1. ## Difference equation

Hi!

Problem: $(1) \; \; y_{n+1}-y_{n}=e^{n} \; , y_{0}=0$

Solution: Characteristic equation: $r-1=0$ .

$r = 1$

$y_{n}^{h} = C_{1}\cdot 1^{n}$

Particular solution: $y_{n}^{p}=An\cdot e^{n}$ <-- Is this correct ?

When I insert this into (1) I get, $e^{n}\left[A(n+1)\cdot e - An\right] = e^{n}$

Thx

2. Wouldn't the the form of the particular solution be just $Ae^{n}$ ?

$Ae^{n+1} - Ae^{n} = e^{n}$

$A = \frac{e^{n}}{e^{n+1}-e^{n}} = \frac{1}{e-1}$

so $y^{p}_{n} = \frac{1}{e-1}e^{n}$

3. Hi!

The book says, if right side is $(polynomial)\cdot k`{n}$ , then you should try $y_{n}^{p}=(polynomial)\cdot k^{n} \cdot n^{m}$

where $m$ is the number of solutions to the charateristic solution that is equal to $k$ .

And I think, we always have $1^{n}$ on the right side to to speak.

They do something like that in a previous example.

Edit: I tried your suggestion, it worked. I guess I misinterpreted the book. Here we only have $e^{n}$ , so we donīt mulitply by any $n$ .