2. The equation
$$
6 y^{4}-35 y^{3}+62 y^{2}-35 y+6=0
$$

has ( ) integer solutions.
(A) 1
(B) 2
(C) 3
(D) 4
2. B.

From the given, we have
$$
\begin{array}{l}
6\left(y+\frac{1}{y}\right)^{2}-35\left(y+\frac{1}{y}\right)+50=0 \\
\Rightarrow y+\frac{1}{y}=2+\frac{1}{2} \text { or } y+\frac{1}{y}=3+\frac{1}{3} \\
\Rightarrow y_{1}=2, y_{2}=\frac{1}{2}, y_{3}=3, y_{4}=\frac{1}{3} .
\end{array}
$$