Using Power Series Methods, the Solution Of $2 x y ^ { \prime \prime } + y ^ { \prime } + 2 y = 0$

question 47

Multiple Choice

Using power series methods, the solution of $2 x y ^ { \prime \prime } + y ^ { \prime } + 2 y = 0$ is

• A) \[\begin{array} { l }
• y = c _ { 0 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) + \\
• c _ { 1 } x ^ { 1 / 2 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) )
• \end{array}\]
• B) \[\begin{array} { l }
• y = c _ { 0 } \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) + \\
• c _ { 1 } x ^ { 1 / 2 } \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]
• \end{array}\]
• C) \[\begin{array} { l }
• y = c _ { 0 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\
• c _ { 1 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]
• \end{array}\]
• D) \[\begin{array} { l }
• y = c _ { 0 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\
• c _ { 1 } x ^ { 1 / 2 } \left[ 1 + \sum _ { n = 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]
• \end{array}\]
• E) \[\begin{array} { l }
• y = \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 1 \cdot 3 \cdots ( 2 n - 1 ) ) ) \right] + \\
• x ^ { 1 / 2 } \left[ 1 + \sum _ { n - 1 } ^ { \infty } ( - 2 ) ^ { n } x ^ { n } / ( n ! ( 3 \cdot 5 \cdots ( 2 n + 1 ) ) ) \right]
• \end{array}\]