Solving $y''-4y=x^2 e^{2x}$. I want to solve the differential equation $$y''-4y = x^2e^{2x}$$
Clearly $y_1 = e^{2x}$ and $y_2 = e^{-2x}$ are linearly independent solutions of the homogeneous equation. I would propose $y = (a+bx)e^{2x}$ as a solution, but since it contains $ae^{2x}$, which is in the span of $y_1$ and $y_2$, it won't do. Then I propose $$y = (ax+bx^2)e^{2x} = axe^{2x} + bx^2e^{2x}$$
as a solution, and we procced to find $a$ and $b$. We have: $$\begin{align} y &= axe^{2x} + bx^2e^{2x}\\ y' &= ae^{2x} + (2a+2b)xe^{2x} + 2bx^2e^{2x}\\ y'' &=(4a+2b)e^{2x} + (4a+8b)xe^{2x} + 4bx^2e^{2x}\end{align}$$
I mean, I could be wrong here... but I don't think so, I triple checked it. Then, forcing it as a solution, I get: $$\left\{\begin{array}{l} 4a + 2b = 0 \\8b = 0 \\0 = 1 \end{array} \right.$$
Can someone explain to me what's going on wrong here? Thanks.
 A: With the help in the comments, I have solved the problem correctly, and checked it with Mathematica. I'll post my solution here, it might be useful to someone who might stumble upon this post. My mistake before was taking initially $(a+bx)e^{2x}$, and then adding a $x$. Since $x^2e^{2x}$ contains a second degree $x^2$, I should have started with a $(a+bx+cx^2)e^{2x}$, and from there, considered $y = (ax+bx^2+cx^3)e^{2x}$ as a possible solution. Now this will work. We have: 
$$\begin{align} y &= axe^{2x} + bx^2e^{2x}+cx^3e^{2x}\\ y' &= ae^{2x} + (2a+2b)xe^{2x} + (2b+3c)x^2e^{2x} + 2cx^3e^{2x}\\ y'' &=(4a+2b)e^{2x} + (4a+8b+6c)xe^{2x} + (4b+12c)x^2e^{2x} + 4cx^3e^{2x}\end{align}$$
Forcing $y$ as a solution, cancelling $e^{2x}$ in both sides, and equalling the coefficients, we obtain:
$$\left\{\begin{array}{l} 4a + 2b = 0 \\8b +6c= 0 \\12c = 1 \end{array} \right.$$
It is easy to check that $a = \frac{1}{32}$, $b = -\frac{1}{16}$ and $c = \frac{1}{12}$ do the job. With this, the general solution to the equation is: $$y = \left(c_1+\frac{x}{32}-\frac{x^2}{16} + \frac{x^3}{12}\right)e^{2x}  + c_2e^{-2x}, \qquad c_1,c_2 \in \Bbb R.$$
