First Order Differential Equations I am having trouble isolating the $x$ and $y$ to separate side in the differential equations below.  Could someone give me a hint as to how to to this.
Equation 1:  $$\frac{dy}{dx} - \frac{x}{y}  = \frac{1}{x}$$ 
Equation 2:  $$xy\frac{dy}{dx} = y^2$$
 A: The first is not separable.
For the second, we have:
$$xy\frac{dy}{dx} = y^2$$
Dividing, we have:
$$\dfrac{y~dy}{y^2} = \dfrac{dx}{x}$$
This is separable and we can now integrate each side as:
$$\int \dfrac{1}{y}~ dy = \int \dfrac{1}{x}~ dx$$
I think you can take it from here.
A: The second one is $\dot y = \frac{y}{x}$, and then you should try to put $y=xz$ and solve for $z$.
A: As far as I can see, only the second one is separable and after simplifying it becomes:
$$\frac{dx}{x}=\frac{dy}{y} $$
Straightforward integration will do the trick.
As a general rule, more often than not it is the multiplicative types of ODE's that are separable.
A: For $\dfrac{dy}{dx}-\dfrac{x}{y}=\dfrac{1}{x}$ , 
$y\dfrac{dy}{dx}=\dfrac{y}{x}+x$
This belongs to an Abel equation of the second kind.
Let $x=e^t$ ,
Then $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{\dfrac{dy}{dt}}{e^t}=e^{-t}\dfrac{dy}{dt}$
$\therefore e^{-t}y\dfrac{dy}{dt}=e^{-t}y+e^t$
$y\dfrac{dy}{dt}-y=e^{2t}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
For $xy\dfrac{dy}{dx}=y^2$ , it simply belongs to a separable ODE.
$\dfrac{dy}{y}=\dfrac{dx}{x}$
$\int\dfrac{dy}{y}=\int\dfrac{dx}{x}$
$\ln y=\ln x+c$
$y=Cx$
