= Implicit factorization can reveal nonunique initial solutions
{title2=$(2\sin y-x)y'=\tan y,\qquad y(0)=0$}
The displayed implicit <ordinary differential equation> has a singular initial point at the origin. On a nonconstant regular branch, treating $x$ as a function of $y$ gives $dx/dy+x\cot y=2\cos y$. The <integrating factor> $\sin y$ yields the <first integral of an ordinary differential equation>
$$
x\sin y-\sin^2y=C.
$$
One can verify this identity without dividing by $\sin y$: its <derivative> along a solution is $\sin y-(2\sin y-x)\cos y\,y'=0$.
The initial point forces $C=0$, so $\sin y(x-\sin y)=0$. Near the origin this gives the two differentiable solutions $y=0$ and $y=\arcsin x$, with initial slopes zero and one. A division by $\sin y$ 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 $y'$ requires division by $2\sin y-x$, 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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