## Calculus: Early Transcendentals 8th Edition

The function $y=Ax^2+Bx+C$ would satisfy the differential equation if $A=-\frac{1}{2}$, $B=-\frac{1}{2}$ and $C=-\frac{3}{4}$
$$y''+y'-2y=x^2$$ Consider the function $y=Ax^2+Bx+C$ - First derivative: $y'=2Ax+B$ - Second derivative: $y''=2A$ To find $A, B, C$, we need to apply the above function and derivatives to the differential equation: $$2A+(2Ax+B)-(2Ax^2+2Bx+2C)=x^2$$ $$2A+2Ax+B-2Ax^2-2Bx-2C=x^2$$ $$(-2Ax^2)+(2A-2B)x+(2A+B-2C)=x^2+0x+0$$ Now we see that, to satisfy the differential equation, the followings must be satisfied: 1) $-2A=1$ So, $A=-\frac{1}{2}$ 2) $(2A-2B)=0$. Therefore, $A-B=0$. So, $B=A=-\frac{1}{2}$ 3) $2A+B-2C=0$. Therefore, $C=A+\frac{B}{2}=-\frac{1}{2}-\frac{1}{4}=-\frac{3}{4}$ These are the values of A, B and C we need to find.