$y'' + y' -2y = x^2$, 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.
 A: If you plug the quadratic polynomial into your differential equation you will get
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2.$$
Compare the coefficients and determine $A$, $B$, $C$.
A: 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.
A: With
$y = Ax^2 + Bx + C, \tag 1$
we may substitute $y$, $y'$, and $y''$ into
$y'' + y' - 2y = x^2, \tag 2$
viz,
$y' = 2Ax + B, \tag 3$
$y'' = 2A, \tag 4$
$y'' + y' - 2y = 2A + (2Ax + B) - 2(Ax^2 + Bx + C) = x^2; \tag 5$
we group together like powers of $x$:
$-2Ax^2 + 2(A - B) x + (2A + B -2C) = x^2, \tag 5$
from which we infer
$-2A = 1, \tag 6$
$2(A - B) = 0, \tag 7$
$2A + B - 2C = 0; \tag 8$
thus
$A = B = -\dfrac{1}{2}, \tag 9$
$C = -\dfrac{3}{4}, \tag{10}$
and of course
$y = -\dfrac{1}{2}x^2 - \dfrac{1}{2}x - \dfrac{3}{4}. \tag{11}$
We Check:
From (11),
$y' = -x - \dfrac{1}{2}, \tag{12}$
$y'' = -1, \tag{13}$
$y'' + y' - 2y$
$= -1 - x - \dfrac{1}{2} - 2(-\dfrac{1}{2}x^2 - \dfrac{1}{2}x - \dfrac{3}{4}) = -1 - x - \dfrac{1}{2} + x^2 + x + \dfrac{3}{2} = x^2. \tag{14}$
A: Consider $z^2+z-2=(z+2)(z-1)=-2(1+\frac{1}{2}z)(1-z)$. Its inverse (as a formal power series) is
$$
-\frac{1}{2}(1+z+z^2+\dotsb)\left(1-\frac{1}{2}z+\frac{1}{4}z^2+\dotsb\right)
=-\dfrac{1}{2}-\dfrac{1}{4}z-\dfrac{3}{8}z^2+\dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we can discard higher order terms):
$$
-\dfrac{1}{2}x^2-\dfrac{1}{4}2x-\dfrac{3}{8}2=-\dfrac{1}{2}x^2-\dfrac{1}{2}x-\dfrac{3}{4}
$$
