I am having trouble rearranging equations $u$ and $v$ into $x$ and $y$ in terms of $u$ and $v$.
The equations are: $u = e^x - y$ and $v = y^2 +4e^{-2x}$.
I wish to find $x = x(u,v)$ and $y = y(u,v)$ from these two equations.
So far I have rearranged $v$ in terms of $x$ for $x = \frac{1}{2}\ln(\frac{4}{v-y^2})$ and simplified substituting this into $u = e^x - y$ for:
$$ y^4+2uy^3+u^2y^2-y^2v-2uvy-u^2v +4 = 0$$
However, I am just at a loss on how to solve this quartic for $y$. What am I doing wrong? Is there a better method for this?
Be careful what you wish for. Here's the solution for just $y$, produced by Mathematica:
$$y = -\frac{1}{2} \sqrt{\frac{1}{3} \left(u^2-v\right)+\frac{\sqrt[3]{2 \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}{3 \sqrt[3]{2}}+\frac{\sqrt[3]{2} \left(u^4-2 u^2 v+v^2+48\right)}{3 \sqrt[3]{2 \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}+v}-\frac{1}{2} \sqrt{\frac{1}{3} \left(v-u^2\right)+u^2-\frac{\sqrt[3]{2 \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}{3 \sqrt[3]{2}}-\frac{\sqrt[3]{2} \left(u^4-2 u^2 v+v^2+48\right)}{3 \sqrt[3]{2 \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}-\frac{8 u^3-8 u \left(u^2-v\right)}{4 \sqrt{\frac{1}{3} \left(u^2-v\right)+\frac{\sqrt[3]{2 \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}{3 \sqrt[3]{2}}+\frac{\sqrt[3]{2} \left(u^4-2 u^2 v+v^2+48\right)}{3 \sqrt[3]{2 \left(u^2-v\right)^3-288 \left(u^2-v\right)+432 u^2+24 \sqrt{3} \sqrt{u^6 v-3 u^4 v^2-4 u^4+3 u^2 v^3+80 u^2 v-v^4+32 v^2-256}}}+v}}+v}-\frac{u}{2}$$