Question

3x^2-2y^2+(1-4xy)y'=0

Answer 1

Let us solve the differential equation $3x^2-2y^2+(1-4xy)y'=0$ which is equivalent to the equation $(3x^2-2y^2)dx+(1-4xy)dy=0.$ Since $\frac{\partial(3x^2-2y^2)}{\partial y}=-4y=\frac{\partial(1-4xy)}{\partial x},$ we conclude that this equation is exact, and hence $U_x=3x^2-2y^2, U_y=1-4xy$ for some function $U=U(x,y).$ It follows that $U=x^3-2y^2x+C(y),$ and hence $\frac{\partial U}{\partial y}=-4yx+C'(y).$ Therefore, $-4yx+C'(y)=1-4xy,$ and hence $C(y)=y+C.$

We conclude that the general solution of the differential equation is

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