1
$\begingroup$

For the equation:

$$y^2 = \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}$$

How might one go about solving for $x$ in terms of $y$? First, I attempted solving the equation by first performing the subsitution $u = xe^{xy^2}$, which then reduces the problem to: $$\frac{\ln(\frac{u}{x})}{x} = \frac{\ln(1 - u)}{1-u}$$

Unfortunately, I could not find where to go from there. Secondly, I tried multiplying both sides of the initial equation by $1 - xe^{xy^2}$ to obtain:

$$y^2-xy^2e^{xy^2} = \ln(1 - xe^{xy^2})$$

From here, I could subtract $y^2$ from both sides and divide by -1 to get the equation in a form where the product logarithm could be applied, but unfortunately the right hand side would contain both $x$ and $y$. Exponentiating both sides with base $e$ only seems to make the problem more convoluted and messy than it already is. So how might I go about actually solving this equation for $x$?

$\endgroup$
0
$\begingroup$

Considering $$y^2 = \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}$$ we could start defining $$z=1 - xe^{xy^2}\implies y^2=\frac{\log(z)}z$$ Solving for $z$ $$z=-\frac{W\left(-y^2\right)}{y^2}$$ where $W(.)$ is Lambert function. So $$1 - xe^{xy^2}=-\frac{W\left(-y^2\right)}{y^2}$$ and, again, Lambert function $$x=\frac{W\left(y^2+W\left(-y^2\right)\right)}{y^2}$$

Do not ask me for $y(x)$, please !

Edit

Since you mention the product logarithm, I suppose that you can run Mathematica. If so,ask for the contour plot of $$f(x,y)=y^2 - \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}=0$$ for $-\frac{1}{\sqrt{e}} \leq y \leq \frac{1}{\sqrt{e}}$ and $-1000 \leq x \leq 0$.

$\endgroup$
0
$\begingroup$

The substitution: $$u=x\;\exp \left( x~y^{2}\right) =x\;\exp \left( x~v\right) ;\;v=y^{2}$$ is the the correct Ansatz here. This leads to: $$\frac{\log \left( \frac{u}{x}\right) }{x}=\frac{\log \left( 1-u\right) }{1-u}$$ Application of the functional equation on the left hand side of the equation leads to: $$\frac{\log \left( u\right) }{x}-\frac{\log \left( x\right) }{x}=\frac{\log \left( 1-u\right) }{1-u}$$ Now we substitute $u$ on the left side of the equation:

$$\frac{\log \left( x\;\exp \left( x~v\right) \right) }{x}-\frac{\log \left( x\right) }{x}=\left( v+\frac{\log \left( x\right) }{x}\right) -\frac{\log \left( x\right) }{x}=v$$ and recognize, that the term: $\frac{\log \left( x\right) }{x}$ vanishes. Now, we can solve for $u$ $$u=\frac{v+W\left( -v\right) }{v}$$ where $W\left( z\right) $ is the Lambert function [Mathematica] (http://functions.wolfram.com/ElementaryFunctions/ProductLog/27/02/) or [Corless] (https://cs.uwaterloo.ca/research/tr/1993/03/W.pdf). In the end we have to substitute $u\rightarrow x\;\exp \left( x~v\right) $ and $v$ and solve the equation for $x$ $$x=\frac{W\left( y^{2}+W\left( -y^{2}\right) \right) }{y^{2}}$$ Numerical comparisons for $y=0.316228$ e.g. $v=0.1$ confirms the result.

$\endgroup$
  • $\begingroup$ So I was a bit to late! $\endgroup$ – stocha Feb 18 '19 at 11:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.