Answer
$y=x-x^{4/3}+0.5$
Work Step by Step
Step 1. Given $y'=1-\frac{4}{3}x^{1/3}$, we have $y=\int(1-\frac{4}{3}x^{1/3})dx=x-x^{4/3}+C$.
Step 2. Using the point on the curve $y(1)=0.5$, we have $1-(1)^{4/3}+C=0.5$ and $C=0.5$.
Step 3. Thus the answer is $y=x-x^{4/3}+0.5$
