Solve by using substitution method $T(n) = T(n-1) + 2T(n-2) + 3$ given $T(0)=3$ and $T(1)=5$ I'm stuck solving by substitution method:
$$T(n) = T(n-1) + 2T(n-2) + 3$$ given $T(0)=3$ and $T(1)=5$
I've tried to turn it into homogeneous by subtracting $T(n+1)$:
$$A: T(n) = T(n-1) + 2T(n-2) + 3$$
$$B: T(n+1) = T(n) + 2T(n-1) +3$$
$$A - B = T(n) - T(n+1) = T(n-1) + 2T(n-2) + 3 - (T(n) + 2T(n-1) + 3)$$
$$2T(n) - T(n+1) - T(n-1) + 2T(n-2) = 0$$
Assume $T(n) = x^{n}$
$$2x^{n} - x^{n+1} - x^{n-1} + 2x^{n-2} = 0$$
Dividing each side by $x^{n-2}$ leaves me with the impossible (beyond the scope of this class) equation:
$$2x^2 - x^3 - x + 2 = 0$$
How can I solve this using substitution method?

Hints
Suppose $T(n) = \alpha$ for some constant $\alpha$. You will see that for a particular  value $ T(n) = \alpha $ will be a particular solution to the equation (maybe this is what is meant by method of substitution).
Now solve the homogeneous equation $ T(n) - T(n- 1) - 2 T(n - 2) = 0  $ and find its general solution $ f(t) $.  Then the general solution to you equation will be $ f(t) + \alpha $ where $ \alpha $ is the particular solution found above.  
