$$y''+y'-2y=x^2$$
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2$$
2A=1
But why do the following equal zero?
2Ax+B=0
$$Ax^2+Bx+C=0$$

Question

$$y''+y'-2y=x^2$$
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2$$
2A=1
But why do the following equal zero?
2Ax+B=0
$$Ax^2+Bx+C=0$$

helder_edwin · Answer

shouldn't u have
$$ \large x^2=2A+(2Ax+B)-2(Ax^2+Bx+C) $$
$$\large =(-2A)x^2+(2A-2B)x+(2A+B-2C) $$
then
$$ \large -2A=1 $$
$$ \large 2A-2B=0 $$
$$ \large 2A+B-2C=0 $$
after equating the coefficients of both polynomials?

anonymous · Answer

yep, but still....why would it be zero though?
$$\large 2A-2B=0
$$

helder_edwin · Answer

because on (my) LHS there's no $x$ so its coefficient is zero.

anonymous · Answer

wouldn't that count for the first one too?

anonymous · Answer

I'm really sorry I'm really trying to understand this....
$$\large =(-2A)x^2+(2A-2B)x+(2A+B-2C)$$

helder_edwin · Answer

the coefficients for $x^2$ are: on LHS 1 and on RHS -2A
they must be equal so 1=-2A.
the same for $x$ and the independent term

anonymous · Answer

RHS and LHS of what? you mean of the whole equation?
$$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2$$
Yeah that makes more sense

helder_edwin · Answer

yes. but u gotta re-arange (order, i mean) your LHS

helder_edwin · Answer

i mean, associate like terms.