Example 4. Find the general solution of the equation

$$
y^{\prime \prime}+10 y^{\prime}+25 y=4 e^{-5 x}
$$
Solution. The characteristic equation $\lambda^{2}+10 \lambda+25=0$ has a double root $\lambda_{1}=\lambda_{2}=-5$, therefore

$$
y_{0.0}=\left(C_{1}+C_{2} x\right) e^{-5 x}
$$

Since $\alpha=-5$ is a root of the characteristic equation of multiplicity $s=2$, we seek the particular solution $y_{\text {p }}$ of the non-homogeneous equation in the form (see Table 1, case II (2))

$$
y_{4 . \mathrm{H}}=B x^{2} e^{-5 x}
$$

then

$$
\begin{aligned}
& y_{4, \mathrm{H}}^{\prime}=B\left(2 x-5 x^{2}\right) e^{-5 x} \\
& y_{4, \mathrm{H}}^{\prime \prime}=B\left(2-20 x+25 x^{2}\right) e^{-5 x}
\end{aligned}
$$

Substituting the expressions for $y_{\text {p }}, y_{4 . . \mathrm{H}}^{\prime}, y_{\text {p }}^{\prime \prime}$ into the original equation, we get $2 B e^{-5 x}=4 e^{-5 x}$, from which $B=2$ and, therefore, $y_{\text {p }}=2 x^{2} e^{-5 x}$. The general solution of the equation is

$$
y(x)=\left(C_{1}+C_{2} x\right) e^{-5 x}+2 x^{2} e^{-5 x}
$$