#### Answer

$$ y= (\frac{4}{3}t^3+c)^{1/4}.$$

#### Work Step by Step

By separation of variables, we have
$$y^3dy=t^2dt $$
then by integration, we get
$$ \frac{1}{4}y^4=\frac{1}{3}t^3+c .$$
So the general solution is given by $$ y= (\frac{4}{3}t^3+c)^{1/4}.$$