Answer
The equation
$$
(y / x+6 x) d x+(\ln x-2) d y=0, \quad x>0
$$
is exact, and its solutions are given implicitly by
$$
y \ln x+3x^{2} -2y =c
$$
where $ c $ is an arbitrary constant
Work Step by Step
$$
(y / x+6 x) d x+(\ln x-2) d y=0, \quad x>0
\quad \quad (i)
$$
Comparing this Equation with the differential form:
$$
M(x,y) d x+N(x,y) d y=0
$$
we observe that
$$
\begin{aligned}M(x,y) &= (y / x+6 x)
\\ N(x,y) &=(\ln x-2) \end{aligned}
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
M_{y}(x, y)=\frac{1}{x}=N_{x}(x, y)
$$
so the given equation is exact. Thus there is a $\psi (x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=M(x,y) = (y / x+6 x) ,
\\ \psi_{y}(x, y) &=N(x,y) =(\ln x-2) \end{aligned} \quad \quad (ii)
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\begin{aligned} \psi(x, y) &=\int{ (y / x+6 x) } dx,
\\ \ & =y \ln x+3x^{2} +h(y) \end{aligned} \quad \quad (iii)
$$
where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (ii), we compute $\psi_{y}(x, y) $ from Eq. (iii) and set it equal to $N$, obtaining
$$
\psi_{y}(x, y)= \ln x+h^{\prime}(y)=\ln x-2
$$
Thus $h^{\prime}(y)=-2 $ and $h(y)=-2y $ The constant of integration can be omitted since any solution of the preceding differential equation is satisfactory.
Now substituting for h(y) in Eq. (iii) gives
$$
\psi(x, y)=y \ln x+3x^{2} -2y
$$
Hence solutions of Eq. (i) are given implicitly by
$$
\psi(x, y) =y \ln x+3x^{2} -2y =c
$$
where $ c $ is an arbitrary constant