Question

solve the equation in exact differentials

(2-9xy²)xdx+(4y²-6x³)ydy=0

Answer 1

Let us solve the equation in exact differentials $(2-9xy^2)xdx+(4y^2-6x^3)ydy=0.$

It follows that there exists the function $u=u(x,y)$ such that

$\frac{\partial u}{\partial x}=(2-9xy^2)x=2x-9x^2y^2, \ \frac{\partial u}{\partial y}=(4y^2-6x^3)y=4y^3-6x^3y.$

Therefore, $u=x^2-3x^3y^2+c(y),$ and hence

$\frac{\partial u}{\partial y}=-6x^3y+c'(y)=4y^3-6x^3y.$

It follows that $c'(y)=4y^3,$ and thus $c(y)=y^4+C.$

We conclude that the general solution of the differential equation is

$x^2-3x^3y^2+y^4=C.$