Example 2 Given $a_{1}=1, a_{n}=\frac{2}{3} a_{n-1}+n^{2}-15(n \geqslant 2)$, find $a_{n}$.
Solve by introducing undetermined constants $a, b, c$ such that
$$
a_{n}+\left(a n^{2}+b n+c\right)=\frac{2}{3}\left\{a_{n-1}+\left[a(n-1)^{2}+b(n-1)+c\right]\right\},
$$
after rearranging, we have
$$
a_{n}=\frac{2}{3} a_{n-1}+\left(-\frac{1}{3} a\right) n^{2}+\left(-\frac{4}{3} a-\frac{1}{3} b\right) n+\frac{2}{3} a-\frac{2}{3} b-\frac{1}{3} c \text {, and compare with } a_{n}=\frac{2}{3} a_{n-1}+
$$
$$
n^{2}-15
$$
by comparing coefficients, we get
$$
\left\{\begin{array} { l } 
{ - \frac { 1 } { 3 } a = 1 , } \\
{ - \frac { 4 } { 3 } a - \frac { 1 } { 3 } b = 0 , } \\
{ \frac { 2 } { 3 } a - \frac { 2 } { 3 } b - \frac { 1 } { 3 } c = - 1 5 , }
\end{array} \text { so } \left\{\begin{array}{l}
a=-3, \\
b=12, \\
c=15 .
\end{array}\right.\right.
$$

Thus, $a_{n}-3 n^{2}+12 n+15=\frac{2}{3}\left[a_{n-1}-3(n-1)^{2}+12(n-1)+15\right]$.
It is clear that $b_{n}=a_{n}-3 n^{2}+12 n+15$ is a geometric sequence with the first term $b_{1}=a_{1}-3+12+15=25$ and common ratio $\frac{2}{3}$, so $b_{n}=25 \cdot\left(\frac{2}{3}\right)^{n-1}$, from which we can derive
$$
a_{n}=25\left(\frac{2}{3}\right)^{n-1}+3 n^{2}-12 n-15 .
$$