Answer
\[
-\frac{1}{2} x^{2}-\frac{1}{2} x-\frac{3}{4}=y
\]
Work Step by Step
Given the general form for $t,$ we know that its derivatives should be:
\[
\begin{array}{c}
2 A x+B =y^{\prime}\\
2 A=y^{\prime \prime}
\end{array}
\]
Plug these into the equation given, we have that
\[
y^{\prime \prime}+y^{\prime}-2 y=(2 A)+(2 A x+B)-2\left(A x^{2}+B x+C\right)=x^{2}
\]
Re-arranging the terms on the right hand side, this implies that
\[
-2 A x^{2}+(2 A-2 B) x+(2 A+B-2 C)=x^{2}
\]
since the coefficients on $x^{2}$ should be equal on both sides, we have $-2 A=1$. Meaning that $A=-\frac{1}{2}$
Also, since the coefficients on $x$ should be equal, we have that
\[
0=2 A-2 B=2\left(-\frac{1}{2}\right)-2 B=-1-2 B
\]
Which gives us $-\frac{1}{2}=B$ also.
Since the constant terms on both sides of the equation should be equal, we have that
\[
\begin{array}{c}
2 A+B-2 C=0 \\
2\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)-2 C=0
\end{array}
\]
Thus, $-\frac{3}{2}=2 C,$ meaning that $-\frac{3}{4}=C$
So, we can conclude that $-\frac{1}{2} x^{2}-\frac{1}{2} x-\frac{3}{4}=y$
