n+1} = \dfrac{x_n}{n}$$ with initial conditions $x_1$ is $$x_n = \dfrac{x_1}{\Gamma(n)} = x(n).$$ Taking the derivative of $x(n)$ gives $$\dfrac{dx}{dn} = x_1\dfrac{d}{dn}\dfrac{1}{\Gamma(n)} = \dfrac{x_1}{\Gamma(n)^2}\dfrac{d\Gamma}{dn} = \dfrac{x^2}{x_1}\dfrac{d\Gamma}{dn},$$ giving us our equivalent differential equation system.

Answer 2

Hint:

Let $f\left(n\right)=x_{n}$. Then Taylor expanding

$$f\left(n+1\right)\approx f\left(n\right)+f^{\prime}\left(n\right)$$

Thus, the recurrence

$$f\left(n+1\right)=\dfrac{f\left(n\right)}{n}$$

is translated to

$$f^{\prime}\left(y\right)\approx\dfrac{f\left(y\right)-yf\left(y\right)}{y}$$

The full solution is given by

$$\int\dfrac{{\rm d}f}{f}\approx\int\dfrac{1-y}{y}{\rm d}y$$ $$\ln\left(\dfrac{f}{f_{0}}\right)\approx\ln\left(\frac{y}{y_{0}}\right)-y+y_{0}$$ $$\dfrac{f\left(y\right)}{f\left(y_{0}\right)}\approx\dfrac{y}{y_{0}}e^{-\left(y-y_{0}\right)}$$