Example 5.   Investigate the initial value problem    with  .   

Solution 5.

The function    does not have a removable singularity at  .  

Remark.  The function      can be considered a multivalued function and does not have a removable singularity, it's singularities are algebraic branch points.

The singularities of the function     are not poles, they are algebraic branch points.

Therefore, the differential equation    does not have the Painlevé property.     

(c) John H. Mathews 2005