Nonhomogenous Linear Equation

Solve the following differential equation:

$\displaystyle y^{''}+y=e^x+x^3$

so starting off with the auxiliary equation I have:

$\displaystyle r^2+1=0$ where the only available solution is $\displaystyle \pm i$

so I would have:

$\displaystyle y_c=c_1 \cos(x)+c_2 \sin(x)$

working on the particular solution I let $\displaystyle y_{p1}(x) = Ae^x$

substituting for $\displaystyle y_{p1}$ I come up $\displaystyle Ae^x+Ae^x=e^x \Rightarrow 2Ae^x=e^x \Rightarrow A=\frac{1}{2} e^x$

(this is the part that I get stuck on) Since $\displaystyle x^3$ is a polynomial of third degree I would have:

$\displaystyle y_{p2}=Bx^3+Cx^2+Dx+E$

where $\displaystyle y_{p2}^{''} = 6Bx+C$

substituting I get:

$\displaystyle (6Bx+C)+ (Bx^3+Cx^2+Dx+E) =x^3$, then

$\displaystyle (6B+C)x^3+ (Bx^3+Cx^2+Dx+E) =x^3$

so $\displaystyle (6B+C)=1$ and $\displaystyle (B+C+D+E)=0$ where I don't think that I'm approaching it correctly since i have 4 unknowns.

any help would be greatly appreciated.