Problems finding particular solution for $y''-4y'+4y=x+3e^{2x}$ 
Solve $$y''-4y'+4y=x+3e^{2x}.$$


I have found that $y_H(x)=Ae^{2x}+Bxe^{2x}$.
My problem is to propose a particular solution.
I have proposed $$y_P(x)=(C+C_1x)+C_2e^{2x}$$ but I end up cancelling the term $C_2e^{2x}$.
What particular solution do we have to propose?
 A: You can reach the gereral solution by integrating the given ODE repeatedly. Indeed, one has
\begin{align*}
& y'' - 4y' + 4 = x + 3e^{2x} \Longleftrightarrow (y' - 2y)' - (2y' - 4y) = x + 3e^{2x} \Longleftrightarrow\\\\
& w' - 2w = x + 3e^{2x} \Longleftrightarrow (we^{-2x})' = xe^{-2x} + 3 \Longleftrightarrow we^{-2x} = \int xe^{-2x}\mathrm{d}x + 3x + c \Longleftrightarrow\\\\
& we^{-2x} = -\frac{xe^{-2x}}{2} + \int \frac{e^{-2x}}{2}\mathrm{d}x + 3x + c = -\frac{xe^{-2x}}{2} - \frac{e^{-2x}}{4} + 3x + c \Longleftrightarrow\\\\
& w = -\frac{(2x+1)}{4} + (3x+c)e^{2x} \Longleftrightarrow y' - 2y = -\frac{(2x+1)}{4} + (3x+c)e^{2x}
\end{align*}
Can you take it from here?
A: For the particular solution you can try:
$$y_p=ax+b+cx^2e^{2x}$$
Or you can do this:
$$y''-4y'+4y=3e^{2x}.$$
$$e^{-2x}y''-4y'e^{-2x}+4ye^{-2x}=3$$
$$e^{-2x}y''-2y'e^{-2x}-2y'e^{-2x}+4ye^{-2x}=3$$
$$(e^{-2x}y')'-(2ye^{-2x})'=3$$
$$(e^{-2x}y)''=3$$
Integrate twice and you get the particular solution with the correct coefficients.
$$y=(\frac 3 2x^2+c_1x+c_2)e^{2x}$$
For the second particular solution
$$y′′−4y′+4y=x.$$
Try $y_p=ax+b$
$$\implies (a,b)=(\frac 1 4, \frac 1 4)$$
So finally:
$$y(x)=(\frac 3 2x^2+c_1x+c_2)e^{2x}+\frac 1 4x+\frac 1 4$$
A: ^^A general method**
First find two linearly independent solutions of the homogeneous part
$$y''-4y'+4y=0 \implies y_1=e^{2x}, y_2= x e^{2x}$$
The Wronskian $W[y_1,y_2]=e^{4x}.$
According to the method of variation of parameters $(C_1,C_2)$ the solution of
$$Y''-4Y'-4Y=f(x)= x+3e^{2x}$$ is
$$Y(x)=C_1(x) y_1(x) + C_2(x) y_2(x),~~~~(1)$$
were $$C_1(x)=-\int \frac{y_2(x) f(x)}{W(x)} dx +D_1, ~~ C_2(x)=\int \frac{y_1(x) f(x)}{W(x)} dx +D_2.$$
Here $$C_1(x)=-\int \frac{x e^{2x} (x+3e^{2x})}{e^{4x}} dx +D_1= -\int [x^2 e^{2x}+3x] +dx +D_1$$ $$\implies C_1(x) =(-1/4-x/2-x^2/2)e^{-2x}-3x^2/2 +D_1~~~~(2)$$
and $$C_2(x)= \int[xe^{-2x}+3] dx+D_2=-(1/4+x/2)e^{-2x}+3x+D_2~~~~(3)$$
By inseting (2) and (3) in (1), we get
$$Y(x)=(D_1+D_2 x) e^{2x}+(1/4+x/4+3x^2 e^{2x}/2)$$
