this is separable!!
#dy/dx=cosx \ e^(y+sinx)#
split the exponent...so
#dy/dx=cosx \ e^y \ e^sinx#
#e^-y \ dy/dx=cosx \ e^sinx#
and this is separated, so next...integrate each side wrt x
#int \ e^-y \ dy/dx \ dx = int \ cosx \ e^sinx \ dx#
#int \ e^-y \ dy = int \ cosx \ e^sinx \ dx#
#- e^-y = color(red)( int \ cosx \ e^sinx \ dx) qquad triangle#
we can do the red RHS in a number of ways, the more laboured involving some kind of sub: #u = sin x#
but before we do that, just note that
#d/dx e^(f(x)) = f'(x) e^(f(x))# and therefore
#int \ d/dx e^(f(x)) \ dx = int \ f'(x) e^(f(x)) \dx#
OR
# e^(f(x)) = int \ f'(x) e^(f(x)) \dx + C#
if you see that, brilliant. the job is done!
if not :-((
well, then we go with #triangle# but with a sub as mentioned. #u = sin x, du = cos x \ dx, color{blue}{u' = cos x}# so #triangle# becomes
#-e^-y = int \ u' \ e^u \ (1)/(u') \ du= int \ e^u \ du#
#implies -e^-y = e^u + C = e^(sin x) + C#
So, to tidy it all up....
#e^-y = C - e^(sin x) #
#ln( e^-y) =ln ( C - e^(sin x) )#
#-y =ln ( C - e^(sin x) )#
#y =ln (1/( C - e^(sin x) ))#
so #y_o = 0# means that
#0 =ln (1/( C - 1 ))# so #C = 2#
#y =ln (1/( 2 - e^(sin x) ))#
