dy/dx = (x*e(to pwr of y) )/ x(to pwr of 2) + 1 given conditions y=0 at x=0.
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$\displaystyle \frac{dy}{dx}=\frac{xe^{y}}{x^{2}+1}\implies e^{-y}\,dy=\frac{x}{x^{2}+1}\,dx.$ Can you take it from there?
$\displaystyle \frac{dy}{dx} = \frac{xe^y}{x^2+1}$ separate variables ... $\displaystyle e^{-y} \, dy = \frac{x}{x^2+1} \, dx$ do the integration.
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.
Yes, and don't forget the constant.
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...
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