solve the equation in exact differentials
(2-9xy2)xdx+(4y2-6x3)ydy=0
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."
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