Solving T(n)=T(n−1)+2T(n−2) using substitution
The secret of effective substitution is to find the magic numbers -1 and 2. Look and behold.

$$ T(n)+T(n-1)=2T(n-1)+2T(n-2)=2(T(n-1)+T(n-2))$$
Let $S(n)=T(n)-(-1)T(n-1)$. Then 
$$S(n)=2S(n-1)=2^2S(n-2)=\cdots=2^{n-2}S(2)\tag{I}$$

$$ T(n)-2T(n-1)=-T(n-1)+2T(n-2)=-(T(n-1)-2T(n-2))$$
Let $R(n)=T(n)-2T(n-1)$. Then 
$$R(n)=-R(n-1)=R(n-2)=\cdots=(-1)^nR(2)\tag{II}$$

Eliminating $T(n-1)$ from formula ($\text{I}$) and ($\text{II}$), we obtain $$T(n)= \frac{2S(n) + R(n)}3=\frac{2^{n-1}S(2) + (-1)^nR(2)}3=\frac{S(2)}62^n + O(1)$$

So, $T(n) = O(2^n)$.

It is assumed in the above that S(2) and R(2) are defined or, what is equivalent, T(2) and T(1) are defined. Their actual values do not affect the conclusion.



In fact, we can be more precise. If $T(2)+T(n-1)&gt;0$, i.e., $S(2) &gt; 0$, we will have $T(n)=\Theta(2^n)$. If $T(2)+T(n-1)=0$, we will have $T(n)=O(1)$. If $T(2)+T(n-1)&lt;0$, we will have $-T(n)=\Theta(2^n)$.



How to find those two magic numbers?  

Consider T(n) as $x^2$, T(n-1) as $x$ and T(n-2) as 1. Then you get $x^2 = x + 2$. Solving that equation, you get $x=-1$ or $x=2$.



(Exercise.) Let $F(n)=F(n-1)+F(n-2)$ for $n\in\Bbb N$, $n\ge2$. Show $F(n)= O(\alpha^n)$, where $\alpha = \frac{1+\sqrt 5}2$.