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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) = xr In x
and therefore the general solution for this case is,
y(x) = c1 xr + c2 xr ln x = xr (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.
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