Question

0. a) Find the solution of the difference equation yk+2 − 4yk+1 + 4yk = ;0 k = ,1,0 K. Also find
the particular solution when 1 y0 = and 6 y1 = .

Answer 1

Answer on Question #60835 – Math – Discrete Mathematics

Question

a) Find the solution of the difference equation $yk + 2 - 4yk + 1 + 4yk = 0$; $0k = K, 1, 0$. Also find the particular solution when $y0 = 1$ and $y1 = 6$.

Solution

The auxiliary equation for the homogeneous difference equation

$$y_{k+2} - 4y_{k+1} + 4y_k = 0$$

is

$$r^2 - 4r + 4 = 0 \quad \text{or} \quad (r - 2)^2 = 0.$$

Therefore, the general solution of the difference equation (1) is given by

$$y_k = A2^k + Bk2^k.$$

To find the particular solution of (1) we should find constants A and B using the conditions

$$y_0 = 1, \quad y_1 = 6.$$

So

$$\left\{ \begin{array}{l} A * 2^0 + B * 0 * 2^0 = 1 \\ A * 2^1 + B * 1 * 2^1 = 6 \end{array} \right. \rightarrow A = 1, \quad B = 2.$$

Thus, the particular solution is

$$y_k = 2^k + 2 * k2^k = 2^k(2k + 1).$$

Answer: $y_k = A2^k + Bk2^k$; $y_k = 2^k(2k + 1)$.

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