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

Question 17

Multiple Choice

In the previous problem, using the notation uij=u(x,t) u _ { i j } = u ( x , \mathrm { t } ) , and letting λ=ck/h2\lambda = c k / h ^ { 2 } , the equation becomes


A) ui,j1=λui+1,j+(1+2λ) ui,j+λui1,ju _ { i , j - 1 } = \lambda u _ { i + 1 , j } + ( 1 + 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
B) ui,j1=λui+1,j+(12λ) ui,j+λui1,ju _ { i , j - 1 } = \lambda u _ { i + 1 , j } + ( 1 - 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
C) ui,j+1=λui+1,j+(1+2λ) ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 + 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
D) ui,j+1=λui+1,j+(12λ) ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 - 2 \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }
E) ui,j+1=λui+1,j+(1λ) ui,j+λui1,ju _ { i , j + 1 } = \lambda u _ { i + 1 , j } + ( 1 - \lambda ) u _ { i , j } + \lambda u _ { i - 1 , j }

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