Answer
See below
Work Step by Step
Given:
$\frac{dy}{dx}=\frac{2}{3}(y-1)^{1/2}\\
\Leftrightarrow\frac{dy}{dx}\frac{1}{\sqrt y-1}=\frac{2}{3}$
Integrate: $\int \frac{1}{\sqrt y-1}dy=\int \frac{2}{3}dx\\
\rightarrow 2\sqrt y-1=\frac{2}{3}x+c\\
\rightarrow \sqrt y-1=\frac{1}{3}x+\frac{1}{2}c\\
\rightarrow \sqrt y-1-\frac{1}{3}x=\frac{1}{2}c$
Substitute: $\rightarrow \sqrt 1-1-\frac{1}{3}(1)=\frac{1}{2}c\\
\rightarrow c=-\frac{1}{3}$
then $\sqrt y-1=\frac{1}{3}x-\frac{1}{3}\\
\rightarrow \sqrt y-1=\frac{x-1}{3}\\
\rightarrow y-1=(\frac{x-1}{3})^2\\
\rightarrow y-1=\frac{x^2-2x+1}{9}\\
\rightarrow y=\frac{x^2-2x+10}{9}$
We can see that the equation has more than one solution at $y=1$, then this equation is an exception to the Uniqueness and Existence Theorem.
