Solve the differential equation?

Question

Solve the differential equation e^x dx+(e^x cot(y)+2y csc(y))dy=0.

M=e^x

N=e^x cot(y)+2y csc(y)

My=0

Nx=e^x cot(y)

Obviously isn't exact.

What to do next?

Answer 1

Multiply your equation by siny. You will obtain:

            e^x sinydx + e^xcosy dy + d(y ²) = 0

or   

                              d(e^x siny) + d(y²) = 0   

Thus, your solution is 

                                   e^x siny + y² = const     (y is not equal to zero)