An exact first-degree first-order ODE is one of the form
$$
A(x, y) d x+B(x, y) d y=0 \quad \text { and for which } \quad \frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}
$$
In this case $A(x, y) d x+B(x, y) d y$ is an exact differential, $d U(x, y)$ . In other words
$$
A d x+B d y=d U=\frac{\partial U}{\partial x} d x+\frac{\partial U}{\partial y} d y
$$
from which we obtain
$$
\begin{aligned}
&A(x, y)=\frac{\partial U}{\partial x} \
&B(x, y)=\frac{\partial U}{\partial y}
\end{aligned}
$$
Since $\partial^{2} U / \partial x \partial y=\partial^{2} U / \partial y \partial x$ we therefore require
$$
\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}
$$
PHS1002 COURSE NOTES ：
Substituting $y=v x$ we obtain
$$
v+x \frac{d v}{d x}=v+\tan v .
$$
Cancelling $v$ on both sides, rearranging and integrating gives
$$
\int \cot v d v=\int \frac{d x}{x}=\ln x+c_{1}
$$
But
$$
\int \cot v d v=\int \frac{\cos v}{\sin v} d v=\ln (\sin v)+c_{2}
$$
so the solution to the $\mathrm{ODE}$ is $y=x \sin ^{-1} A x$, where $A$ is a constant.