1. ## First-order nonlinear DE

Hello,
I tried searching similar DE-problems, but I couldn't find one. The problem is following differential equation:
$\displaystyle e^{x}y'+xe^{-y}=0$

So, it seems to be first-order nonlinear DE. But, I don't know how to solve it. I couldn't separe it, though I managed to separate simpler equations, but I think that's the right way to go.

Just got it to form,
$\displaystyle e^{-y}(e^{y+x}y'+x)=0$,

which obviously doesn't help anything...

Any help is appreciated. Thank you!

2. $\displaystyle y'=-\dfrac{xe^{-x}}{e^x}=-\dfrac{x}{e^{2x}}$

3. Originally Posted by Greg98
Hello,
I tried searching similar DE-problems, but I couldn't find one. The problem is following differential equation:
$\displaystyle e^{x}y'+xe^{-x}=0$

So, it seems to be first-order nonlinear DE. But, I don't know how to solve it. I couldn't separe it, though I managed to separate simpler equations, but I think that's the right way to go.

Just got it to form,
$\displaystyle e^{-y}(e^{y+x}y'+x)=0$,

which obviously doesn't help anything...

Any help is appreciated. Thank you!
Dear Greg98,

$\displaystyle e^{x}y'+xe^{-x}=0$

$\displaystyle \frac{dy}{dx}=-xe^{-2x}$

$\displaystyle y=-\int{xe^{-2x}}dx$

This integration could be done by integration by part method.....

Hope this helps.

4. Thanks for the help! Because of my omnipotent typing skills, I still need some advice (see original post). Sorry... I tried some logarithms to eliminate $\displaystyle e^{-y}$, but that didn't help.

5. $\displaystyle e^{x}y'+xe^{-y}=0$

$\displaystyle e^{x}y'+\dfrac{x}{e^{y}}=0$

divide both sides by $\displaystyle e^x$

$\displaystyle y'+\dfrac{x}{e^{x}\;e^{y}}=0$

multiply both sides by $\displaystyle e^y$

$\displaystyle y'\;e^y +\dfrac{x}{e^x}=0$

$\displaystyle e^y \frac{dy}{dx}=-\dfrac{x}{e^x}$

$\displaystyle e^y\;dy=-xe^{-x}\;dx$

finish...