In the Previous Problem, Using the Notation $u _ { i j } = u ( x , \mathrm { t } )$

Multiple Choice

In the previous problem, using the notation $u _ { i j } = u ( x , \mathrm { t } )$ , and letting $\lambda = c k / h$ , the equation becomes

A) $u _ { i , j - 1 } = \lambda ^ { 2 } u _ { i + 1 , j } + 2 \left( 1 + \lambda ^ { 2 } \right) u _ { i j } + \lambda ^ { 2 } u _ { i - 1 , j } + u _ { i , j - 1 }$
B) $u _ { i , j - 1 } = \lambda u _ { i + 1 , j } + 2 ( 1 - \lambda ) u _ { i j } + \lambda u _ { i - 1 , j } + u _ { i , j - 1 }$
C) $u _ { i , j + 1 } = \lambda ^ { 2 } u _ { i + 1 , j } + 2 \left( 1 + \lambda ^ { 2 } \right) u _ { i j } + \lambda ^ { 2 } u _ { i - 1 , j } - u _ { i , j - 1 }$
D) $u _ { i , j + 1 } = \lambda ^ { 2 } u _ { i + 1 , j } + 2 \left( 1 - \lambda ^ { 2 } \right) u _ { i j } + \lambda ^ { 2 } u _ { i - 1 , j } - u _ { i , j - 1 }$
E) $u _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 - \lambda ) u _ { i j } + \lambda u _ { i - 1 , j } - u _ { i , j - 1 }$