Answer
$A=-\frac{1}{2}$
$B=-\frac{1}{2}$
$C=-\frac{3}{4}$
Work Step by Step
$y=Ax^2+Bx+C$
Find $y'$ and $y''$:
$y'=A\cdot 2x^{2-1}+B+0$
$y'=2Ax+B$
and
$y''=2A$
Substituting $y$, $y'$, and $y''$ into the given differential equation,
$2A+(2Ax+B)-2(Ax^2+Bx+C)=x^2$
$2A+2Ax+B-2Ax^2-2Bx-2C=x^2$
$-2Ax^2+(2A-2B)x+(2A+B-2C)=1x^2+0x+0$
It should be:
$-2A=1\Rightarrow A=-\frac{1}{2}$
$2A-2B=0\Rightarrow B=A=-\frac{1}{2}$
$2A+B-2C=0\Rightarrow 2\cdot (-\frac{1}{2})+(-\frac{1}{2})-2C=0\Rightarrow -\frac{3}{2}-2C=0\Rightarrow C=-\frac{3}{4}$
Thus, $A=-\frac{1}{2}$, $B=-\frac{1}{2}$, and $C=-\frac{3}{4}$.
