$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation. I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:

$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.

I understand how to find derivatives if I know the function, but this is stumping me.

Assuming that $y = y(x)$ is of the given form, we need only substitute $y$ into the differential equation 
$$
y^{\prime\prime} + y^\prime  - 2y = x^2
$$
and solve for $A,B$ and $C$. Now, we have
\begin{align}\label{eq:1}\tag{1}
y^\prime(x) = 2Ax + B \quad \text{and} \quad y^{\prime\prime}(x) = 2A.
\end{align}
So, if $y$ solves the differential equation, then
\begin{align}
x^2 = y^{\prime\prime} + y^\prime  - 2y &= 2A + (2Ax+B) -2\left(Ax^2 + Bx + C \right)\label{eq:2}\tag{2}\\
&= -2Ax^2 + (2A-2B)x + \left( 2A + B -2C\right)\label{eq:3}\tag{3}.
\end{align}
Comparing the coefficients of this equation, we find that
\begin{align*}
\begin{cases}
-2A = 1,\\
2A-2B = 0,\\
2A+B-2C = 0.
\end{cases}
\end{align*}
Solving this system will give the explicit values of $A,B$ and $C$ required.
