Question

dy/dx=-cos(xy)+xy sin(xy)/2y-x^2 sin(xy)

Answer 1

Given differential Equation is $\frac{dy}{dx} = \frac{-cos(xy)+xysin(xy)}{2y-x^2sin(xy)}$

It can be written as, $(cos(xy)-xysin(xy))dx +(2y-x^2sin(xy))dy = 0$

Comparing with $Mdx +Ndy = 0$

For exact differential equation,

$\frac{\partial N}{\partial x} =\frac{\partial M}{\partial y}$

Then

$\frac{\partial N}{\partial x} = -2xsin(xy) -x^2ycos(xy)$

$\frac{\partial M}{\partial y} = -2xsin(xy) -x^2ycos(xy)$

Hence equation is exact differential equation.

Integrating both sides,

$\int (cos(xy)-xysin(xy))dx +\int (2y-x^2sin(xy))dy = C$

$\frac{sin(xy)}{y} - \frac{sin(xy)}{y} + xcos(xy) + y^2 = C$

$xcos(xy) + y^2 = C$