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I solved a differential equation some time ago and I need to solve for $y$. How can we solve for $y$ using the Lambert W function?

$$C_1+x = e^y+Cy$$

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...related problem here, interesting how things evolve :) – hhh Feb 21 '12 at 0:00
@hhh Infinite loop! – Peter Tamaroff Feb 21 '12 at 0:01

1 Answer

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Hint

$$\large C^{-1}\exp\left(C^{-1}e^y+y\right)=(C^{-1} e^y)e^{(C^{-1} e^y)} $$

Answer

$$\hskip 2in \displaystyle \begin{array}{} A+x=e^y+Cy \\ \frac{A+x}{C}=\frac{e^y}{C}+y \\ \frac{1}{C}\exp\left(\frac{A+x}{C}\right)=\frac{e^y}{C}\exp\left(\frac{e^y}{C}\right) \\ W\left(\frac{1}{C}e^{(A+x)/C}\right)=\frac{e^y}{C} \\ y=\log\left[ C\; W\left(\frac{1}{C}e^{(A+x)/C}\right)\right] \end{array}$$

Note also that $\log W(z)=\log z-W(z)$, if you want to compare with what W|A gives.

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So you have $$y = \log W\left( {\frac{{{C_1} + x}}{{{C^2}}}} \right) + \log C$$ ? – Peter Tamaroff Feb 20 '12 at 4:53
@Peter: No, you forgot to exponentiate the LHS of the original equation. – anon Feb 20 '12 at 5:02
I was too focused on the RHS. How do you code that spoiler? It's really cool to make answers. – Peter Tamaroff Feb 20 '12 at 5:34
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@Peter: It's problematic putting $$\LaTeX$$ in spoilers so you'll have to check what markup I used there, but the spoiler code is >! text. You may be able to see an "Edit" button and see what I wrote down. – anon Feb 20 '12 at 5:38
Thanks. I mostly use MathType's "translator" to code my equations, but I know a little of LaTeX so I think I can manage. – Peter Tamaroff Feb 20 '12 at 5:45

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