Problem 05 | 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