Question

e^x dx+(e^x coty +2ycscy)dy=0

Answer 1

The given differential equation is

$$e^xdx+(e^xcoty+2ycosecy)dy = 0.$$

Now if we multiply both sides by $siny , we \space get$

$e^xsiny \space dx + (e^xcoty *siny + 2y *cosecy*siny)\space dy =0\\or, e^xsiny \space dx + (e^xcosy + 2y) \space dy=0\\or, e^x siny \space dx +e^xcosy \space dy+ 2ydy = 0\\$

Now this is much more simplified form of the given differential equation and can be written as

$$d(e^xsiny)+ d(y^2) = 0$$

Now integrating botth sides we get,

$\int d(e^xsiny) + \int d(y^2) = C$ ( where C is a constant of integration)

$e^xsiny + y^2 = C$

$\therefore$ The required solution is

$$e^xsiny + y^2 = C$$