Functional equation which might be related to some DE

I have been trying to solve the following functional equation

$$f(x+2)+af(x+1)+bf(x)=0$$ for all real values of $$x$$.

My guess and intuition leans towards $$f(x)$$ being some exponential function. I started out with $$f(x)=\lambda a^x$$ but ended up with a quadratic which seemed like a dead end.

It does look similar to the general differential equation $${d^2y \over dx^2} + a {dy\over dx} + b=0$$ which also leads to a quadratic but I'm getting messy expressions for the functional equation.

• The problem is simple for integer $x$ Commented May 29, 2021 at 8:39
• @ClaudeLeibovici I wish to solve $f(x)$ for all real values of $x$ Commented May 29, 2021 at 10:03
• Why would the quadratic seem like a dead end? It can give you some solutions.
– Sil
Commented May 29, 2021 at 11:37
• Essentially, the result for the integers is all you can get. More precisely, for any $g : [ 0 , 2 ) \to \mathbb R$, you can extend $g$ to an $f : \mathbb R \to \mathbb R$ satisfying the desired equation; just treat every $x \in [ 0 , 1 )$ separately, and find the value of $f ( x + n )$ for all $n \in \mathbb Z$. Even further assumptions on $f$ like continuity/smoothness doesn't make the solution set very small (you'd only need $g$ to be continuous/smooth with the additional assumption that its limit or the limits of its derivatives near $2$ equal to the value at $0$). Commented May 29, 2021 at 22:26
• The situation is different if you restrict the case more than that, and for example require $f$ to be real analytic. in that case, the only solutions are those that are (linear) combinations of functions of the form you've already mentioned. Commented May 29, 2021 at 22:27

$$\mathrm{f(x) = k_1\alpha ^x + k_2\beta ^x}$$ is the solution for the given functional equation, where $$\mathrm{\alpha}$$ and $${\beta}$$ are the roots of the equation $$\mathrm{x^2+ax+b=0}$$.