1. **State the problem:** Find the differential $dy$ of the function $$y = \frac{x + 2}{2x - 3}.$$\n\n2. **Recall the formula:** For a function $y = \frac{u}{v}$, the differential $dy$ is given by the quotient rule:\n$$dy = \frac{v \, du - u \, dv}{v^2}$$\nwhere $du$ and $dv$ are differentials of $u$ and $v$ respectively.\n\n3. **Identify $u$ and $v$:**\n$$u = x + 2, \quad v = 2x - 3.$$\n\n4. **Find differentials $du$ and $dv$:**\nSince $du = d(x + 2) = dx$ and $dv = d(2x - 3) = 2 \, dx$.\n\n5. **Apply the quotient rule:**\n$$dy = \frac{(2x - 3) \, dx - (x + 2)(2 \, dx)}{(2x - 3)^2}.$$\n\n6. **Simplify the numerator:**\n$$ (2x - 3) \, dx - 2(x + 2) \, dx = \left(2x - 3 - 2x - 4\right) dx = (-7) \, dx.$$\n\n7. **Write the final expression for $dy$:**\n$$dy = \frac{-7 \, dx}{(2x - 3)^2}.$$\n\n**Answer:**\n$$\boxed{dy = \frac{-7}{(2x - 3)^2} \, dx}.$$