I want to solve $$\frac{\, \mathrm dy}{\, \mathrm dx}=e^{x^{2}}.$$ i using variable separable method to solve this but after some stage i stuck with the integration of $\int e^{x^{2}}\, \mathrm dx$. i dont know what is the integration of $\int e^{x^{2}}\, \mathrm dx$. Please help me out!
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There is no closed form integral of $e^{x^2}$. Are you sure you don't have a typo? – Emily Jul 20 '12 at 04:03
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3@Ed: it has a closed form. It's not elementary, however. – J. M. ain't a mathematician Jul 20 '12 at 04:10
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Er, yes. Thanks for that correction. I should have been more precise. – Emily Jul 20 '12 at 04:11
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See, for instance, this or this. – J. M. ain't a mathematician Jul 20 '12 at 04:23
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1Why in this green earth was this question downvoted? – J. M. ain't a mathematician Jul 20 '12 at 05:39
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1@J.M. Maybe someone was on the blue part of Earth. – Pedro Jul 20 '12 at 07:04
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@J.M. Honest question: what does closed form mean if not elementary? – Jul 20 '12 at 07:38
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1@Rahul: I wrote a long-ish answer to that question many moons ago, but the gist is, I consider the error function as a "known quantity", and I thus treat it as a closed form. – J. M. ain't a mathematician Jul 20 '12 at 07:43
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$$ \frac{dy}{dx}=e^{x^{2}} $$ has no elementary solution. The error function (also called the Gauss error function) is a special function (non-elementary) of sigmoid shape which occurs in probability, statistics and partial differential equations. It is defined as: $$ \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2} dt. $$ See the link for reference and more information and thus, J.M. ...?
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3...and thus, the solution of the DE is $y=-\frac{i\sqrt\pi}{2}\mathrm{erf}(ix)+C$. (the imaginary error function) – J. M. ain't a mathematician Jul 20 '12 at 05:41