Don't underestimate this functional equation!

By the functional equation, $f(x+2)-5f(x+1)+6f(x)=0$

Putting $x=0$ , we have

$f(2)-5f(1)+6f(0)=0 \implies f(2)-5f(1)+6 \times 2=0 \implies f(2)-5f(1)+12=0 (1)$

Putting $x=1$ , we have

$f(3)-5f(2)+6f(1)=0 \implies 35-5f(2)+6f(1)=0 (2)$

Eliminating equations $(1)$ and $(2)$ , we have $f(1)=5$ and $\boxed{f(2)=13}$

Let's express the above functional equation in terms of the characteristic equation:

$f(x+2) - 5 \cdot f(x+1) + 6 \cdot f(x) = 0 \Rightarrow r^2 - 5r + 6 = 0 \Rightarrow (r-2)(r-3) = 0 \Rightarrow r = 2, 3$

This in turns yields:

$f(x) = A\cdot2^x + B\cdot3^x; f(0) = 2, f(3) = 35$

for real constants A and B. We now solve for A & B using the supplied initial conditions:

$2 = A + B; 35 = 8A + 27B \Rightarrow A = B = 1$

hence, we finally obtain $f(x) = 2^x + 3^x$ (which $f(1) = 5$ is a prime number!) Thus, $f(2) = 2^2 + 3^2 = \boxed{13}.$

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