Hints on how to solve $(x+y^2)dy = ydx$? I'm looking for hints on how to solve the differential equation: $(x+y^2)dy = ydx$ .
I tried finding an integrating factor and dividing both sides by $y$ but that didn't work. 

In $(x+y^2)dy=ydx$ you can indeed divide by $y^2$ to get
$$
d\left(\frac{x}{y}\right)=\frac{y\,dx-x\,dy}{y^2}=dy
$$
which is directly integrable.

For the first version of the question, $(x^2+y^2)dy=ydx$, observe that
$$
\frac{dx}{dy}=y+\frac{x^2}{y}
$$
looks like a Riccati equation. Set $x(y)=-\dfrac{yu'(y)}{u(y)}$ to get a second order linear ODE in $u$
$$
y+\frac{yu'^2}{u^2}=x'=-\frac{yu''+u'}{u}+\frac{yu'^2}{u^2}
\\\iff\\
yu''+u'+yu=0.
$$
Then apply power series expansion or identify the special function type that solves this equation.
