Find the general solution of the difference equation.
-Given that the signal $y _ { k } = 3 k - 5$ is a solution of the given difference equation, find a description of all solutions of the equation. $y _ { k + 2 } - 6 y _ { k + 1 } + 5 y _ { k } = - 12$ for all $k$
A) $3 \mathrm { k } - 5 + \mathrm { c } _ { 1 } + \mathrm { c } _ { 2 } ( 6 ) ^ { \mathrm { k } }$
B) $3 \mathrm { k } - 5 + \mathrm { c } _ { 1 } ( - 1 ) ^ { \mathrm { k } } + \mathrm { c } _ { 2 } ( - 5 ) ^ { \mathrm { k } }$
C) $\mathrm { c } _ { 1 } + \mathrm { c } _ { 2 } ( 5 ) ^ { \mathrm { k } }$
D) $3 \mathrm { k } - 5 + \mathrm { c } _ { 1 } + \mathrm { c } _ { 2 } ( 5 ) ^ { \mathrm { k } }$

Question

Quizplus · Accepted Answer

The given signal is a solution to the difference equation, so it must satisfy the equation. Substituting the signal into the equation gives:$$(3(k+2) - 5) - 6(3(k+1) - 5) + 5(3k - 5) = -12$$Simplifying this equation gives:$$9k + 1 - 54k + 30 + 15k - 25 = -12$$$$-30k - 12 = -12$$$$-30k = 0$$$$k = 0$$Therefore, the given signal is a solution to the difference equation when k = 0.The general solution to the difference equation is:$$y_k = 3k - 5 + c_1 + c_2(6)^k$$where c1 and c2 are constants.

Find the General Solution of the Difference Equation yk=3k−5y _ { k } = 3 k - 5yk​=3k−5

Find the General Solution of the Difference Equation $y _ { k } = 3 k - 5$