What particular solution should I guess for $y'' - 4y' + 4y = (6x - 2)e^{2x}$? I have the following initial value problem
$$ y'' - 4y' + 4y = (6x-2)e^{2x},\ y(0)=2,\ y'(0)=1 $$
to which I have obtained the following solution for the homogenous ODE
$$ y_{h} = e^{2x}(2 - 3x) $$
Now I am trying to figure out the particular solution but am lost as to what I should guess. My initial thought was to simply guess $y_{p} = Ce^{2x}$ but this quickly leads to nothing. I then guessed $y_{p} = (Ax+B)e^{2x}$ and it too lead to nothing. How do I continue?

EDIT: I have solved the problem with the help of the comments, see my solution down below.

After advice from the comments I managed to solve the problem.
\begin{split}
y'' - 4y' + 4y &= 2(3x-1)e^{2x} \\
y(0) &= 2\\
y'(0) &= 1
\end{split}
$$ y'' - 4y' + 4y = 0 $$
$y_{h} = e^{rx}$, $y'_{h} = re^{rx}$, $y''_{h} = r^{2}e^{rx}$.
\begin{split}
r^{2}e^{rx} - 4re^{rx} + 4e^{rx} &= 0 \\
e^{rx}(r^{2} - 4r + 4) &= 0
\end{split}
Solving for the roots of the characteristic equation
$$ r^{2} - 4r + 4 = 0 $$
yields a double root $r=2$. General solution becomes
$$ y_{h} = C_{1}xe^{2x} + C_{2}e^{2x} $$

Now for the particular solution I make the assumption
$$y_{p} = e^{2x}(Ax^{3} + Bx^{2}) $$
Finding expressions for $y'_{p}$ and$y''_{p}$ and then substituting them into the ODE we obtain
\begin{split}
2e^{2x}(Ax(2x^{2}+6x+3) + B(2x^{2}+4x+1)) + 4e^{2x}(Ax^{3}+Bx^{2}) - 4e^{2x}x(Ax(2x+3)+2B(x+1)) &= (6x-2)e^{2x} \\
e^{2x}(4Ax^{3} + 12Ax^{2} + 6Ax + 4Bx^{2} + 8Bx + 2B + 4Ax^{3} + 4Bx^{2}) - 4e^{2x}x(2Ax^{2}+3Ax + 2Bx + 2B) &= (6x-2)e^{2x} \\
e^{2x}(4Ax^{3} + 12Ax^{2} + 6Ax + 4Bx^{2} + 8Bx + 2B + 4Ax^{3} + 4Bx^{2} - 8Ax^{3} - 12Ax^{2} - 8Bx^{2} - 8Bx) &= (6x-2)e^{2x} \\
e^{2x}( 6A + 2B ) &= (6x-2)e^{2x} \\
\end{split}
\begin{split}
6A = 6 &\implies A = 1 \\
2B = -2 &\implies B = -1
\end{split}
$$ y_{p} = e^{2x}(x^{3} - x^{2}) $$

Now for the final solution we get
$$ y = y_{h} + y_{p} = C_{1}xe^{2x} + C_{2}e^{2x} + e^{2x}(x^{3} - x^{2})$$
What is left is to determine the coefficients $C_{1}$ and $C_{2}$ which is easily done by applying our initial values.
\begin{split}
y &= C_{1}xe^{2x} + C_{2}e^{2x} + e^{2x}(x^{3} - x^{2}) \\
y' &= e^{2x}(2C_{1}x + C_{1} + 2C_{2} + x(2x^{2} + x -2))
\end{split}
Solving for the coefficients yields $C_{1} = -3$ and $C_{2} = 2$ which gives us the final complete solution
$$ y(x) = e^{2x}(x^{3} - x^{2} -3x + 2) $$

Edit: This way of solving this initial value problem seems highly tedious and annoying, is there any better faster way to do it?
