$f: \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2)=5$ The function $f : \mathbb{R} \to \mathbb{R}$ satisfies $(x-2)f(x)-(x+1)f(x-1) = 3$. Evaluate $f(2013)$, given that $f(2) = 5$.

Hint 1: $3=(x+1)-(x-2)$

$(x-2)f(x)-(x+1)f(x-1)=3\Leftrightarrow (x-2)f(x)-(x+1)f(x-1)=(x+1)-(x-2)\Leftrightarrow (x-2)(f(x)+1)-(x+1)(f(x-1)+1)=0$


Hint 2: Is there a function closely related to $f$ that would verify a simpler equation?

 $g(x)=f(x)+1$

$ $

 $(x-2)g(x) - (x+1)g(x-1)=0 \Leftrightarrow (x-2)g(x)=(x+1)g(x-1) \Leftrightarrow g(x)=\cfrac{x+1}{x-2}g(x-1)$



 $g(x)=\cfrac{x+1}{x-2}g(x-1)=\cfrac{x+1}{x-2}\cfrac{x+1-1}{x-2-1}g(x-1-1) = \left(\prod\limits_{k=0}^{n-1} \cfrac{x+1-k}{x-2-k}\right) g(x-n)$

Now give the good value to $n$, simplify the product and you'll have the expression of $g$ you need.
