1. **State the problem:** Solve the differential equation $$y''' - 2y'' - y' + 2y = 4e^{3x} + 6x$$. 2. **Find the complementary solution (homogeneous equation):** Solve $$y''' - 2y'' - y' + 2y = 0$$. 3. **Characteristic equation:** $$r^3 - 2r^2 - r + 2 = 0$$. 4. **Factor the characteristic polynomial:** Try possible roots using Rational Root Theorem: test $r=1$: $$1 - 2 - 1 + 2 = 0$$ so $r=1$ is a root. 5. Divide polynomial by $(r-1)$: $$r^3 - 2r^2 - r + 2 = (r-1)(r^2 - r - 2)$$ 6. Factor quadratic: $$r^2 - r - 2 = (r-2)(r+1)$$ 7. Roots are $r=1, 2, -1$. 8. **Complementary solution:** $$y_c = C_1 e^{x} + C_2 e^{2x} + C_3 e^{-x}$$ 9. **Find particular solution $y_p$:** The right side is $4e^{3x} + 6x$. 10. For $4e^{3x}$, try $y_{p1} = Ae^{3x}$. 11. For $6x$, try polynomial $y_{p2} = Bx + C$. 12. Compute derivatives for $y_{p1}$: $$y_{p1} = Ae^{3x}, y'_{p1} = 3Ae^{3x}, y''_{p1} = 9Ae^{3x}, y'''_{p1} = 27Ae^{3x}$$ 13. Substitute $y_{p1}$ into left side: $$27Ae^{3x} - 2(9Ae^{3x}) - 3Ae^{3x} + 2Ae^{3x} = (27 - 18 - 3 + 2)Ae^{3x} = 8Ae^{3x}$$ 14. Set equal to $4e^{3x}$: $$8A e^{3x} = 4 e^{3x} \\ 8A = 4 \\ A = \frac{1}{2}$$ 15. Compute derivatives for $y_{p2} = Bx + C$: $$y'_{p2} = B, y''_{p2} = 0, y'''_{p2} = 0$$ 16. Substitute $y_{p2}$ into left side: $$0 - 2(0) - B + 2(Bx + C) = -B + 2Bx + 2C$$ 17. Set equal to $6x$: $$-B + 2Bx + 2C = 6x$$ 18. Equate coefficients: For $x$: $2B = 6 \Rightarrow B = 3$ Constant term: $-B + 2C = 0 \Rightarrow -3 + 2C = 0 \Rightarrow 2C = 3 \Rightarrow C = \frac{3}{2}$ 19. **Particular solution:** $$y_p = \frac{1}{2} e^{3x} + 3x + \frac{3}{2}$$ 20. **General solution:** $$y = y_c + y_p = C_1 e^{x} + C_2 e^{2x} + C_3 e^{-x} + \frac{1}{2} e^{3x} + 3x + \frac{3}{2}$$