First assume y₂=ux², and y₂’=2ux+u’x² and y₂’’=u’’x²+4u’x+2u.

Substituting into x²y’’+2xy’-6y=0 we get x²(u’’x²+4u’x+2u)+2x(2ux+u’x²)-6y=0 and simplifying by adding

like terms we get: u’’x⁴+6u’x³=0.

We reduce the order by w=u’ to get

w’x⁴+6wx³=0. Now dividing by wx⁴ and rearranging, we get

and integrating both sides, or , therefore we get u=C₁x^-5+C₂.

If we let C₁=1, C₂=0 we get u=x^-5, hence our second solution y₂=x²x^-5=x^-3. 

Our fundamental set is {x²,x^-3}and the general solution is  y=C₁x²+C₂x^-3

Rewrite x²y’’+2xy’-6y=0 (dividing by x²) as,

and so use the formula  to get .

If we let the C=0, we get y₂=x^-3