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.
 A: 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?
A: The "guessing" method does have a theory behind it and here it goes. First, your equation is:
$$(D-2)^2[y] = 6xe^{2x} - 2e^{2x},$$
where $D = \dfrac{d}{dx}.$
Now, you know to solve a homogeneous version of this ODE, so we should aim to transform your equation to one you can immediately solve. To do this, ask yourself the question - what differential operators annihilate the RHS? For $2e^{2x}$, it's easy since we know $D(e^{2x}) = 2e^{2x}\implies (D-2)[2e^{2x}] = 0.$ For $xe^{2x},$ it's the same idea:
$$D[xe^{2x}] = e^{2x} + 2xe^{2x}\implies (D-2)[xe^{2x}] = e^{2x}\implies (D-2)^2[xe^{2x}] = 0.$$
Therefore, if $y$ solves your inhomogeneous ODE, then we got:
$$(D-2)^4[y] = 0\implies y_p = e^{2x}(a+bx+cx^2+dx^3).$$
This guess is guaranteed to work but you can forget the constant and linear term, because they will be annihilated by the given ODE anyway. So this is how, you know for sure the right guess is:
$$y_p = (cx^2+dx^3)e^{2x}.$$
A: Hint:
$$y'' - 4y' + 4y = (6x-2)e^{2x}$$
$$ y(0)=2, y'(0)=1$$
Rewrite the DE as :
$$(ye^{-2x})'' = 6x-2$$
$$u'' = 6x-2$$
$$ u(0)=2,\ u'(0)=-3$$
Then the particular solution is easier to find.
