Example 4 Factorize:
$$
x^{2}+x y-6 y^{2}+x+13 y-6 \text{. }
$$
(10th Jinyun Cup Junior High School Mathematics Invitational Competition)
Solution: Since the original expression is a quadratic in two variables and can only be factored into the product of two linear expressions in two variables, considering
$$
x^{2}+x y-6 y^{2}=(x+3 y)(x-2 y) \text {, }
$$

we can assume
$$
\begin{array}{l} 
x^{2}+x y-6 y^{2}+x+13 y-6 \\
=(x+3 y+a)(x-2 y+b) \\
= x^{2}+x y-6 y^{2}+(a+b) x+(3 b- \\
2 a) y+a b .
\end{array}
$$

By comparing the coefficients of like terms on both sides of the identity, we get
$$
\left\{\begin{array}{l}
a+b=1, \\
3 b-2 a=13, \\
a_{u}^{\prime}=-6 .
\end{array}\right.
$$

Solving (1) and (2), we get $a=-2, b=3$, substituting into (3), it fits.
$$
\begin{array}{r}
\therefore x^{2}+x y-6 y^{2}+x+13 y-6 \\
\quad=(x+3 y-2)(x-2 y+3) .
\end{array}
$$