#### Answer

$$ y =\frac{-t^2}{2 +ct^2}.$$

#### Work Step by Step

We have
$$ t^3 y'+4y^2=0\Longrightarrow y^{-2}dy=-4t^{3}dt\\
\Longrightarrow\int y^{-2}dy=\int -4t^{3}dt+c\\
\Longrightarrow -y^{-1}=2t^{-2}+c
\Longrightarrow y =\frac{-1}{2t^{-2}+c}.$$
Hence, the general solution is
$$ y =\frac{-t^2}{2 +ct^2}.$$