differential eq, conditions

**fiksi** · February 25th 2009, 02:33 PM

dy/dx = (x*e(to pwr of y) )/ x(to pwr of 2) + 1
given conditions y=0 at x=0.

**Krizalid** · February 25th 2009, 02:37 PM

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

Can you take it from there?

**skeeter** · February 25th 2009, 02:39 PM

$\frac{dy}{dx} = \frac{xe^y}{x^2+1}$

separate variables ...

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

do the integration.

**fiksi** · February 25th 2009, 02:44 PM

Originally Posted by Krizalid

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

Can you take it from there?

Is that separation of variables now? Hm, so we get I of e to -y = I of right side, right?
Not sure, am new to this... thx for help.

**Krizalid** · February 25th 2009, 02:45 PM

Yes, and don't forget the constant.

**fiksi** · February 25th 2009, 02:59 PM

Originally Posted by Krizalid

Yes, and don't forget the constant.

I get e to -y = 1/2 log (x2 +1 ) + c

So -y = log (1/2 log (x2 + 1) +c ) and y=-log of said? Afraidn i messed smth again...

Thread: differential eq, conditions