Dear scholars,(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums - The Fusion of Science and Community**

# Lambert W function

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Lambert W function

Loading...

**Physics Forums - The Fusion of Science and Community**