Answer
$y=\sqrt[3] {x^3+8}$
Work Step by Step
We are given that $\dfrac{dy}{dx}=\dfrac{x^2}{y^2}$
We will separate the variables to obtain:
$y^2 \ dy=x^2 \ dx$
Integrate to obtain:
$\int y^2 \ dy=\int x^2 \ dx$
This implies that $\dfrac{y^3}{3}=\dfrac{x^3}{3}+C$
After applying the initial conditions, $y=2$ when $x=0$, we get $C=\dfrac{8}{3}$
Therefore, we have: $ \dfrac{y^3}{3}=\dfrac{x^3}{3}+\dfrac{8}{3} \implies y=\sqrt[3] {x^3+8}$
