This case will lead to similar problem that we've had all other time we've run in double roots or double eigenvalues. We only find a single solution and will require a second solution. In this case this can be demonstrated as the second solution will be,

y₂(x) = x^r In x

and therefore the general solution for this case is,

 y(x) = c₁ x^r + c₂ x^r ln x = x^r (c₁ + c₂ ln x)

 We can again see a purpose for requiring x>0. If we didn't we'd contain all sorts of problems along with this logarithm.