1. **State the problem:** Solve the differential equation $$y'' - 4y' - 5y = 0$$ using the method of undetermined coefficients. 2. **Identify the type of equation:** This is a linear homogeneous differential equation with constant coefficients. 3. **Characteristic equation:** Assume a solution of the form $$y = e^{rx}$$. Substitute into the differential equation: $$r^2 e^{rx} - 4r e^{rx} - 5 e^{rx} = 0$$ Divide both sides by $$e^{rx}$$ (which is never zero): $$r^2 - 4r - 5 = 0$$ 4. **Solve the characteristic equation:** Use the quadratic formula: $$r = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-5)}}{2 \times 1} = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2}$$ So, $$r = \frac{4 \pm 6}{2}$$ 5. **Find roots:** - $$r_1 = \frac{4 + 6}{2} = 5$$ - $$r_2 = \frac{4 - 6}{2} = -1$$ 6. **General solution:** Since roots are real and distinct, the general solution is: $$y = C_1 e^{5x} + C_2 e^{-x}$$ where $$C_1$$ and $$C_2$$ are arbitrary constants determined by initial conditions. **Final answer:** $$y = C_1 e^{5x} + C_2 e^{-x}$$