differential eq, conditions

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• Feb 25th 2009, 02:33 PM
fiksi
differential 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 PM
Krizalid
$\displaystyle \frac{dy}{dx}=\frac{xe^{y}}{x^{2}+1}\implies e^{-y}\,dy=\frac{x}{x^{2}+1}\,dx.$

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

separate variables ...

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

do the integration.
• Feb 25th 2009, 02:44 PM
fiksi
Quote:

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

Can you take it from there?

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.
• Feb 25th 2009, 02:45 PM
Krizalid
Yes, and don't forget the constant.
• Feb 25th 2009, 02:59 PM
fiksi
Quote:

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...