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How do you isolate for y when 0 = 2y + e^y

by MathewsMD
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MathewsMD
#1
Sep12-13, 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!
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johnqwertyful
#2
Sep12-13, 09:07 PM
P: 332
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.
johnqwertyful
#3
Sep12-13, 09:18 PM
P: 332
http://www.wolframalpha.com/input/?i...5Ey+-+4x+%2B+3

Yeah, it's impossible.

MathewsMD
#4
Sep12-13, 09:30 PM
P: 295
How do you isolate for y when 0 = 2y + e^y

:(

10char
johnqwertyful
#5
Sep12-13, 09:44 PM
P: 332
In a lot of situations, it's not necessary to get analytic solutions. What is this for?
HallsofIvy
#6
Sep13-13, 07:54 AM
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Thanks
PF Gold
P: 39,344
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.
jackmell
#7
Sep13-13, 11:28 AM
P: 1,666
Quote Quote by MathewsMD View Post
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!
Well, when we have these mixed exponential equations, we try to put it in Lambert-W form. That is, in the form:

[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=2x-3/2-y[/tex]
[tex]1/2 e^{2x-3/2}=e^{-y} e^{2x-3/2}(2x-3/2-y)[/tex]

or:

[tex](2x-3/2-y)e^{2x-3/2-y}=1/2 e^{2/x-3/2}[/tex]

I'll let you finish it to isolate y in terms of the perfectly valid (multi-valued) function of x in terms of the Lambert W-function.


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