If [tex]y' = e^y \sin(x)[/tex] and [tex]y(-\pi)=0[/tex], separate variables in the differential equation to get
[tex]e^{-y} \, dy = \sin(x) \, dx[/tex]
Integrate both sides:
[tex]\displaystyle \int e^{-y} \, dy = \int \sin(x) \, dx \implies -e^{-y} = -\cos(x) + C[/tex]
Use the initial condition to solve for [tex]C[/tex] :
[tex]-e^{-0} = -\cos(-\pi) + C \implies -1 = 1 + C \implies C = -2[/tex]
Then the particular solution to the initial value problem is
[tex]-e^{-y} = -\cos(x) - 2 \implies e^{-y} = \cos(x) + 2[/tex]
(A)
