Solve $\frac{dy}{dx}+\frac{y}{x-2}=5\left(x-2\right)\sqrt{y}$ Based on the equation it looks like I need an integrating factor:
$\frac{dy}{dx}+\frac{y}{x-2}=5\left(x-2\right)\sqrt{y}$
Integrating factor
$e^{\int\frac{1}{x-2}\;dx}$, which when integrated gives 
$\ln\left|x-2\right|$ 
or 
$x-2$ 
But this is the part where I get very confused with the integrating factor. I believe (though I don't understand why) the left side simplifies to:
$\frac{d}{dx}\left(x-2\right)y$
and when writing out the entire equation I get:
$\frac{d}{dx}\left(x-2\right)y = \left(x-2\right)\left(5\left(x-2\right)\sqrt{y}\right)$
Any helps as to where I go from here?  The math after this point doesn't make sense to me, which leads me to believe I made a mistake somewhere else.

Let $x=u+2$; then
$$u \frac{dy}{du} + y = 5 u^2 \sqrt{y}$$
or
$$\frac{d}{du} (u y) = 5 u^{3/2} \sqrt{u y}$$
or
$$\frac{d(u y)}{\sqrt{u y}} = 5 u^{3/2} du \implies 2 \sqrt{u y} = 2 u^{5/2}+ 2 C $$
Finally,
$$y(x) = \frac{\left [(x-2)^{5/2}+C \right ]^2}{x-2} = (x-2)^4+ 2 C (x-2)^{3/2}+\frac{C^2}{x-2}$$
