Prove using some form of induction that $T(n) = O(n^2)$ given recurrence relation 
Here is a recursively defined function where $c \ge 0$.
$T(n) = c$ if $n = 0$
$T(n) = c$ if $n = 1$
$T(n) = 2T(n-1) - T(n-2) + 2$ if $ \ n \geq  2$
Prove using some form of induction that $ \ T(n) = O(n^2)$

My attempt:
We have to show there exist positive constants $ \ c, n_{0}$ such that $ \ T(n) \le cn^2 \  \forall n \ge n_{0}$. Pick $ \ n_{0} = 1$.
Base cases:
$n=1$: $ \ T(1) = c $ by definition and $ \ c \le c(1)^2 = c$ is true
$n=2$. $ \ T(2) = c $ by definition and $ \ c \le c(2)^2 = 4c$ is true
I.H:
Suppose that $ \ n\ge 2$ and that $ \ P(1),P(2), ...., P(n)$ all hold
$ \ T(n+1) = 2T(n-1) -T(n-2) +2 \le 2c(n-1)^2 -c(n-2)^2 + 2$, by I.H.
Then, $ \ 2c(n-1)^2 -c(n-2)^2 + 2 = cn^2 -2c +2 \le cn^2 +2$ since $  -2c \le 0$ and $ \ cn^2 + 2 \le cn^2 + 2cn + c = c(n^2 + 2n +1) = c(n+1)^2$

Some corrections are needed. In the induction hypothesis you used the inequality $-T(n-2) \le -c(n-2)^2$ or $T(n-2) \ge c(n-2)^2$ which is not true. 
Hint. For $n\geq 1$, let $S(n):=T(n) -T(n-1)$. Then $S(1)=0$ and for $n\geq 2$
$$S(n)=T(n) -T(n-1)=T(n-1) -T(n-2)+2=S(n-1) + 2.$$ 
Hence (by induction?), for $n\geq 1$, $S(n)=2(n-1)$ that is 
$$T(n)=T(n-1) +2(n-1).$$
Now, by using the above recurrence, show that there exist a positive integer $n_0$ and a constant $C$ such that $T(n)\leq Cn^2$ for all $n\geq n_0$ (for example you can take $n_0=1$ and $C=c+1$).
P.S. You can also prove by induction that $T(n)=c+n(n-1)$. Then it is easy to  show that $T(n) = O(n^2)$.
