\begin{aligned} \text { Example 2-63 } a_{n}-3 a_{n-1}-10 a_{n-2}=28 \cdot 5^{n} \\ a_{0}=25, \quad a_{1}=120\end{aligned}
Let
$$
\begin{array}{l}
G(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots \\
D(x) \triangleq 1-3 x-10 x^{2} \\
D(x) G(x)=\left(1-3 x-10 x^{2}\right)\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots\right) \\
=a_{0}+\left(a_{1}-3 a_{0}\right) x+\left(a_{2}-3 a_{1}-10 a_{0}\right) x^{2} \\
+\left(a_{3}-3 a_{2}-10 a_{1}\right) x^{2}+\cdots \\
=25+(120-75) x+28 \cdot 5^{2} x^{2}+28 \cdot 5^{3} x^{3}+\cdots \\
=25+40 x+\frac{28 \cdot 5^{2} x^{2}}{1-5 x}=\frac{25-80 x+475 x^{2}}{1-5 x} \\
G(x)=\frac{25-80 x+475 x^{2}}{(1-5 x)^{2}(1+2 x)} \\
=\frac{A}{(1-5 x)^{2}}+\frac{B}{1-5 x}+\frac{C}{1+2 x} \\
\end{array}
$$

Multiplying both sides by $1+2 x$ gives
$$
\begin{array}{c}
\frac{25-80 x+475 x^{2}}{(1-5 x)^{2}}=\left[\frac{A}{(1-5 x)^{2}}+\frac{B}{1-5 x}\right](1+2 x)+C \\
C=\left.\frac{25-80 x+475 x^{2}}{(1-5 x)^{2}}\right|_{x=-\frac{1}{2}}=15
\end{array}
$$

Then we have
Similarly, we can find $A=20, B=-10$.
Thus,
$$
\begin{aligned}
G(x) & =\frac{20}{(1-5 x)^{2}}-\frac{10}{1-5 x}+\frac{15}{1+2 x} \\
& =\sum_{n=0}^{\infty}\left[20(n+1) \cdot 5^{n}-10(5)^{n}+20(-2)^{n}\right] x^{n} \\
a_{n} & =[20(n+1)-10] 5^{n}+20 \cdot(-2)^{n} \\
& =(20 n+10) \cdot 5^{n}+20(-2)^{n}
\end{aligned}
$$