Find the solution to the given IVP.
\[ -2 \, {y}^{2} {y'} = -\frac{4 \, t}{{y}} - \frac{4}{{y}} \hspace{2em} y( 3 )= 3 \]
Answer:
\[ {y} = {\left(t^{2} + 2 \, t + 12\right)}^{\frac{1}{3}} \]