1. **State the problem:** Solve the differential equation $$d^4y/dx^4 + d^3y/dx^3 = 1 - e^{-x}$$. 2. **Rewrite the equation:** Let us denote derivatives as $$y^{(n)} = \frac{d^n y}{dx^n}$$. The equation becomes: $$y^{(4)} + y^{(3)} = 1 - e^{-x}$$. 3. **Solve the homogeneous equation:** $$y^{(4)} + y^{(3)} = 0$$. The characteristic equation is: $$r^4 + r^3 = 0$$ Factor out $$r^3$$: $$r^3(r + 1) = 0$$ So the roots are: $$r = 0$$ (with multiplicity 3) and $$r = -1$$. 4. **Write the homogeneous solution:** $$y_h = C_1 + C_2 x + C_3 x^2 + C_4 e^{-x}$$ where $$C_1, C_2, C_3, C_4$$ are constants. 5. **Find a particular solution $$y_p$$:** The right side is $$1 - e^{-x}$$. Try a particular solution of the form: $$y_p = A x^4 + B x^3 e^{-x}$$ 6. **Compute derivatives of $$y_p$$:** Calculate $$y_p^{(3)}$$ and $$y_p^{(4)}$$ and substitute into the left side. 7. **Substitute and equate coefficients:** After substitution and simplification, solve for $$A$$ and $$B$$ to satisfy: $$y_p^{(4)} + y_p^{(3)} = 1 - e^{-x}$$. 8. **Solution for $$A$$:** Since the constant term on the right is 1, and the polynomial part of the left must match it, we find: $$A = \frac{1}{24}$$ 9. **Solution for $$B$$:** Matching the $$e^{-x}$$ terms gives: $$B = -1$$ 10. **Write the particular solution:** $$y_p = \frac{x^4}{24} - x^3 e^{-x}$$ 11. **Write the general solution:** $$y = y_h + y_p = C_1 + C_2 x + C_3 x^2 + C_4 e^{-x} + \frac{x^4}{24} - x^3 e^{-x}$$ **Final answer:** $$\boxed{y = C_1 + C_2 x + C_3 x^2 + C_4 e^{-x} + \frac{x^4}{24} - x^3 e^{-x}}$$