1. **State the problem:** Solve the differential equation $$y'' - 6y' + 3y = \cos x$$. 2. **Identify the type of equation:** This is a non-homogeneous linear second-order differential equation with constant coefficients. 3. **Solve the homogeneous equation:** $$y'' - 6y' + 3y = 0$$ The characteristic equation is: $$r^2 - 6r + 3 = 0$$ Use the quadratic formula: $$r = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 3}}{2} = \frac{6 \pm \sqrt{36 - 12}}{2} = \frac{6 \pm \sqrt{24}}{2} = \frac{6 \pm 2\sqrt{6}}{2} = 3 \pm \sqrt{6}$$ So the general solution to the homogeneous equation is: $$y_h = C_1 e^{(3 + \sqrt{6})x} + C_2 e^{(3 - \sqrt{6})x}$$ 4. **Find a particular solution $y_p$:** Since the right side is $\cos x$, try a particular solution of the form: $$y_p = A \cos x + B \sin x$$ Compute derivatives: $$y_p' = -A \sin x + B \cos x$$ $$y_p'' = -A \cos x - B \sin x$$ Substitute into the differential equation: $$y_p'' - 6 y_p' + 3 y_p = (-A \cos x - B \sin x) - 6(-A \sin x + B \cos x) + 3(A \cos x + B \sin x)$$ Simplify: $$= (-A + 3A) \cos x + (-B + 3B) \sin x + 6A \sin x - 6B \cos x$$ $$= (2A - 6B) \cos x + (2B + 6A) \sin x$$ Set equal to the right side $\cos x$: $$2A - 6B = 1$$ $$2B + 6A = 0$$ 5. **Solve the system:** From the second equation: $$2B = -6A \implies B = -3A$$ Substitute into the first: $$2A - 6(-3A) = 2A + 18A = 20A = 1 \implies A = \frac{1}{20}$$ Then: $$B = -3 \times \frac{1}{20} = -\frac{3}{20}$$ 6. **Write the general solution:** $$y = y_h + y_p = C_1 e^{(3 + \sqrt{6})x} + C_2 e^{(3 - \sqrt{6})x} + \frac{1}{20} \cos x - \frac{3}{20} \sin x$$ This is the complete solution to the differential equation.