# Lambert W function

1. Jul 8, 2014

### Nandan_78

Dear scholars,

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.

2. Jul 8, 2014

Correct.

3. Jul 8, 2014

### HallsofIvy

Staff Emeritus
Yes, very good!