dy/dx = (x*e(to pwr of y) )/ x(to pwr of 2) + 1

given conditions y=0 at x=0.

Printable View

- Feb 25th 2009, 02:33 PMfiksidifferential eq, conditions
dy/dx = (x*e(to pwr of y) )/ x(to pwr of 2) + 1

given conditions y=0 at x=0. - Feb 25th 2009, 02:37 PMKrizalid
$\displaystyle \frac{dy}{dx}=\frac{xe^{y}}{x^{2}+1}\implies e^{-y}\,dy=\frac{x}{x^{2}+1}\,dx.$

Can you take it from there? - Feb 25th 2009, 02:39 PMskeeter
$\displaystyle \frac{dy}{dx} = \frac{xe^y}{x^2+1}$

separate variables ...

$\displaystyle e^{-y} \, dy = \frac{x}{x^2+1} \, dx$

do the integration. - Feb 25th 2009, 02:44 PMfiksi
- Feb 25th 2009, 02:45 PMKrizalid
Yes, and don't forget the constant.

- Feb 25th 2009, 02:59 PMfiksi