1. **State the problem:** We have a discrete-time system described by the difference equation $$y(k+2) + 5y(k+1) + 6y(k) = U(k)$$ with input $$u(k) = 1$$ for $$k \geq 0$$ and initial conditions $$y(0) = 0$$, $$y(1) = 0$$. We want to find the output $$y(k)$$. 2. **Identify the type of equation:** This is a non-homogeneous linear difference equation with constant coefficients. 3. **Solve the homogeneous equation:** First solve $$y(k+2) + 5y(k+1) + 6y(k) = 0$$. The characteristic equation is $$r^2 + 5r + 6 = 0$$. 4. **Find roots:** Factor or use quadratic formula: $$r^2 + 5r + 6 = (r+2)(r+3) = 0$$ So, $$r = -2$$ or $$r = -3$$. 5. **General solution of homogeneous equation:** $$y_h(k) = C_1(-2)^k + C_2(-3)^k$$ 6. **Find particular solution:** Since $$u(k) = 1$$ (a constant), try a constant particular solution $$y_p(k) = A$$. Substitute into the difference equation: $$A + 5A + 6A = 1$$ $$12A = 1$$ $$A = \frac{1}{12}$$ 7. **General solution:** $$y(k) = y_h(k) + y_p(k) = C_1(-2)^k + C_2(-3)^k + \frac{1}{12}$$ 8. **Apply initial conditions:** - For $$k=0$$: $$y(0) = C_1(-2)^0 + C_2(-3)^0 + \frac{1}{12} = C_1 + C_2 + \frac{1}{12} = 0$$ - For $$k=1$$: $$y(1) = C_1(-2)^1 + C_2(-3)^1 + \frac{1}{12} = -2C_1 - 3C_2 + \frac{1}{12} = 0$$ 9. **Solve system:** From $$k=0$$: $$C_1 + C_2 = -\frac{1}{12}$$ From $$k=1$$: $$-2C_1 - 3C_2 = -\frac{1}{12}$$ Multiply first equation by 2: $$2C_1 + 2C_2 = -\frac{1}{6}$$ Add to second equation: $$(-2C_1 - 3C_2) + (2C_1 + 2C_2) = -\frac{1}{12} - \frac{1}{6}$$ $$-C_2 = -\frac{1}{12} - \frac{2}{12} = -\frac{3}{12} = -\frac{1}{4}$$ $$C_2 = \frac{1}{4}$$ Substitute back: $$C_1 + \frac{1}{4} = -\frac{1}{12}$$ $$C_1 = -\frac{1}{12} - \frac{1}{4} = -\frac{1}{12} - \frac{3}{12} = -\frac{4}{12} = -\frac{1}{3}$$ 10. **Final solution:** $$y(k) = -\frac{1}{3}(-2)^k + \frac{1}{4}(-3)^k + \frac{1}{12}$$ This expression gives the output $$y(k)$$ for $$k \geq 0$$ with the given input and initial conditions.