u_(n+1) = u_n +2u_(n−1) +2u_0 u_0 =1 ⇒u


Question and Answers Forum

All Questions Topic List
Logic Questions
Previous in All Question Next in All Question
Previous in Logic Next in Logic
Question Number 130782 by EDWIN88 last updated on 28/Jan/21

$u_{n + 1} = u_{n} + 2 u_{n - 1} + 2 u_{0}$ $u_{0} = 1 \Rightarrow u_{n} = ?$
Commented by prakash jain last updated on 29/Jan/21

$u_{n + 1} = u_{n} + 2 u_{n - 1} + 2$ $u_{n + 1} - u_{n} + 2 u_{n - 1} = 2$ $For homegeneous solution we$ $can take$ $x^{n + 1} - x^{n} - 2 x^{n - 1} = 0$ $x^{2} - x - 2 = 0$ $u_{2} = u_{1} + 4 u_{0}$
Commented by prakash jain last updated on 29/Jan/21

$u_{n + 1} = u_{n} + 2 u_{n - 1} + 2$ $u_{n + 1} - u_{2} - 2 u_{n - 1} = 2$ $Homigenous solution$ $x^{2} - x - 2 = 0 \Rightarrow x = - 1, 2$ $u_{n} = c_{1} {(- 1)}^{n} + c_{2} 2^{n}$ $General Solution$ $u_{n} = c_{1} {(- 1)}^{n} + c_{2} 2^{n} - 1$ $you need to know at least two$ $values to determine sequence$
Commented by EDWIN88 last updated on 29/Jan/21

$t h a n k y o u$
	Answered by mr W last updated on 29/Jan/21

	$s a y u_{1} = 2, u_{0} = 1$ $u_{n + 1} = u_{n} + 2 u_{n - 1} + 2$ $l e t u_{n} = a_{n} + c$ $a_{n + 1} + c = a_{n} + c + 2 (a_{n - 1} + c) + 2$ $a_{n + 1} = a_{n} + 2 a_{n - 1} + 2 c + 2$ $l e t 2 c + 2 = 0, i . e . c = - 1$ $\Rightarrow u_{n} = a_{n} - 1$ $\Rightarrow a_{n + 1} - a_{n} - 2 a_{n - 1} = 0$ $x^{2} - x - 2 = 0$ $(x + 1) (x - 2) = 0$ $\Rightarrow x = - 1, 2$ $a_{n} = A {(- 1)}^{n} + B 2^{n}$ $a_{0} = A + B = u_{0} + 1 = 1 + 1 = 2$ $a_{1} = - A + 2 B = u_{1} + 1 = 2 + 1 = 3$ $\Rightarrow B = \frac{5}{3}, A = \frac{1}{3}$ $\Rightarrow u_{n} = \frac{1}{3} [5 \times 2^{n} + {(- 1)}^{n}] - 1$
	Commented by EDWIN88 last updated on 29/Jan/21

	$t h a n k y o u$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum