= Solution
{c}
Near $(1,1)$ the <solution of a differential equation> has $y>0$, so the <separable differential equation> gives
$$
\frac{y}{y+1}\,dy=x\,dx,\qquad
y-\log(1+y)=\frac{x^2}{2}+\frac12-\log2.
$$
Equivalently its trajectory lies on the <level set>
$$
\boxed{h(x,y):=y-\log(1+y)-\frac{x^2}{2}=\frac12-\log2.}
$$
On $y>0$, the <function> $G(y)=y-\log(1+y)$ is <strictly increasing> because $G'(y)=y/(1+y)>0$, and its range is $(0,\infty)$. Thus the positive branch exists exactly where
$$
\frac{x^2}{2}+\frac12-\log2>0.
$$
The <connected component> containing $x=1$ is
$$
\boxed{x>x_*:=\sqrt{2\log2-1}\approx0.621526.}
$$
The <implicit function theorem> gives a smooth <solution of a differential equation> at every such $x$. It remains finite at every finite $x>x_*$, because $G$ increases without bound. As $x\downarrow x_*$, we have $y\downarrow0$, and the original right side $x+x/y$ becomes singular. Hence this is the maximal interval $(x_*,\infty)$ through the prescribed <initial condition>.
In particular, \b[the solution is not defined for all $x\in(0,2)$]. It cannot pass through $x=x_*$ as a <classical solution>, since its required value there is $y=0$.
Back to article page
