Answer
$y=-ln [c-\dfrac{2}{5} (x-2)^{5/2}-\dfrac{4}{3} (x-2)^{3/2}]$
Work Step by Step
Re-arrange the given equation as: $y'=xe^y(\sqrt {x-2})$
Now, $\int e^{-y} dy=\int x\sqrt {x-2} dx$
Consider $a=x-2$ or, $ da=dx$
$ e^{-y} +c=\int (a+2) \sqrt a da$
or, $ e^{-y} =c-(\dfrac{2}{5}) a^{5/2}-(\dfrac{4}{3}) a^{3/2}$
This implies that
$ \ln [e^{-y}] =\ln [c-(\dfrac{2}{5}) a^{5/2}-(\dfrac{4}{3}) a^{3/2}]$
Hence, $y=-ln [c-\dfrac{2}{5} (x-2)^{5/2}-\dfrac{4}{3} (x-2)^{3/2}]$
