1. **State the problem:** Find the general solution to the differential equation $$y'' - 2y' + y = \frac{1}{6} e^{5x} \cos(3x).$$ 2. **Solve the homogeneous equation:** The associated homogeneous equation is $$y'' - 2y' + y = 0.$$ The characteristic equation is $$r^2 - 2r + 1 = 0,$$ which factors as $$(r-1)^2 = 0.$$ So, the repeated root is $$r=1.$$ 3. **Write the homogeneous solution:** For a repeated root $r=1$, the homogeneous solution is $$y_h = (C_1 + C_2 x) e^{x}.$$ 4. **Find a particular solution:** The right side is $$\frac{1}{6} e^{5x} \cos(3x).$$ We try a particular solution of the form $$y_p = e^{5x} (A \cos(3x) + B \sin(3x)).$$ 5. **Compute derivatives:** $$y_p' = e^{5x} \big((5A + 3B) \cos(3x) + (5B - 3A) \sin(3x)\big),$$ $$y_p'' = e^{5x} \big((16A + 30B) \cos(3x) + (16B - 30A) \sin(3x)\big).$$ 6. **Substitute into the left side:** $$y_p'' - 2y_p' + y_p = e^{5x} \big((16A + 30B) - 2(5A + 3B) + A\big) \cos(3x) + e^{5x} \big((16B - 30A) - 2(5B - 3A) + B\big) \sin(3x).$$ Simplify coefficients: Cosine term: $$16A + 30B - 10A - 6B + A = 7A + 24B,$$ Sine term: $$16B - 30A - 10B + 6A + B = 7B - 24A.$$ 7. **Set equal to right side:** $$7A + 24B = \frac{1}{6},$$ $$7B - 24A = 0.$$ 8. **Solve the system:** From the second equation, $$7B = 24A \Rightarrow B = \frac{24}{7} A.$$ Substitute into the first: $$7A + 24 \times \frac{24}{7} A = \frac{1}{6} \Rightarrow 7A + \frac{576}{7} A = \frac{1}{6}.$$ Multiply both sides by 7: $$49A + 576A = \frac{7}{6} \Rightarrow 625A = \frac{7}{6} \Rightarrow A = \frac{7}{3750}.$$ Then, $$B = \frac{24}{7} \times \frac{7}{3750} = \frac{24}{3750} = \frac{4}{625}.$$ 9. **Write the particular solution:** $$y_p = e^{5x} \left( \frac{7}{3750} \cos(3x) + \frac{4}{625} \sin(3x) \right).$$ 10. **Write the general solution:** $$y = y_h + y_p = (C_1 + C_2 x) e^{x} + e^{5x} \left( \frac{7}{3750} \cos(3x) + \frac{4}{625} \sin(3x) \right).$$