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)=-\frac32x^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)=-\frac32x^2y^2+xy+C=C`

or

`F(x,y)=\boxed{-\frac32x^2y^2+xy=C}`