Solve the initial-value problem. $( 1 + \cos x ) \frac { d y } { d x } = \left( 2 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$
A) $y = \ln ( 2 - \cos x ) ( 2 + \cos x )$
B) $y = \frac { ( 7 - \cos x ) } { ( 2 + \cos x ) }$
C) $y = \ln \frac { \cos x } { ( 2 + \cos x ) }$
D) $y = \frac { \ln ( 2 + \cos x ) } { \ln ( 2 - \cos x ) }$
E) $y = \ln ( 7 - \cos x ) - \ln ( 1 + \cos x )$

Question

Quizplus · Accepted Answer

The differential equation can be solved by separating variables and integrating both sides. Applying the initial condition $y(0) = 0$ helps determine the constant of integration, leading to the solution $y = \ln ( 7 - \cos x ) - \ln ( 1 + \cos x )$.

Solve the Initial-Value Problem (1+cos⁡x)dydx=(2+e−y)sin⁡x,y(0)=0( 1 + \cos x ) \frac { d y } { d x } = \left( 2 + e ^ { - y } \right) \sin x , y ( 0 ) = 0(1+cosx)dxdy​=(2+e−y)sinx,y(0)=0

Solve the Initial-Value Problem $( 1 + \cos x ) \frac { d y } { d x } = \left( 2 + e ^ { - y } \right) \sin x , y ( 0 ) = 0$