Question

(secx)dy/dx=e^(y+sinx), please help me solve the differential equation. Thanks :)

Answer 1

What is that and were the abcds at

Answer 2

First, you must know these formula  d(e^f(x) = f'(x)e^x dx, e^a+b=e^a.e^b, and d(sinx) = cosxdx, secx = 1/ cosx

(secx)dy/dx=e^(y+sinx), implies  dy/dx=cosx .e^(y+sinx), and then 
dy=cosx .e^(y+sinx).dx, integdy=integ(cosx .e^(y+sinx).dx, equivalent of 
integdy=integ(cosx .e^y.e^sinx)dx, integdy=e^y.integ.(cosx e^sinx)dx, but we know that   d(e^sinx) =cosx e^sinx dx,
so integ.d(e^sinx) =integ.cosx e^sinx dx,
and e^sinx + C=integ.cosx e^sinxdx
 finally, integdy=e^y.integ.(cosx e^sinx)dx=e^2. (e^sinx) +C
the answer is 
y = e^2. (e^sinx) +C, you can check this answer to calculate dy/dx