What was my mistake solving $y''-2y'=-2x$? I somehow missed the negative numbers on the last two terms of this problem, and we were required to use variation of parameters.
Here is the work I did:
$y''-2y'=-2x$
First I solved the characteristic equation:
$\begin{align*}
    r^2-2r&=0 \\
    r(r-2)&=0 \\
    r&=\{0,2\}
\end{align*}$
Therefore the complementary function $y_c$ is
$y_c=c_1+c_2e^{2x}$ for some arbitrary constants $c_1$ and $c_2$.
Then, to find out the particular function $y_p$, I solved the following system of equations for $v_1$ and $v_2$ given that $y_c=c_1y_1+c_2y_2$ and $ay''+by'+cy=G(x)$:
$\Bigg\{\begin{array}{c}
    v_1'y_1+v_2'y_2=0 \\
    v_1'y_1'+v_2'y_2'=\displaystyle\frac{G(x)}{a}
\end{array}$
$\Rightarrow\Bigg\{\begin{array}{c}
    v_1'+v_2'e^{2x}=0 \\
    2v_2'e^{2x}=-2x
\end{array}$
Using substitution to solve the system and subsequently integrating $v_1'$ and $v_2'$:
$v_2'=-xe^{-2x}$
$\begin{align*}
    v_1'+(-xe^{-2x})e^{2x}&=0\\
    v_1'-x&=0\\
    v_1'&=x
\end{align*}$
$v_1=\int v_1' \,dx=\int x\,dx=\frac{1}{2}x^2$
$v_2=\int v_2'\,dx=\int -xe^{-2x}\,dx$
$\begin{array}{cc} 
    u=-x & v=-\frac{1}{2}e^{-2x}\\
    du=-dx & dv = e^{-2x}
\end{array}$
$\begin{align*}
    v_2&=\frac{1}{2}xe^{-2x}-\int\frac{1}{2}e^{-2x}\,dx\\
    &=\frac{1}{2}xe^{-2x}+\frac{1}{4}e^{-2x}
\end{align*}$
Since $y_p=v_1y_1+v_2y_2$:
$\begin{align*}
    y_p&=\frac{1}{2}x^2(1)+\left(\frac{1}{2}xe^{-2x}+\frac{1}{4}e^{-2x}\right)e^{2x}\\
    &=\frac{1}{2}x^2+\frac{1}{2}x+\frac{1}{4}
\end{align*}$
The solution to the differential equation is the sum of $y_c$ and $y_p$: $\boxed{y=c_1+c_2e^{2x}+\frac{1}{2}x^2+\frac{1}{2}x+\frac{1}{4}}$
However, the correct answer that was given in the multiple choice answer choices was:
$y=c_1+c_2e^{2x}+\frac{1}{2}x^2+\frac{1}{4}(-2x-1).$
I'm not sure how I missed the negatives on the $x$ term and the final constant, since otherwise my answer was the same. What I'm equally as confused about is why the $-\frac{1}{4}$ doesn't "absorb" into $c_1$.
Thank you for your help!

Note that the DE is:
$$(y'e^{-2x})'=-2xe^{-2x}$$
$$y'e^{-2x}=xe^{-2x}+\dfrac 12e^{-2x}+C$$
$$y'=x+\dfrac 12+Ce^{2x}$$
$$y(x)=\dfrac {x^2}2+\dfrac x2+C_1e^{2x}+C_2$$
The constant $\dfrac 14$ is abosorbed by $C_2$.
