1. **State the problem:** Solve the differential equation $$y''' - 3y'' + 3y' - y = 4e^t$$ with initial conditions $$y(0) = 1$$, $$y'(0) = 1$$, and $$y''(0) = -1$$. 2. **Identify the type of equation:** This is a third-order linear nonhomogeneous differential equation with constant coefficients. 3. **Solve the homogeneous equation:** The associated homogeneous equation is $$y''' - 3y'' + 3y' - y = 0$$. 4. **Find the characteristic equation:** $$r^3 - 3r^2 + 3r - 1 = 0$$. 5. **Factor the characteristic polynomial:** Recognize this as a perfect cube: $$r^3 - 3r^2 + 3r - 1 = (r - 1)^3 = 0$$. 6. **Find the roots:** The root is $$r = 1$$ with multiplicity 3. 7. **Write the general solution to the homogeneous equation:** $$y_h = (C_1 + C_2 t + C_3 t^2) e^t$$. 8. **Find a particular solution:** Since the right side is $$4e^t$$ and $$e^t$$ is a solution to the homogeneous equation with multiplicity 3, multiply by $$t^3$$ to find a particular solution of the form: $$y_p = A t^3 e^t$$. 9. **Compute derivatives of $$y_p$$:** $$y_p = A t^3 e^t$$ $$y_p' = A e^t (3 t^2 + t^3)$$ $$y_p'' = A e^t (6 t + 6 t^2 + t^3)$$ $$y_p''' = A e^t (6 + 18 t + 9 t^2 + t^3)$$. 10. **Substitute into the left side of the differential equation:** $$y_p''' - 3 y_p'' + 3 y_p' - y_p = A e^t [ (6 + 18 t + 9 t^2 + t^3) - 3(6 t + 6 t^2 + t^3) + 3(3 t^2 + t^3) - t^3 ]$$ Simplify inside the bracket: $$6 + 18 t + 9 t^2 + t^3 - 18 t - 18 t^2 - 3 t^3 + 9 t^2 + 3 t^3 - t^3 = 6$$ So the expression reduces to: $$A e^t imes 6 = 6 A e^t$$. 11. **Set equal to the right side:** $$6 A e^t = 4 e^t$$ Divide both sides by $$e^t$$: $$6 A = 4$$ Solve for $$A$$: $$A = \frac{4}{6} = \frac{2}{3}$$. 12. **Write the general solution:** $$y = y_h + y_p = (C_1 + C_2 t + C_3 t^2) e^t + \frac{2}{3} t^3 e^t$$. 13. **Apply initial conditions:** Calculate $$y(0)$$: $$y(0) = (C_1 + 0 + 0) e^0 + 0 = C_1 = 1$$. Calculate $$y'(t)$$: $$y' = \frac{d}{dt} \left[ (C_1 + C_2 t + C_3 t^2) e^t + \frac{2}{3} t^3 e^t \right]$$ Use product rule: $$y' = \left(C_2 + 2 C_3 t \right) e^t + (C_1 + C_2 t + C_3 t^2) e^t + \frac{2}{3} (3 t^2 e^t + t^3 e^t)$$ Simplify: $$y' = (C_2 + 2 C_3 t) e^t + (C_1 + C_2 t + C_3 t^2) e^t + 2 t^2 e^t + \frac{2}{3} t^3 e^t$$ At $$t=0$$: $$y'(0) = (C_2 + 0) + C_1 + 0 + 0 + 0 = C_2 + C_1 = 1$$ Since $$C_1 = 1$$, then $$C_2 = 0$$. 14. **Calculate $$y''(t)$$:** $$y'' = \frac{d}{dt} y'$$ Derive each term carefully: $$y' = (C_2 + 2 C_3 t) e^t + (C_1 + C_2 t + C_3 t^2) e^t + 2 t^2 e^t + \frac{2}{3} t^3 e^t$$ Derivatives: - $$\frac{d}{dt}[(C_2 + 2 C_3 t) e^t] = 2 C_3 e^t + (C_2 + 2 C_3 t) e^t$$ - $$\frac{d}{dt}[(C_1 + C_2 t + C_3 t^2) e^t] = (C_2 + 2 C_3 t) e^t + (C_1 + C_2 t + C_3 t^2) e^t$$ - $$\frac{d}{dt}[2 t^2 e^t] = 4 t e^t + 2 t^2 e^t$$ - $$\frac{d}{dt}[\frac{2}{3} t^3 e^t] = 2 t^2 e^t + \frac{2}{3} t^3 e^t$$ Sum all: $$y'' = 2 C_3 e^t + (C_2 + 2 C_3 t) e^t + (C_2 + 2 C_3 t) e^t + (C_1 + C_2 t + C_3 t^2) e^t + 4 t e^t + 2 t^2 e^t + 2 t^2 e^t + \frac{2}{3} t^3 e^t$$ Group terms: $$y'' = e^t [2 C_3 + 2 C_2 + 4 C_3 t + C_1 + C_2 t + C_3 t^2 + 4 t + 4 t^2 + \frac{2}{3} t^3]$$ At $$t=0$$: $$y''(0) = 2 C_3 + 2 C_2 + C_1 = -1$$ Recall $$C_1 = 1$$ and $$C_2 = 0$$, so: $$2 C_3 + 0 + 1 = -1$$ $$2 C_3 = -2$$ $$C_3 = -1$$. 15. **Final solution:** $$y = (1 + 0 imes t - 1 imes t^2) e^t + \frac{2}{3} t^3 e^t = (1 - t^2) e^t + \frac{2}{3} t^3 e^t$$. **Answer:** $$\boxed{y = (1 - t^2) e^t + \frac{2}{3} t^3 e^t}$$