Can I Solve y in y^2=ln(1-y^2)?

[NODARman]

TL;DR Summary

How can I find the extrema of the function?

Hi, I'm differentiating the "z" function to find extreme points but after solving the first partial derivatives with respect to "x" and "y" and also the "x" variable of the system, I can't find "y" (still in the system) using "ln" (natural logarithm).

The question is can I differentiate both sides and easily write what "y" is or I can't use that method to solve it?

The problem: y^2=ln(1-y^2)

This is how I've got here (picture):

 

[pasmith]

You have $z = (x^2 + y^2)e^{-(x^2 + y^2)} - (x^2 + y^2)$. It would make sense here to use polar coordinates instead, and find the value of $r$ at which $z$ is maximal. This leads to $$1 - r^2 = e^{r^2}$$ which is what you obtained for $y$ after concluding that $x = 0$.

Now by inspection $r = 0$ is a solution, and since the left hand side is less than 1 for $r > 0$ and the right hand side is greater than 1 for $r > 0$ the only solution is $r = 0$. Thus the extremum is at $x = y = 0$.

That one could be solved analytically, but in general equations of the form $$\mbox{(polynomial)} = \mbox{(exponential)}$$ have to be solved numerically.

 

[vela]

The problem: y^2=ln(1-y^2)

You can solve this particular equation similarly to the way @pasmith did above for $r$. The lefthand side is non-negative while the righthand side is non-positive. The only possible solution is when both sides equal 0, which is at $y=0$.