1. **State the problem:** Solve the differential equation $$y''' - 2y'' - 4y' + 8y = 6xe^{2x}$$ using the method of undetermined coefficients. 2. **Find the complementary solution ($y_c$):** Solve the homogeneous equation $$y''' - 2y'' - 4y' + 8y = 0$$. The characteristic equation is: $$r^3 - 2r^2 - 4r + 8 = 0$$ 3. **Solve the characteristic equation:** Try possible roots using Rational Root Theorem. Test $r=2$: $$2^3 - 2(2^2) - 4(2) + 8 = 8 - 8 - 8 + 8 = 0$$ So, $r=2$ is a root. Divide polynomial by $(r-2)$: $$r^3 - 2r^2 - 4r + 8 = (r-2)(r^2 - 4)$$ Factor $r^2 - 4$: $$(r-2)(r-2)(r+2) = (r-2)^2 (r+2)$$ 4. **Write complementary solution:** Since $r=2$ is a repeated root of multiplicity 2 and $r=-2$ is a simple root, $$y_c = (C_1 + C_2 x)e^{2x} + C_3 e^{-2x}$$ 5. **Find particular solution ($y_p$):** The right side is $6xe^{2x}$. Since $e^{2x}$ is part of the complementary solution with multiplicity 2, multiply trial by $x^2$. Try: $$y_p = x^2 (A x + B) e^{2x} = (A x^3 + B x^2) e^{2x}$$ 6. **Compute derivatives of $y_p$:** Let $u = A x^3 + B x^2$, then $$y_p = u e^{2x}$$ First derivative: $$y_p' = (u' + 2u) e^{2x}$$ Second derivative: $$y_p'' = (u'' + 4u' + 4u) e^{2x}$$ Third derivative: $$y_p''' = (u''' + 6u'' + 12u' + 8u) e^{2x}$$ Calculate derivatives of $u$: $$u = A x^3 + B x^2$$ $$u' = 3A x^2 + 2B x$$ $$u'' = 6A x + 2B$$ $$u''' = 6A$$ 7. **Substitute into left side:** $$y''' - 2y'' - 4y' + 8y = (u''' + 6u'' + 12u' + 8u) e^{2x} - 2(u'' + 4u' + 4u) e^{2x} - 4(u' + 2u) e^{2x} + 8 u e^{2x}$$ Simplify coefficients: $$= [u''' + 6u'' + 12u' + 8u - 2u'' - 8u' - 8u - 4u' - 8u + 8u] e^{2x}$$ Combine like terms: $$= [u''' + (6u'' - 2u'') + (12u' - 8u' - 4u') + (8u - 8u - 8u + 8u)] e^{2x}$$ $$= [u''' + 4u'' + 0 u' + 0] e^{2x} = (u''' + 4u'') e^{2x}$$ 8. **Plug in $u'''$ and $u''$:** $$u''' + 4u'' = 6A + 4(6A x + 2B) = 6A + 24A x + 8B = 24A x + (6A + 8B)$$ 9. **Set equal to right side:** $$24A x + (6A + 8B) = 6 x$$ Match coefficients: For $x$: $$24A = 6 \\ A = \frac{6}{24} = \frac{1}{4}$$ For constant term: $$6A + 8B = 0 \\ 6 \times \frac{1}{4} + 8B = 0 \\ \frac{3}{2} + 8B = 0 \\ 8B = -\frac{3}{2} \\ B = -\frac{3}{16}$$ 10. **Write particular solution:** $$y_p = (\frac{1}{4} x^3 - \frac{3}{16} x^2) e^{2x}$$ 11. **Write general solution:** $$y = y_c + y_p = (C_1 + C_2 x) e^{2x} + C_3 e^{-2x} + \left(\frac{1}{4} x^3 - \frac{3}{16} x^2\right) e^{2x}$$