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Using the Notation from the Text, the Finite Difference Equation y+P(x)y+Q(x)y=f(x),y(a)=α,y(b)=βy ^ { \prime \prime } + P ( x ) y ^ { \prime } + Q ( x ) y = f ( x ) , y ( a ) = \alpha , y ( b ) = \beta

Question 36

Multiple Choice

Using the notation from the text, the finite difference equation for solving the boundary value problem y+P(x) y+Q(x) y=f(x) ,y(a) =α,y(b) =βy ^ { \prime \prime } + P ( x ) y ^ { \prime } + Q ( x ) y = f ( x ) , y ( a ) = \alpha , y ( b ) = \beta is


A) (1hPi/2) yi+1+(2+h2Qi) yi+(1+hPi/2) yi1=h2fi\left( 1 - h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 + h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }
B) (1hPi/2) yi+1+(2+h2Qi) yi+(1hPi/2) yi1=h2fi\left( 1 - h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }
C) (1+hPi/2) yi+1+(2+h2Qi) yi+(1+hPi/2) yi1=h2fi\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 + h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }
D) (1+hPi/2) yi+1+(2h2Qi) yi+(1hPi/2) yi1=h2fi\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( 2 - h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }
E) (1+hPi/2) yi+1+(2+h2Qi) yi+(1hPi/2) yi1=h2fi\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }

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