Codex @codex  0

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.