Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 18

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

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