Solve the Initial-Value Problem $( 1 + \cos x ) \frac { d y } { d x } = \left( 6 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$

question 34

Multiple Choice

Solve the initial-value problem.
$( 1 + \cos x ) \frac { d y } { d x } = \left( 6 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$

• A) \(y = \frac { \ln ( 6 + \cos x ) } { \ln ( 6 - \cos x ) }\)
• B)
• \(y = \ln \left( \frac { 1 } { 6 } \left( \sec ^ { 12 } \left( \frac { x } { 2 } \right) - 1 \right) \right.\)
• C) \(y = \ln ( 6 - \cos x ) ( 6 + \cos x )\)
• D)
• \(y = \ln \left( \frac { 1 } { 6 } \left( 7 \sec ^ { 12 } \left( \frac { x } { 2 } \right) - 1 \right) \right.\)
• E)
• \(y = \ln \left( 6 \sec \left( \frac { x } { 2 } \right) - 1 \right)\)

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