You can realize that the derivative of [math]y’/y[/math] is [math](yy”-(y’)^2)/y^2[/math], so setting [math]z=y’/y[/math] might help.
Indeed, you find, due to the previous remark,
[math]y^2z’-y^2y’=0[/math]
and therefore either [math]y=0[/math] or [math]z’=y'[/math]. The latter becomes
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