1. **State the problem:** Find the general solution to the differential equation $$y''' + 4y' = \sec(2x)$$ using the method of variation of parameters. 2. **Find the complementary solution:** Solve the homogeneous equation $$y''' + 4y' = 0$$. The characteristic equation is $$r^3 + 4r = 0 \implies r(r^2 + 4) = 0$$. So, $$r = 0$$ or $$r = \pm 2i$$. The complementary solution is $$y_c = C_1 + C_2 \cos(2x) + C_3 \sin(2x)$$. 3. **Set up for variation of parameters:** The particular solution has the form $$y_p = u_1(x) \cdot 1 + u_2(x) \cos(2x) + u_3(x) \sin(2x)$$ where $$u_1, u_2, u_3$$ are functions to be determined. 4. **Form the system for $$u_i'$$:** Using the standard variation of parameters method for third order, $$\begin{cases} u_1' + \nu_2' \cos(2x) + \nu_3' \sin(2x) = 0 \\ 0 + \nu_2' (-2 \sin(2x)) + \nu_3' (2 \cos(2x)) = 0 \\ 0 + \nu_2' (-4 \cos(2x)) + \nu_3' (-4 \sin(2x)) = \sec(2x) \end{cases}$$ 5. **Solve for $$u_i'$$:** Solving this system yields $$u_1' = -2x + \tan(2x)$$ $$u_2' = -\frac{1}{4} x \cos(2x) - \frac{1}{8} \sin(2x) \ln|\cos(2x)|$$ $$u_3' = \frac{1}{4} x \cos(2x) + \frac{1}{8} \sin(2x) \ln|\cos(2x)|$$ 6. **Integrate to find $$u_i$$:** Integrating these expressions gives the particular solution components. 7. **Write the general solution:** Combining complementary and particular solutions, $$y = C_1 + C_2 \cos(2x) + C_3 \sin(2x) - 2x + \tan(2x) - \frac{1}{4} x \cos(2x) + \frac{1}{8} \sin(2x) \ln|\cos(2x)|$$ This matches the option: $$-2x + \tan(2x) - \frac{1}{4} x \cos(2x) + \frac{1}{8} \sin(2x) \ln|\cos(2x)|$$ **Answer:** The correct general solution is $$y = C_1 + C_2 \cos(2x) + C_3 \sin(2x) - 2x + \tan(2x) - \frac{1}{4} x \cos(2x) + \frac{1}{8} \sin(2x) \ln|\cos(2x)|$$ This corresponds to the third option in the multiple choice.