First-order nonlinear DE

Printable View

November 25th 2010, 05:26 AM
Greg98

First-order nonlinear DE

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

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,
$e^{-y}(e^{y+x}y'+x)=0$ ,

which obviously doesn't help anything...

Any help is appreciated. Thank you!
November 25th 2010, 05:35 AM
harish21

$y'=-\dfrac{xe^{-x}}{e^x}=-\dfrac{x}{e^{2x}}$
November 25th 2010, 05:40 AM
Sudharaka

Quote:

Originally Posted by Greg98

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

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,
$e^{-y}(e^{y+x}y'+x)=0$ ,

which obviously doesn't help anything...

Any help is appreciated. Thank you!

Dear Greg98,

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

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

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

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

Hope this helps.
November 25th 2010, 08:00 AM
Greg98

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 $e^{-y}$ , but that didn't help.
November 25th 2010, 08:29 AM
harish21

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

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

divide both sides by $e^x$

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

multiply both sides by $e^y$

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

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

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

finish...