Question

Use the "mixed partials" check to see if the following differential equation is exact. If it is exact find a function F(x,y) whose level curves are solutions to the differential equation. (−3xy^2+y)dx + (−3x^2y+x)dy =0

Answer 1

We have

$(-3xy^2+y)_y=--6xy+1$

and

$(-3x^2y+x)_x=-6xy+1$

so the equation is indeed exact. So we want to find a function $F(x,y)=C$ such that

$F_x=-3xy^2+y$

$F_y=-3x^2y+x$

Integrating both sides of the first equation wrt $x$ gives

$F(x,y)=-\dfrac32x^2y^2+xy+f(y)$

Differentiating both sides wrt $y$ gives

$F_y=-3x^2y+x=-3x^2y+x+f_y\implies f_y=0\implies f(y)=C$

So we have

$F(x,y)=-\dfrac32x^2y^2+xy+C=C$

or

$F(x,y)=\boxed{-\dfrac32x^2y^2+xy=C}$