Question

solve: y^-1 * dy+y e^cosx * sinx * dx=0

Answer 1

\[y = -\frac{1}{e^{\cos x}}+ C\]

Answer 2

big thanks (:

Answer 3

dividing by dx on both sides of the equations
y^-1*(dy/dx)+ye^cosx*sinx=0
=> y^-1(dy/dx) = -y e^cosx*sinx
=> dy/dx= -y^2 * e^cosx*sinx