1. **Problem statement:** Solve the difference equation $$y_{n+2} - 7y_{n+1} + 12y_n = 2^n$$ with initial conditions $$y_0 = 0$$ and $$y_1 = 0$$ using the Z-transform. 2. **Recall the Z-transform definition:** For a sequence $$y_n$$, the Z-transform is $$Y(z) = \sum_{n=0}^\infty y_n z^{-n}$$. 3. **Apply the Z-transform to each term:** - $$Z\{y_{n+2}\} = z^2 Y(z) - z^2 y_0 - z y_1$$ - $$Z\{y_{n+1}\} = z Y(z) - z y_0$$ - $$Z\{y_n\} = Y(z)$$ 4. **Apply the Z-transform to the right side:** - $$Z\{2^n\} = \frac{z}{z-2}$$ for $$|z| > 2$$. 5. **Substitute into the equation:** $$z^2 Y(z) - z^2 y_0 - z y_1 - 7(z Y(z) - z y_0) + 12 Y(z) = \frac{z}{z-2}$$ 6. **Plug in initial conditions $$y_0=0$$ and $$y_1=0$$:** $$z^2 Y(z) - 0 - 0 - 7(z Y(z) - 0) + 12 Y(z) = \frac{z}{z-2}$$ 7. **Simplify:** $$z^2 Y(z) - 7 z Y(z) + 12 Y(z) = \frac{z}{z-2}$$ 8. **Factor out $$Y(z)$$:** $$Y(z)(z^2 - 7 z + 12) = \frac{z}{z-2}$$ 9. **Factor the quadratic:** $$z^2 - 7 z + 12 = (z-3)(z-4)$$ 10. **Solve for $$Y(z)$$:** $$Y(z) = \frac{z}{(z-2)(z-3)(z-4)}$$ 11. **Partial fraction decomposition:** Assume $$\frac{z}{(z-2)(z-3)(z-4)} = \frac{A}{z-2} + \frac{B}{z-3} + \frac{C}{z-4}$$ Multiply both sides by denominator: $$z = A(z-3)(z-4) + B(z-2)(z-4) + C(z-2)(z-3)$$ 12. **Find coefficients by plugging values:** - For $$z=2$$: $$2 = A(2-3)(2-4) = A(-1)(-2) = 2A \Rightarrow A=1$$ - For $$z=3$$: $$3 = B(3-2)(3-4) = B(1)(-1) = -B \Rightarrow B = -3$$ - For $$z=4$$: $$4 = C(4-2)(4-3) = C(2)(1) = 2C \Rightarrow C=2$$ 13. **Rewrite $$Y(z)$$:** $$Y(z) = \frac{1}{z-2} - \frac{3}{z-3} + \frac{2}{z-4}$$ 14. **Inverse Z-transform:** Recall $$Z^{-1}\left\{ \frac{1}{z-a} \right\} = a^n u_n$$ where $$u_n$$ is the unit step. So, $$y_n = 1 \cdot 2^n - 3 \cdot 3^n + 2 \cdot 4^n$$ 15. **Check initial conditions:** - $$y_0 = 1 - 3 + 2 = 0$$ - $$y_1 = 2 - 9 + 8 = 1$$ but initial condition is 0, so we must consider the initial conditions carefully. 16. **Adjust for initial conditions:** The solution above is the particular solution. The homogeneous solution is: $$y_n^h = K_1 3^n + K_2 4^n$$ General solution: $$y_n = y_n^p + y_n^h = A 2^n + B 3^n + C 4^n$$ From the partial fractions, particular solution is: $$y_n^p = 2^n$$ So, $$y_n = 2^n + K_1 3^n + K_2 4^n$$ 17. **Apply initial conditions:** - $$y_0 = 2^0 + K_1 3^0 + K_2 4^0 = 1 + K_1 + K_2 = 0$$ - $$y_1 = 2^1 + K_1 3^1 + K_2 4^1 = 2 + 3 K_1 + 4 K_2 = 0$$ 18. **Solve system:** From first: $$K_1 + K_2 = -1$$ From second: $$3 K_1 + 4 K_2 = -2$$ Multiply first by 3: $$3 K_1 + 3 K_2 = -3$$ Subtract from second: $$(3 K_1 + 4 K_2) - (3 K_1 + 3 K_2) = -2 - (-3) \Rightarrow K_2 = 1$$ Then, $$K_1 = -1 - K_2 = -1 - 1 = -2$$ 19. **Final solution:** $$y_n = 2^n - 2 \cdot 3^n + 1 \cdot 4^n = 2^n - 2 \cdot 3^n + 4^n$$