Solved

Use Power Series to Solve the Differential Equation y′′+x2y=0,y(0)=6,y′(0)=0y ^ { \prime \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ { \prime } ( 0 ) = 0

Question 42

Multiple Choice

Use power series to solve the differential equation. y′′+x2y=0,y(0) =6,y′(0) =0y ^ { \prime \prime } + x ^ { 2 } y = 0 , y ( 0 ) = 6 , y ^ { \prime } ( 0 ) = 0


A) y(x) =6−∑n=1∞(−1) nx4n4n(4n−1) ⋅(4n−4) (4n−5) ⋯4⋅3y ( x ) = 6 - \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
B) y(x) =∑n=1∞(−1) n6x4n4n(4n−1) ⋅(4n−4) (4n−5) ⋯4⋅3y ( x ) = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
C) y(x) =6+∑n=1∞(−1) n6x4n4n(4n−1) ⋅(4n−4) (4n−5) ⋯4⋅3y ( x ) = 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { 6 x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
D) y(x) =−6+∑n=1∞(−1) nx4n4n(4n−1) ⋅(4n−4) (4n−5) ⋯4⋅3y ( x ) = - 6 + \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }
E) y(x) =∑n=1∞(−1) nx4n4n(4n−1) ⋅(4n−4) (4n−5) ⋯4⋅3y ( x ) = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } \frac { x ^ { 4 n } } { 4 n ( 4 n - 1 ) \cdot ( 4 n - 4 ) ( 4 n - 5 ) \cdots 4 \cdot 3 }

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