2. In $\triangle A B C$, it is known that the three interior angles $\angle A$, $\angle B$, $\angle C$ are opposite to the sides $a$, $b$, $c$ respectively, and satisfy $a \sin A \cdot \sin B + b \cos ^{2} A = \sqrt{2} a$. Then $\frac{b}{a}=$ . $\qquad$
2. $\sqrt{2}$.

From the given and the Law of Sines, we have
$$
\begin{array}{l}
\sqrt{2} a=a \sin A \cdot \sin B+b\left(1-\sin ^{2} A\right) \\
=b+\sin A(a \sin B-b \sin A)=b . \\
\text { Therefore, } \frac{b}{a}=\sqrt{2} .
\end{array}
$$