1. **State the problem:** We are given the function $$y = Ae^{2x} + Bxe^{2x}$$ where $A$ and $B$ are arbitrary constants. The goal is to eliminate the arbitrary constants $A$ and $B$ to find a differential equation satisfied by $y$. 2. **Recall the formula and rules:** Since $y$ is a linear combination of $e^{2x}$ and $xe^{2x}$, which are solutions to a second-order linear differential equation, we can find this equation by differentiating $y$ and eliminating $A$ and $B$. 3. **Calculate the first derivative:** $$y = Ae^{2x} + Bxe^{2x}$$ Using the product rule on $Bxe^{2x}$: $$y' = 2Ae^{2x} + B(e^{2x} + 2xe^{2x}) = 2Ae^{2x} + Be^{2x} + 2Bxe^{2x}$$ 4. **Calculate the second derivative:** Differentiate $y'$: $$y'' = 4Ae^{2x} + B(2e^{2x} + 2e^{2x} + 4xe^{2x}) = 4Ae^{2x} + 4Be^{2x} + 4Bxe^{2x}$$ 5. **Rewrite $y$, $y'$, and $y''$ in terms of $e^{2x}$ and $xe^{2x}$:** $$y = Ae^{2x} + Bxe^{2x}$$ $$y' = 2Ae^{2x} + Be^{2x} + 2Bxe^{2x}$$ $$y'' = 4Ae^{2x} + 4Be^{2x} + 4Bxe^{2x}$$ 6. **Form a system to eliminate $A$ and $B$:** Express $y$, $y'$, and $y''$ as: $$y = Ae^{2x} + Bxe^{2x}$$ $$y' = 2Ae^{2x} + Be^{2x} + 2Bxe^{2x}$$ $$y'' = 4Ae^{2x} + 4Be^{2x} + 4Bxe^{2x}$$ 7. **Use the fact that $y$, $y'$, and $y''$ are linear combinations of $e^{2x}$ and $xe^{2x}$:** We can write: $$y'' - 4y' + 4y = 0$$ Check this: $$y'' - 4y' + 4y = (4Ae^{2x} + 4Be^{2x} + 4Bxe^{2x}) - 4(2Ae^{2x} + Be^{2x} + 2Bxe^{2x}) + 4(Ae^{2x} + Bxe^{2x})$$ Simplify: $$= 4Ae^{2x} + 4Be^{2x} + 4Bxe^{2x} - 8Ae^{2x} - 4Be^{2x} - 8Bxe^{2x} + 4Ae^{2x} + 4Bxe^{2x}$$ $$= (4A - 8A + 4A)e^{2x} + (4B - 4B + 0) e^{2x} + (4B - 8B + 4B) xe^{2x} = 0$$ All terms cancel, confirming the differential equation. **Final answer:** $$\boxed{y'' - 4y' + 4y = 0}$$ This is the differential equation satisfied by the function $y = Ae^{2x} + Bxe^{2x}$, eliminating the arbitrary constants $A$ and $B$.