## Calculus (3rd Edition)

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