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,

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