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Lambert W function 
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#1
Jul814, 09:44 AM

P: 1

Dear scholars,
I am working on the following equation and wonder whether my solution is correct. The actual problem is more complex but the example below captures the main features. I have not a lot of experience with the Lambert W function. Thanks in advance for your comments! Equation: $$Y=x \exp(x^2)$$ Substitute: $$u=x^2$$ Then: $$Y=\sqrt{u} \exp(u)$$ $$Y^2=u \exp(2u)$$ $$2Y^2=2u \exp(2u)$$ Then using W function: $$2u=W(2Y^2)$$ From which $$x$$ follows. 


#2
Jul814, 10:32 AM

P: 759

Correct.



#3
Jul814, 10:45 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,491

Yes, very good!



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