Problem 06 | Separation of Variables

From Solution 04,
$\dfrac{x^2}{y^3} = c$
 

when x = 2, y = -1
$\dfrac{2^2}{(-1)^3} = c$

$c = -4$
 

Thus,
$\dfrac{x^2}{y^3} = -4$

$y^3 = -\dfrac{x^2}{4}$

$y = -\sqrt[3]{\left( \dfrac{x}{2} \right)^2}$

$y = -(x / 2)^{2/3}$           answer