1. **State the problem:** Solve the differential equation $$(D^3 + D^2 - 4D - 4)y = 3e^{-x} - 4x - 6$$ where $D$ denotes differentiation with respect to $x$. 2. **Rewrite the operator:** The differential operator is $$D^3 + D^2 - 4D - 4 = (D^2 - 4)(D + 1)$$ by factoring. 3. **Solve the homogeneous equation:** Set the right side to zero: $$ (D^3 + D^2 - 4D - 4)y = 0 $$ The characteristic equation is: $$ r^3 + r^2 - 4r - 4 = 0 $$ Factoring: $$ (r^2 - 4)(r + 1) = 0 $$ So roots are: $$ r = 2, -2, -1 $$ The homogeneous solution is: $$ y_h = C_1 e^{2x} + C_2 e^{-2x} + C_3 e^{-x} $$ 4. **Find a particular solution $y_p$:** The right side is: $$ 3e^{-x} - 4x - 6 $$ Since $e^{-x}$ is a solution to the homogeneous equation (root $r=-1$), multiply by $x$ to find a particular solution for that term: $$ y_{p1} = A x e^{-x} $$ For the polynomial $-4x - 6$, try a polynomial of degree 1: $$ y_{p2} = Bx + C $$ 5. **Apply the operator to $y_p = y_{p1} + y_{p2}$:** Calculate: $$ (D^3 + D^2 - 4D - 4)(A x e^{-x} + Bx + C) = 3e^{-x} - 4x - 6 $$ 6. **Compute derivatives:** - For $y_{p1} = A x e^{-x}$: $$ y_{p1}' = A(e^{-x} - x e^{-x}) = A(1 - x)e^{-x} $$ $$ y_{p1}'' = A(-e^{-x} - (1 - x)e^{-x}) = A(-1 - 1 + x)e^{-x} = A(x - 2)e^{-x} $$ $$ y_{p1}''' = A(e^{-x} + (x - 2)(-e^{-x})) = A(1 - x + 2)e^{-x} = A(3 - x)e^{-x} $$ - For $y_{p2} = Bx + C$: $$ y_{p2}' = B $$ $$ y_{p2}'' = 0 $$ $$ y_{p2}''' = 0 $$ 7. **Apply operator to $y_{p1}$:** $$ D^3 y_{p1} = A(3 - x)e^{-x} $$ $$ D^2 y_{p1} = A(x - 2)e^{-x} $$ $$ -4 D y_{p1} = -4 A(1 - x)e^{-x} = -4A e^{-x} + 4A x e^{-x} $$ $$ -4 y_{p1} = -4 A x e^{-x} $$ Sum: $$ A(3 - x)e^{-x} + A(x - 2)e^{-x} - 4A e^{-x} + 4A x e^{-x} - 4 A x e^{-x} $$ Simplify terms: $$ A(3 - x + x - 2 - 4) e^{-x} + (4A x - 4A x) e^{-x} = A(3 - 2 - 4) e^{-x} + 0 = A(-3) e^{-x} $$ 8. **Apply operator to $y_{p2}$:** $$ D^3 y_{p2} = 0 $$ $$ D^2 y_{p2} = 0 $$ $$ -4 D y_{p2} = -4 B $$ $$ -4 y_{p2} = -4 (B x + C) = -4 B x - 4 C $$ Sum: $$ -4 B - 4 B x - 4 C = -4 B x - 4 (B + C) $$ 9. **Combine results:** $$ (D^3 + D^2 - 4D - 4) y_p = -3 A e^{-x} - 4 B x - 4 (B + C) $$ Set equal to right side: $$ -3 A e^{-x} - 4 B x - 4 (B + C) = 3 e^{-x} - 4 x - 6 $$ 10. **Match coefficients:** - For $e^{-x}$: $$ -3 A = 3 \\ A = -1 $$ - For $x$: $$ -4 B = -4 \\ B = 1 $$ - For constant term: $$ -4 (B + C) = -6 \\ -4 (1 + C) = -6 \\ 1 + C = \frac{6}{4} = 1.5 \\ C = 0.5 $$ 11. **Write the particular solution:** $$ y_p = -x e^{-x} + x + 0.5 $$ 12. **Write the general solution:** $$ y = y_h + y_p = C_1 e^{2x} + C_2 e^{-2x} + C_3 e^{-x} - x e^{-x} + x + 0.5 $$