1. ## Exact differential.

(1+e^x/y)dx + e^x/y ( 1-x/y )dy =0

2. Is this

$\left(1 + \frac{e^x}{y}\right)\,dx + \frac{e^x}{y}\left(1 - \frac{x}{y}\right)\,dy = 0$

or

$\left(1 + e^{\frac{x}{y}}\right)\,dx + e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)\,dy = 0$?

3. 2nd one

4. $\displaystyle{\left(1 + e^{\frac{x}{y}}\right)\,dx + e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)\,dy = 0}$

If this is an exact equation of the form $M\,dx + N\,dy = 0$ then $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.

$\frac{\partial M}{\partial y} = -\frac{x}{y^2}e^{\frac{x}{y}}$.

$\frac{\partial N}{\partial x} = -\frac{1}{y}e^{\frac{x}{y}} + \frac{1}{y}e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)$

$= -\frac{1}{y}e^{\frac{x}{y}} + \frac{1}{y}e^{\frac{x}{y}} - \frac{x}{y^2}e^{\frac{x}{y}}$

$= -\frac{x}{y^2}e^{\frac{x}{y}}$

$= \frac{\partial M}{\partial y}$ as required.

So it's an exact equation.

That means that the solution $F(x, y)$ has the property that

$\frac{\partial F}{\partial x} = M$ and $\frac{\partial F}{\partial y} = N$.

Thus $\frac{\partial F}{\partial x} = 1 + e^{\frac{x}{y}}$

$F = \int{1 + e^{\frac{x}{y}}\,dx}$

$= x + y\,e^{\frac{x}{y}} + g(y)$.

$\frac{\partial F}{\partial y} = e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)$

$F = \int{e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)\,dy}$

$= y\,e^{\frac{x}{y}} + h(x)$ (according to Wolfram)...

Comparing these two results, it's easy to see that $h(x) = x + C$ and $g(y) = C$.

Therefore $F =x + y\,e^{\frac{x}{y}} + C$.

5. Thank you very much