1. **State the problem:** Find the general solution of the differential equation $$y''' - 4y'' - 5y' = 0$$. 2. **Identify the type of equation:** This is a linear homogeneous differential equation with constant coefficients. 3. **Write the characteristic equation:** Replace derivatives by powers of $r$: $$r^3 - 4r^2 - 5r = 0$$ 4. **Factor the characteristic equation:** Factor out $r$: $$r(r^2 - 4r - 5) = 0$$ 5. **Solve for roots:** - From $r=0$ - Solve quadratic $r^2 - 4r - 5 = 0$ using quadratic formula: $$r = \frac{4 \pm \sqrt{16 + 20}}{2} = \frac{4 \pm \sqrt{36}}{2} = \frac{4 \pm 6}{2}$$ So roots are: $$r = 5 \quad \text{and} \quad r = -1$$ 6. **Write the general solution:** Since roots are $r=0, 5, -1$ (all distinct real roots), the general solution is: $$y = C_1 e^{0x} + C_2 e^{5x} + C_3 e^{-x} = C_1 + C_2 e^{5x} + C_3 e^{-x}$$ where $C_1, C_2, C_3$ are arbitrary constants. **Final answer:** $$y = C_1 + C_2 e^{5x} + C_3 e^{-x}$$