How to look for a solution of type $xg (x) $ in a second order DE?

Question

$$y''- \frac{1}{x} y' + \frac{1}{x^2}y= 0$$

with $x \in [1,3]$

I learned how to use contraction mapping and find a fixed point for finding a solution of $F'(x)=(x,F (x)).$ But how should I "look" for a solution of type $xg (x) $ by showing g' satisfies first order DE and that solution is vector Valued?

Answer 1

Compute the derivatives of $y(x)=xg(x)$, $$ y'=xg'+g\\ y''=xg''+2g' $$ and insert $$ 0=(xg''+2g')-\frac1x(xg'+g)+\frac1{x^2}xg. $$where you will find that some reductions in the order are trivially possible.