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

[MathewsMD]

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!

 

[johnqwertyful]

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]

http://www.wolframalpha.com/input/?i=0+=+2y+++e^y+-+4x+++3

Yeah, it's impossible.

 

[MathewsMD]

:(

10char

 

[johnqwertyful]

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

 

[HallsofIvy]

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

 

[jackmell]

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:

g(x,y)e^{g(x,y)}=h(x)

Then by definition of the Lambert W function which you can look up, we take the W function of both sides and obtain:

g(x,y)=W(h)

Now, doing a little moving around of your equation:

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

or:

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

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.