Example 2 Find the range of the function $y=\sqrt{2 x-3}-\sqrt{x-2}$.
The domain of the function is $[2,+\infty)$, so $y>0$. From the second conclusion above, we get
$$\begin{array}{l}
y^{2}=(\sqrt{2 x-3}-\sqrt{x-2})^{2}=\left(\sqrt{2} \cdot \sqrt{x-\frac{3}{2}}-1 \cdot \sqrt{x-2}\right)^{2} \\
\geqslant(2-1)\left[\left(x-\frac{3}{2}\right)-(x-2)\right]=\frac{1}{2}
\end{array}$$

That is, $y^{2} \geqslant \frac{1}{2}$ (when and only when $\sqrt{2} \cdot \sqrt{x-2}=\sqrt{1} \cdot \sqrt{x-\frac{3}{2}}$, i.e., $x=\frac{5}{2}$, $y^{2}=\frac{1}{2}$), and since $y>0$, we have $y \geqslant \frac{\sqrt{2}}{2}$, so the range is $\left[\frac{\sqrt{2}}{2},+\infty\right)$.