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Work over the reals with $x\ne 0$, and put $p=dy/dx$. The algebraic equation implies $p\ne 0$ andDifferentiate this identity and use $y^{\prime }=p$; cancellation gives $dp/dx=p/x$. Hence $p=Cx$, with nonzero real $C$, and substitution gives the complete familyThere is no extra real singular branch lost in the cancellation, since $p^2+1\ne 0$. At a point $(x_0,y_0)$ with $x_0\ne 0$, the permitted constants satisfy $x_0^2{C}^2-2y_{0}C-1=0$, givingThese are distinct, real and nonzero, so exactly two integral curves pass through each such point. This does not contradict an Picard-Lindelöf theorem: the implicit equation permits two different slopes and is not one specified explicit first-order ordinary differential equation. Each choice of the smooth slope branch separately defines a locally unique curve.

Separate the variables in this first-order ordinary differential equation:Integration gives $\frac{1}{2}\ln (y^2+2)=\frac{1}{2}\ln |x^2-3|+C$, so $y^2+2=K|x^2-3|$. On the interval containing the prescribed point $x=3$, we have $x^2-3>0$. The initial condition $y(3)=4$ therefore gives $18=6K$, hence $K=3$. Choosing the positive square root to match that initial condition yieldsThe maximal interval containing $3$ on which this solves the explicit ordinary differential equation is $\boxed{x>\sqrt{11/3}}$. At the left endpoint $y=0$, the original denominator vanishes and the derivative becomes unbounded. Directly, $y^{\prime }=3x/y$ and $x(2+y^2)/[y(x^2-3)]=3x/y$, verifying the equation throughout this interval.

For $y_1=(1-x{)}^{-1}$, the derivatives are $y_1^{\prime }=(1-x{)}^{-2}$ and $y_1^{\prime \prime }=2(1-x{)}^{-3}$. Consequently the left-hand side of the ordinary differential equation iswhere $x\ne 1$.

Use reduction of order, writing $y=u/(1-x)$. A direct application of the product rule givesFor $v=u^{\prime }$ the resulting first-order ordinary differential equation is $v^{\prime }/v=2/(1-x)$ whenever $v\ne 0$. Integration yields $v=C(1-x{)}^{-2}$ and $u=A+B/(1-x)$; the case $v=0$ is already included. Henceon either interval $x<1$ or $x>1$. The two displayed solutions of a differential equation are linearly independent, so they span the solution space of a homogeneous linear differential equation on either interval. The point $x=1$ is a singular point of a differential equation and cannot be crossed by a nonzero member of this family that remains finite there.