Example 3. Find the general solution of the equation $y^{\prime \prime}+y^{\prime}=4 x^{2} e^{x}$.
Solution. The characteristic equation $\lambda^{2}+\lambda=0$ has roots $\lambda_{1}=0$, $\lambda_{2}=-1$. Therefore, the general solution $y_{0.0}$ of the corresponding homogeneous equation is

$$
y_{0.0}=C_{1}+C_{2} e^{-x}
$$

Since $\alpha=1$ is not a root of the characteristic equation, the particular solution $y_{\text {p.n }}$ of the non-homogeneous equation is of the form (see Table 1, Case II (1))

$$
y_{\text {p.n }}=\left(A_{1} x^{2}+A_{2} x+A_{3}\right) e^{x}
$$

Substituting it into the original equation and canceling $e^{x}$ from both sides, we get

$$
2 A_{1} x^{2}+\left(6 A_{1}+2 A_{2}\right) x+2 A_{1}+3 A_{2}+2 A_{3}=4 x^{2}
$$

Equating the coefficients of like powers of $x$ on both sides of the equation, we obtain a system of linear equations for finding the coefficients $A_{1}, A_{2}, A_{3}$:

$$
\left\{\begin{aligned}
2 A_{1} & =4 \\
6 A_{1}+2 A_{2} & =0 \\
2 A_{1}+3 A_{2}+2 A_{3} & =0
\end{aligned}\right.
$$

Solving this system, we find $A_{1}=2, A_{2}=-6, A_{3}=7$, so that

$$
y_{4, n}=\left(2 x^{2}-6 x+7\right) e^{x}
$$

The general solution of the given equation is

$$
y(x)=C_{1}+C_{2} e^{-x}+\left(2 x^{2}-6 x+7\right) e^{x}
$$