Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 20

$$ y =\left(\frac{8t+c}{2}\right)^2.$$

We have $$\frac{dy}{dt}=8\sqrt{y}\Longrightarrow y^{-1/2}dy=8dt\\ \Longrightarrow\int y^{-1/2}dy=8\int dt+c\\ \Longrightarrow 2y^{1/2}=8t+c \Longrightarrow y =\left(\frac{8t+c}{2}\right)^2.$$ Hence, the general solution is $$ y =\left(\frac{8t+c}{2}\right)^2.$$