1. **State the problem:** Solve the differential equation $$y'' + y' - 6y = 2x$$ using the method of undetermined coefficients. 2. **General approach:** The equation is a nonhomogeneous linear differential equation with constant coefficients. The general solution is $$y = y_h + y_p$$ where $$y_h$$ is the homogeneous solution and $$y_p$$ is a particular solution. 3. **Solve the homogeneous equation:** $$y'' + y' - 6y = 0$$ The characteristic equation is $$r^2 + r - 6 = 0$$. 4. **Find roots:** Solve $$r^2 + r - 6 = 0$$ using the quadratic formula: $$r = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$$ So, $$r_1 = 2$$ and $$r_2 = -3$$. 5. **Write homogeneous solution:** $$y_h = C_1 e^{2x} + C_2 e^{-3x}$$ where $$C_1, C_2$$ are constants. 6. **Find particular solution:** Since the right side is a polynomial $$2x$$, try a polynomial of degree 1: $$y_p = Ax + B$$. 7. **Compute derivatives:** $$y_p' = A$$ $$y_p'' = 0$$. 8. **Substitute into the original equation:** $$0 + A - 6(Ax + B) = 2x$$ Simplify: $$A - 6Ax - 6B = 2x$$ Group terms: $$(-6A)x + (A - 6B) = 2x + 0$$ 9. **Equate coefficients:** For $$x$$ terms: $$-6A = 2 \implies A = -\frac{1}{3}$$ For constants: $$A - 6B = 0 \implies -\frac{1}{3} - 6B = 0 \implies B = -\frac{1}{18}$$ 10. **Write particular solution:** $$y_p = -\frac{1}{3}x - \frac{1}{18}$$ 11. **Write general solution:** $$y = C_1 e^{2x} + C_2 e^{-3x} - \frac{1}{3}x - \frac{1}{18}$$ This is the complete solution to the differential equation.