Divide through by x, then substitute $\displaystyle \begin{align*} u = y' \end{align*}$. The equation becomes first order linear, and so can be solved with an integrating factor.
Yes, though I can't imagine why you would use a complicated numerical method when it is easy to solve for an exact solution.
What I would do is this: let u= y' so we get the first order equation . That is linear so we can separate it, first solving which in turn can be separated to . Integrating which is equivalent to .
To find a "particular solution", use "variation of parameters": seek a solution of the form . Then [tex]u'= v'x^2+ 2xv[tex] and the differential equation becomes so .