Solving equation $y^2-2ln(y)=x^2$

Mathematics Asked by Majdgh on January 5, 2022

enter image description hereCan someone help me to solve this equation:
$$y^2-2ln(y)=x^2$$
I want to find y.
I have tried to solve the problem but I couldn’t.

Thanks in advance.

One Answer

(Assuming you only care about real solutions).

The Lambert $${W}$$ function is the inverse to $${xe^{x}}$$ (well, that depends on the value of $$x$$ - you will need to change the branch accordingly. If $${xgeq -1}$$, we take the $${W_0}$$ branch, otherwise - you take the $${W_{-1}}$$ branch. For simplicity, for now I will refer to the whole thing as just $${W}$$). That is,

$${W(xe^{x})=x}$$

Notice that your equation

$${y^2 - 2ln(y)= x^2}$$

is equivalent to

$${2ln(y) -y^2 = -x^2}$$

raising $$e$$ to both sides yields

$${e^{2ln(y) - y^2}=e^{-x^2}=y^2e^{-y^2}}$$

Multiply both sides by $${-1}$$ once again to get

$${Rightarrow -y^2e^{-y^2}=-e^{-x^2}}$$

And now use Lambert $${W}$$

$${W(-y^2e^{-y^2})=W(-e^{-x^2})=-y^2}$$

And so you get

$${y^2 = -W(-e^{-x^2})}$$

which implies that

$${y = pmsqrt{-W(-e^{-x^2})}}$$

But which branch of $${W}$$ do you take? - it turns out in this context you need to take both one at a time, since both will give you valid solutions. Going back to

$${-y^2e^{-y^2}=-e^{-x^2}}$$

A quick cheap way of seeing this is that by looking at the graph of $${-y^2e^{-y^2}}$$ you can see it has a range of $${left[-frac{1}{e},0right]}$$, and every value in it's range is hit by some $$y$$ such that $${(-y^2) < -1}$$, and also get's hit by some $${y}$$ such that $${(-y^2)geq -1}$$ also. And so if you take the $${W_0}$$ branch - it'll give you the $$y$$ solution such that $${(-y^2)geq -1}$$, and if you take the $${W_{-1}}$$ branch it'll give you the solution such that $${(-y^2) < -1}$$. Both solutions are important.

In the context of real solutions, notice that the range of $${-y^2e^{-y^2}}$$ being $${left[-frac{1}{e},0right]}$$ also forces our $${x}$$ to be in the domain of $${(-infty,-1]cup [1,infty)}$$ (since this domain ensures the range of $${-e^{-x^2}}$$ matches accordingly).

Answered by Riemann'sPointyNose on January 5, 2022

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