1. **State the problem:** Solve the differential equation $$y'' + y' - 2y = 5$$ 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:** $$y'' + y' - 2y = 0$$ The characteristic equation is $$r^2 + r - 2 = 0$$. 4. **Find roots:** Solve $$r^2 + r - 2 = 0$$ using the quadratic formula: $$r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}$$ So, $$r_1 = 1$$ and $$r_2 = -2$$. 5. **Write homogeneous solution:** $$y_h = C_1 e^{t} + C_2 e^{-2t}$$ where $$C_1, C_2$$ are constants. 6. **Find particular solution:** Since the right side is a constant (5), try a constant solution $$y_p = A$$. Substitute into the differential equation: $$0 + 0 - 2A = 5 \implies -2A = 5 \implies A = -\frac{5}{2}$$. 7. **Write general solution:** $$y = C_1 e^{t} + C_2 e^{-2t} - \frac{5}{2}$$. This is the complete solution using the method of undetermined coefficients.