<s>[INST] Solving functional equation $f(x)=3f(x+1)-3f(x+2)$ It is given that a function f(x) satisfy:
$$f(x)=3f(x+1)-3f(x+2)\quad \text{ and } \quad f(3)=3^{1000}$$ then find value of $f(2019)$.
I further wanted to ask that is there some general method to solve such equation. The method that I know to solve such questions is to substitute $x$ with $x+1$ in equation and there by making new equation which is
$$ f(x+1)=3f(x+2)-3f(x+3)$$ then again substitute $x$ with $x+2$ in original equation and make new equation $$f(x+2)=3f(x+3)-3f(x+4)$$ Do this for a couple of times and then on combining the equations, in most of the question we get some relation like f(x) = f(x+a) but that does not work here. Please share your ideas on how to solve such questions.
 [/INST] $$ \begin{align} f(x)&=f(x-1)-\frac{f(x-2)}{3}, f(3)=3^{1000}\\
f(x)&=f(x-1)-\frac{1}{3}f(x-2)\\
f(x+1)&=f(x)-\frac{1}{3}f(x-1)=(f(x-1)-\frac{1}{3}f(x-2))-\frac{1}{3}f(x-1)\\
f(x+1) &= \frac{2}{3}f(x-1)-\frac{1}{3}f(x-2)\\
f(x+2)&=f(x+1)-\frac{1}{3}f(x)=(\frac{2}{3}f(x-1)-\frac{1}{3}f(x-2))-\frac{1}{3}(f(x-1)-\frac{1}{3}f(x-2))\\ 
f(x+2)&=\frac{1}{3}f(x-1)-\frac{2}{9}f(x-2)\\
f(x+3)&=f(x+2)-\frac{1}{3}f(x+1)=(\frac{1}{3}f(x-1)-\frac{2}{9}f(x-2))-\frac{1}{3}(\frac{2}{3}f(x-1)-\frac{1}{3}f(x-2))\\
f(x+3) &= \frac{1}{9}f(x-1)-\frac{1}{9}f(x-2)
f(x+4)=(\frac{1}{9}f(x-1)-\frac{1}{9}f(x-2))-\frac{1}{3}(\frac{1}{3}f(x-1)-\frac{2}{9}f(x-2))\\
f(x+4)&=(-\frac{1}{9}+\frac{2}{27})f(x-2)=-\frac{1}{27}f(x-2)\end{align}$$
$$\therefore f(x+4)=-\frac{1}{27}f(x-2) \implies f(x+6)=-\frac{1}{27}f(x)$$
$$\implies f(x+6k)=(-\frac{1}{27})^k f(x)  \forall k \in N$$
$$f(2019)=f(3+6\cdot 336)=(-\frac{1}{27})^{336} f(3)=\frac{1}{3^{1008}}3^{1000} = \frac{1}{3^8}$$
$$\therefore f(2019) = \frac{1}{3^8}$$
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