Problem 05 $2y \, dx = 3x \, dy$, when $x = -2$, $y = 1$.
Solution 05
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
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