A Question on First Order Exact Differential Equations. As I learned , an exact differential equation is given by the following form
$A(x,y)+B(x,y)\frac{dy}{dx}=0$
Following an algebric manipulation it is possible to reach the following :
$A(x,y)dx+B(x,y)dy=0$
The “test” to be sure it it is an exact differential equation is to check the following equality:
$\frac{\partial A}{\partial y}=\frac{\partial B}{\partial x}$
Then to find the solution I need to find a “scalar potential” function U
such that :
$\frac{\partial U}{\partial x}=A,\;\frac{\partial U}{\partial y}=B,$
Then the solutions are $U=C_{(const)}$
(implicit function)
Few things I have trouble to understand.

*

*As I know about finding a scalar potential of a vector field using
$-\vec{\nabla}F=U$
how does that make sense here ? I will appreciate if you can explain in terms of gradient if possible this is the best way for me to understand.


*Say I have a function $U=x^{5}y+7xy^{3}$
, then the differential of $U$
will be $dU=(5x^{4}y+7y^{3})dx+(x^{5}+21xy^{2})dy$
How this differential is different from the Gradient of the function? I feel that it has connection to the topic of exact differential equations, how does this can help me ?
Thanks for the help, I will appreciate if you could explain this to me as simple as possible.
 A: From the chain rule, we have that the differential of a potential field is its gradient dotted with the differential displacement:
$$
\begin{align}
\mathrm{d}U
&= \frac{\partial U}{\partial x} \mathrm{d}x 
  +\frac{\partial U}{\partial y} \mathrm{d}y \\
&= \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y} \right) \cdot
   \left( \mathrm{d}x, \mathrm{d}y \right) \\
&= \vec{\nabla} U \cdot \mathrm{d}\vec{x}.
\end{align}
$$
So $\vec{\nabla} U$ is the gradient, while $\vec{\nabla} U \cdot \mathrm{d}\vec{x}$ is the differential.
For your specific example, the gradient is $(5x^4y+7y^3,x^5+21xy^2)$ while the differential is $(5x^4y+7y^3)\mathrm{d}x+(x^5+21xy^2)\mathrm{d}y$.
Once we find a potential $U$ such that $\frac{\partial U}{\partial x} = A$ and $\frac{\partial U}{\partial y} = B$, then the exact equation
$$ A \,\mathrm{d}x + B \,\mathrm{d}y = 0$$
becomes 
$$
\frac{\partial U}{\partial x} \mathrm{d}x 
+\frac{\partial U}{\partial y} \mathrm{d}y = 0
$$
so that $\mathrm{d}U = 0$, from which it follows that $U$ equals a constant.
