Answer
The function $y=Ax^2+Bx+C$ would satisfy the differential equation if $A=-\frac{1}{2}$, $B=-\frac{1}{2}$ and $C=-\frac{3}{4}$
Work Step by Step
$$y''+y'-2y=x^2$$
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.
