Using the Value Of $y _ { n + 1 } ^ { * }$

question 31

Multiple Choice

Using the value of $y _ { n + 1 } ^ { * }$ from the previous problem, the Adams-Moulton corrector value for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

• A) \(y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } - 19 y _ { n } ^ { \prime } + 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)\)
• B) \(y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } + 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 34 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)\)
• C) \(y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } - 5 y _ { n - 1 } ^ { \prime } - y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)\)
• D) \(y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } - 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)\)
• E) none of the above

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