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}

The Solution Of yy′′=(y′)2y y ^ { \prime \prime } = \left( y ^ { \prime } \right) ^ { 2 }yy′′=(y′)2

Multiple Choice