1. The problem is to eliminate the arbitrary constants $A$ and $B$ from the function $$y = Ax^2 + Be^{2x}$$. 2. This function is a linear combination of two independent functions: $x^2$ and $e^{2x}$. 3. To eliminate $A$ and $B$, we differentiate $y$ twice to get a differential equation that $y$ satisfies. 4. First derivative: $$y' = \frac{d}{dx}(Ax^2) + \frac{d}{dx}(Be^{2x}) = 2Ax + 2Be^{2x}$$ 5. Second derivative: $$y'' = \frac{d}{dx}(2Ax) + \frac{d}{dx}(2Be^{2x}) = 2A + 4Be^{2x}$$ 6. Now, express $A$ and $B e^{2x}$ from $y$, $y'$, and $y''$: From $y$: $$y = Ax^2 + Be^{2x}$$ From $y'$: $$y' = 2Ax + 2Be^{2x}$$ From $y''$: $$y'' = 2A + 4Be^{2x}$$ 7. Multiply $y$ by 2: $$2y = 2Ax^2 + 2Be^{2x}$$ 8. Multiply $y'$ by $-x$: $$-x y' = -2Ax^2 - 2x Be^{2x}$$ 9. Add the results of steps 7 and 8: $$2y - x y' = 2Ax^2 + 2Be^{2x} - 2Ax^2 - 2x Be^{2x} = 2Be^{2x} - 2x Be^{2x} = 2Be^{2x}(1 - x)$$ 10. Multiply $y''$ by $(x - 1)$: $$ (x - 1) y'' = (x - 1)(2A + 4Be^{2x}) = 2A(x - 1) + 4Be^{2x}(x - 1)$$ 11. Subtract step 9 from step 10: $$ (x - 1) y'' - (2y - x y') = 2A(x - 1) + 4Be^{2x}(x - 1) - 2Be^{2x}(1 - x)$$ Note that $4Be^{2x}(x - 1) - 2Be^{2x}(1 - x) = 4Be^{2x}(x - 1) + 2Be^{2x}(x - 1) = 6Be^{2x}(x - 1)$ 12. So, $$ (x - 1) y'' - (2y - x y') = 2A(x - 1) + 6Be^{2x}(x - 1) = (x - 1)(2A + 6Be^{2x})$$ 13. For this to be zero for all $x$, the term in parentheses must be zero: $$2A + 6Be^{2x} = 0$$ 14. Since $A$ and $B$ are arbitrary constants, the only way to eliminate them is to find a differential equation satisfied by $y$. 15. Alternatively, we can find a linear differential equation satisfied by $y$ by combining derivatives: 16. From step 4 and 5, note that: $$y'' - 2 y' = 2A + 4Be^{2x} - 2(2Ax + 2Be^{2x}) = 2A + 4Be^{2x} - 4Ax - 4Be^{2x} = 2A - 4Ax = 2A(1 - 2x)$$ 17. Similarly, from $y$ and $y'$: $$y' - 2x y = 2Ax + 2Be^{2x} - 2x (Ax^2 + Be^{2x}) = 2Ax + 2Be^{2x} - 2Ax^3 - 2x Be^{2x} = 2Ax (1 - x^2) + 2Be^{2x} (1 - x)$$ 18. The key is to find a differential equation involving $y$, $y'$, and $y''$ that eliminates $A$ and $B$. 19. The function $y = Ax^2 + Be^{2x}$ satisfies the differential equation: $$y'' - 4 y' + 4 y = 0$$ 20. Verify: $$y'' = 2A + 4Be^{2x}$$ $$4 y' = 4(2Ax + 2Be^{2x}) = 8Ax + 8Be^{2x}$$ $$4 y = 4(Ax^2 + Be^{2x}) = 4Ax^2 + 4Be^{2x}$$ 21. Substitute into the equation: $$y'' - 4 y' + 4 y = (2A + 4Be^{2x}) - (8Ax + 8Be^{2x}) + (4Ax^2 + 4Be^{2x})$$ $$= 2A + 4Be^{2x} - 8Ax - 8Be^{2x} + 4Ax^2 + 4Be^{2x}$$ $$= 2A - 8Ax + 4Ax^2 + (4Be^{2x} - 8Be^{2x} + 4Be^{2x})$$ $$= 2A - 8Ax + 4Ax^2 + 0 = 2A - 8Ax + 4Ax^2$$ 22. This is not zero for all $x$ unless $A=0$, so this is not the correct differential equation. 23. Instead, note that $x^2$ satisfies $y'''=0$ and $e^{2x}$ satisfies $y'' - 4 y' + 4 y=0$. 24. Since $y$ is a linear combination of $x^2$ and $e^{2x}$, the differential equation satisfied by $y$ is the product of the differential operators: $$(D^3)(D^2 - 4D + 4) y = 0$$ where $D = \frac{d}{dx}$. 25. Therefore, the minimal differential equation eliminating $A$ and $B$ is: $$y^{(5)} - 4 y^{(4)} + 4 y^{(3)} = 0$$ 26. This is the differential equation that eliminates the arbitrary constants $A$ and $B$ from the original function. **Final answer:** $$y^{(5)} - 4 y^{(4)} + 4 y^{(3)} = 0$$