Get explicit, vector-valued function for a curve defined by an implicit expression.

Question

$ \mathcal {B} = \{(x,y) \in \mathbb R^2 $ such that $ 0 = x^2 + y^2 + xy \exp (-x^2) \} $

$ \mathbf f(x,y) = f_1(x,y) \mathbf i + f_2 (x,y) \mathbf j = \mathcal {B} $

Please show explicit expressions of the functions $f_1(x,y)$ and $f_2(x,y)$.

Answer 1

With due credit to both commenters, here is what I was seeking.

Start from the implicit expression:

$$ 0 = x^2 + y^2 + xy \exp (-x^2) $$

Complete a square.

$$ \left (y + \frac 12 x \exp (-x^2) \right )^2 = y^2 + xy \exp (-x^2) + \frac 14 x^2 \exp (-2x^2) $$

Therefore

$$\sqrt {\frac 14 x^2 \exp (-2x^2) -x^2} = y + \frac 12 x \exp (-x^2) $$ $$ $$ $$ y = \frac {-1}2 x \exp (-x^2) + x \sqrt {\frac 14 \exp (-2x^2) - 1} $$