Differential equation: $Ay'' + By' + Cy = h(x)$ I'm stuck solving the equation $y'' - 3y' + 2y = 2x^3-30$.
The auxiliary equation is $k^2 - 3k + 2 = 0$ where $k_1 = 1, k_2=3$. Thus the general solution is: $$y_g = C_1e^x + C_2e^{3x}$$
Then, I tried to find the particular solution taking into consideration that $h(x) = Q_n(x) \cdot e^{\alpha x}$ where $\alpha$ is zero and doesn't equal one of auxiliary equation's roots, and $n$ is the order of $h(x)$ and equals 3. So, I get the equations as:
$$ y_p = Ax^3 + Bx^2 +Cx + D, \\ 
y'_p = 3Ax^2 + 2Bx + C, \\
y''_p = 6Ax + 2B$$
Having substituted they in the initial equation I get the system:
$$ \begin{cases} 
A = 2, \\
B + 3A = 0, \\
C + 2B + 6A = 0, \\
D + C + 2B = -30
\end{cases} \implies 
\begin{cases} 
A = 2, \\
B = -6, \\
C = 0, \\
D = -18
\end{cases}
$$
Thus my particular solution is $y_p = 2x^3 - 6x^2 - 18$ but the correct answer is: $$y_p = x^3 + \frac{9}{2}x^2 + \frac{21}{2}x - \frac{15}{4}$$
Where was I wrong?

The equations you should have are: $2A=2 \\ -9A+2B=0 \\  6A-6B+2C=0 \\ 2B-3C+2D=-30 \\ \text{ Solving this system should give you your solution. }$
