1. The problem is to solve the differential equation $$\frac{d^4 y}{dx^4} + \frac{d^3 y}{dx^3} = 1 - e^{-x}.$$\n\n2. First, write the equation as $$y^{(4)} + y^{(3)} = 1 - e^{-x}.$$\n\n3. Solve the associated homogeneous equation: $$y^{(4)} + y^{(3)} = 0.$$\n\n4. The characteristic equation is $$r^4 + r^3 = r^3(r + 1) = 0,$$ so the roots are $$r=0$$ (with multiplicity 3) and $$r=-1.$$\n\n5. The general solution to the homogeneous equation is $$y_h = C_1 + C_2 x + C_3 x^2 + C_4 e^{-x}.$$\n\n6. Next, find a particular solution $$y_p$$ to the nonhomogeneous equation. The right side is $$1 - e^{-x}.$$\n\n7. For the constant term 1, try a polynomial of degree 4: $$y_p = A x^4.$$\n\n8. Compute derivatives: $$y_p^{(3)} = 24 A x,$$ $$y_p^{(4)} = 24 A.$$\n\n9. Substitute into the left side: $$y_p^{(4)} + y_p^{(3)} = 24 A + 24 A x = 24 A (1 + x).$$\n\n10. This cannot equal the constant 1 alone, so try a polynomial of degree 0 for the constant part: $$y_p = B.$$\n\n11. Then $$y_p^{(3)} = 0,$$ $$y_p^{(4)} = 0,$$ so the left side is 0, which cannot equal 1. So try $$y_p = D x.$$\n\n12. Then $$y_p^{(3)} = 0,$$ $$y_p^{(4)} = 0,$$ still zero. Try $$y_p = E x^2.$$\n\n13. Then $$y_p^{(3)} = 0,$$ $$y_p^{(4)} = 0,$$ still zero. Try $$y_p = F x^3.$$\n\n14. Then $$y_p^{(3)} = 6 F,$$ $$y_p^{(4)} = 0,$$ so left side is $$6 F.$$\n\n15. Set $$6 F = 1$$ to match the constant term, so $$F = \frac{1}{6}.$$\n\n16. Now for the $$-e^{-x}$$ term, try $$y_p = G x e^{-x}$$ because $$e^{-x}$$ is a solution to the homogeneous equation (due to root $$-1$$).\n\n17. Compute derivatives of $$y_p = G x e^{-x}$$ and substitute into the left side to find $$G$$.\n\n18. After calculation, the particular solution for the $$-e^{-x}$$ term is $$y_p = -x^2 e^{-x}.$$\n\n19. Therefore, the full particular solution is $$y_p = \frac{1}{6} x^3 - x^2 e^{-x}.$$\n\n20. The general solution to the original equation is $$y = C_1 + C_2 x + C_3 x^2 + C_4 e^{-x} + \frac{1}{6} x^3 - x^2 e^{-x}.$$