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How do you isolate for y when 0 = 2y + e^yby MathewsMD
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#1
Sep1213, 08:55 PM

P: 295

How do you isolate for y when you have the equation 0 = 2y + e^y  4x + 3?
Any tips, useful links or solutions and an explanation would be greatly appreciated! Thanks! 


#2
Sep1213, 09:07 PM

P: 351

Numerics seems to be the only way. Usually when you have to isolate a variable that's acted on by different types of functions, it's very difficult or impossible to do analytically.



#3
Sep1213, 09:18 PM

P: 351




#4
Sep1213, 09:30 PM

P: 295

How do you isolate for y when 0 = 2y + e^y
:(
10char 


#5
Sep1213, 09:44 PM

P: 351

In a lot of situations, it's not necessary to get analytic solutions. What is this for?



#6
Sep1313, 07:54 AM

Math
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PF Gold
P: 39,488

The "Wolfram Alpha" solution that johnqwertyful links to use the "Lambert W function" which is defined as the inverse function to [itex]f(x)= xe^x[/itex]. It cannot be written in terms of any simpler function.



#7
Sep1313, 11:28 AM

P: 1,666

[tex]g(x,y)e^{g(x,y)}=h(x)[/tex] Then by definition of the Lambert W function which you can look up, we take the W function of both sides and obtain: [tex]g(x,y)=W(h)[/tex] Now, doing a little moving around of your equation: [tex]1/2 e^y=2x3/2y[/tex] [tex]1/2 e^{2x3/2}=e^{y} e^{2x3/2}(2x3/2y)[/tex] or: [tex](2x3/2y)e^{2x3/2y}=1/2 e^{2/x3/2}[/tex] I'll let you finish it to isolate y in terms of the perfectly valid (multivalued) function of x in terms of the Lambert Wfunction. 


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