1. **State the problem:** Find the differential equation of the family of curves given by $$y = Ae^{2x} + Be^{-2x}$$ where $A$ and $B$ are arbitrary constants. 2. **Differentiate the given function:** Calculate the first derivative: $$y' = \frac{d}{dx}\left(Ae^{2x} + Be^{-2x}\right) = 2Ae^{2x} - 2Be^{-2x}$$ 3. **Differentiate again to find the second derivative:** $$y'' = \frac{d}{dx}(y') = \frac{d}{dx}\left(2Ae^{2x} - 2Be^{-2x}\right) = 4Ae^{2x} + 4Be^{-2x}$$ 4. **Express $y''$ in terms of $y$:** Recall that $$y = Ae^{2x} + Be^{-2x}$$ Multiply $y$ by 4: $$4y = 4Ae^{2x} + 4Be^{-2x}$$ Notice that this equals $y''$: $$y'' = 4y$$ 5. **Write the differential equation:** Rearranging gives: $$y'' - 4y = 0$$ **Final answer:** The differential equation of the family of curves is $$y'' - 4y = 0$$.