## Calculus 10th Edition

$$y=\frac{1}{2}x^{\frac{2}{3}}$$
$y'=\frac{2y}{3x}$ $\frac{1}{2y}y'=\frac{1}{3x}$ $\int\frac{1}{2y}dy=\int\frac{1}{3x}dx$ $\frac{1}{2}\ln{y}=\frac{1}{3}\ln{x}+C$ $\ln{y}=\frac{2}{3}\ln{x}+C$ $y=e^{\frac{2}{3}\ln{x}+C}=e^Cx^{\frac{2}{3}}$ Let $k=e^C$ so $y=kx^{\frac{2}{3}}$ We need this graph to pass through the point $(8,2)$ so we will apply it as an initial condition. $2=k8^{\frac{2}{3}}=4k$ so $k=\frac{1}{2}$ and therefore $$y=\frac{1}{2}x^{\frac{2}{3}}$$