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Direct calculation givesWhere $1+\cos x\ne 0$, Abel identity gives $p=-W^{\prime }/W=\sin x/(1+\cos x)$, and substitution of either solution gives $q=1/(1+\cos x)$. ThusThese coefficients have singularities at odd multiples of $\pi$. This does not contradict the Wronskian criterion: its continuous-coefficient interval hypothesis fails at those points.

On $0\le x<\pi$, the initial conditions uniquely select $y=\sin x$. The normalized differential equation is not defined on the whole interval $0\le x<\infty$, so the usual global uniqueness theorem for ordinary differential equations cannot give the global initial-value assertion as posed. No choice of finite coefficients at $\pi$ can repair the normalized equation while retaining both displayed solutions: at that point $1+\cos x$ and its first derivative vanish, but its second derivative equals one.

The continuation through a zero of the leading ODE coefficient is a different problem: consider the multiplied, degenerate equationFor this different equation, a globally $C^2$ solution with the given initial data is uniquely $\sin x$. On each regular interval every solution is $A_j(1+\cos x)+B_j\sin x$; continuity of $y^{\prime }$ and $y^{\prime \prime }$ at an odd multiple of $\pi$ forces both constants to agree across it. If only $C^1$ piecewise solutions away from the singular points are required, the $A_j$ may change and uniqueness fails. Thus the coefficient singularity and the intended solution regularity must be distinguished.

Where $\sin x\ne 0$, divide by it to obtain $y^{\prime }+2\cot xy=\csc x$. The integrating factor is ${\sin }^{2}x$, and the product rule givesThe initial condition gives $C=1$. HenceThe simplified expression has a differentiable extension through $x=0$, with $y(0)=1/2$. Direct substitution verifies the original undivided differential equation there as well: its left side is $2y(0)=1$. Thus the maximal interval containing $\pi /2$ on which this solution is finite is $\boxed{-\pi <x<\pi }$; the apparent exclusion of $0$ arose only from division by $\sin x$. This is continuation through a zero of the leading ODE coefficient.

On an interval with $x\ne 0$, division by $x^2$ gives the separable differential equation $y^{\prime }=(4/x^2-1/x)y^2$. For $y\ne 0$, take antiderivatives:Renaming the constant yields the explicit nonzero familyFor $x\ne 0$, the interval of definition must avoid zeros of the denominator. Division by $y^2$ excluded the additional solution $\boxed{y(x)\equiv 0}$, which must also be included.

The original undivided differential equation also makes sense at $x=0$, where it requires $y(0)=0$. For continuation through a zero of the leading ODE coefficient, rewrite the nonzero family asIt extends differentiably with $y(0)=0$ and $y^{\prime }(0)=1/4$, since $y(x)/x\to 1/4$. One may choose the integration constants independently on the two sides of zero and obtain a differentiable solution through zero. The zero solution also extends through zero, with derivative $0$; joining a nonzero branch on just one side to a zero branch on the other would fail differentiability.