Let $y_n=x_n-1.$ Then $ y_{n}=2y_{n-1}-4y_{n-2} $ so $y_{n+6}=2^6y_n$ and so $y_{2011}=2^{2011}...$

The problem seems rather vaguely phrased, but let me attempt it either way. Note that $x_n =2x_{n-1} -4x_{n-2} +3$ is a non-homogeneous recurrence relation. We make the substitution $y_n =x_n -1$ which transforms the recurrence into $y_n =2y_{n-1} -4y_{n-2}$, which is homogeneous. Beware that $y_0=a-1$ and $y_1 =2$. Applying directly the formulae found here we find that the solution of the recurrence is $y_n =(a-1)2^{n} +n(2-a)2^{n}$. For $n=2011$, we have that $y_{2011}=2^{2011}\left(a-1 +2011(2-a)\right)$, so $k_a =2011$.