Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 84

Find constants $A,B$ and $C$ such that the function $y= Ax^2+Bx+C$
satisfies the differential equation $y''+y'-2y=x^2$.


$
\begin{equation}
\begin{aligned}
y &= Ax^2+Bx+C\\
y' &= A \frac{d}{dx} (x^2) + B \frac{d}{dx} (x) + \frac{d}{dx} (C)\\
y' &= A(2x) + B(1) + 0\\
y' &= 2Ax+B\\
y'' &= 2A \frac{d}{dx}(x) + \frac{d}{dx}(B)\\
y'' &= 2A(1)+0\\
y'' &= 2A
\end{aligned}
\end{equation}
$


Substituting these values to the differential equation,

$ y'' + y' - 2y =x^2$


$
\begin{equation}
\begin{aligned}
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) = x^2
\end{aligned}
\end{equation}
$

Equating each power we get,
For $x^2$:

$
\begin{equation}
\begin{aligned}
-2A &= 1\\
A &= \frac{-1}{2}
\end{aligned}
\end{equation}
$


For $x$:

$
\begin{equation}
\begin{aligned}
2A-2B &= 0\\
\cancel{2}A &= \cancel{2}B\\
B = A &= \frac{-1}{2}
\end{aligned}
\end{equation}
$


For constant,

$
\begin{equation}
\begin{aligned}
2A + B - 2C &= 0\\
2\left(\frac{-1}{2}\right) + \left(\frac{-1}{2}\right) -2C &= 0\\
C &= \frac{-3}{4}
\end{aligned}
\end{equation}
$