<s>[INST] $y'' + y' -2y = x^2$, 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.
 [/INST] 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),
$y' = -x - \dfrac{1}{2}, \tag{12}$
$y'' = -1, \tag{13}$
$y'' + y' - 2y$
$= -1 - x - \dfrac{1}{2} - 2(-\dfrac{1}{2}x^2 - \dfrac{1}{2}x - \dfrac{3}{4}) = -1 - x - \dfrac{1}{2} + x^2 + x + \dfrac{3}{2} = x^2. \tag{14}$
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