The displayed implicit ordinary differential equation has a singular initial point at the origin. On a nonconstant regular branch, treating as a function of gives . The integrating factor yields the first integral of an ordinary differential equation
One can verify this identity without dividing by : its derivative along a solution is .
The initial point forces , so . Near the origin this gives the two differentiable solutions and , with initial slopes zero and one. A division by would lose the first solution. A piecewise switch between the two at zero would fail differentiability because their one-sided slopes differ.
There is no conflict with the Picard-Lindelöf theorem: solving explicitly for requires division by , which vanishes at the initial point. A regular implicit equation or a local uniqueness theorem must be checked before claiming a unique solution through a singular point.

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