<s>[INST] Discussion on the prolongation of the solution I have the following

$$(PC)\begin{cases}y'=\dfrac{1}{x+1}y-\dfrac{4x}{x+1}y^2\\y(3)=\dfrac{2}{b} \end{cases}\text{ with }b\ne 0$$
Determine for which values of $b$, it exists at least one solution defined in $I=[0,4]$.

$\textbf{My attempt:}$ I noticed that the function $f(x,y):=\dfrac{1}{x+1}y-\dfrac{4x}{x+1}y^2$ is defined in $(-\infty,-1)\cup (-1,+\infty)$ and, since $3\in(-1,+\infty)$, I studied the Cauchy problem for $x>-1$.
The differential equation admits a unique local solution and a unique prolongation in its maximal domain.
We have a Bernoulli equation so I substituted $z(x)=\dfrac{1}{y(x)}$ and this substitution defines the new Cauchy problem:
$$(PC')\begin{cases}z'+\dfrac{1}{x+1}z=\dfrac{4x}{x+1}\\z(3)=\dfrac{b}{2} \end{cases}$$
and I got the solution $y(x)=\dfrac{x+1}{2x^2+2b-18}$, so I imposed $2x^2+2b-18\ne 0$ and $y(x)\ne 0$ for the substitution that I did$\implies|x|\ne\sqrt{9-b}$ and $x\ne -1$. Then I considered the solution in
$$(-\infty,-\sqrt{9-b})\cup (-\sqrt{9-b},\sqrt{9-b})\cup (\sqrt{9-b},+\infty)\bigcap (-1,+\infty).$$
If we want $I\subseteq I_{\text{SOL}}$ I think that it's correct to impose that $(-\sqrt{9-b},\sqrt{9-b})$ must contain $[0,4]\implies$ $$\begin{cases} -\sqrt{9-b}<0\text{ or}\\\sqrt{9-b}>4\end{cases}\implies\begin{cases}b<9\text{ or}\\b<-7\end{cases}.$$ For the other case I imposed that $[0,4]\subseteq (\sqrt{9-b},+\infty)$, so
$\sqrt{9-b}> 0$ which defines only the condition $b< 9$.
I always have difficulties in the study of the prolongation of the solution on a certain interval... :-(
 [/INST] You found correctly that
$$
(1+x)z(x)=2(x^2+C)\implies 4·\frac{b}{2}=2·(9+C)
$$
The solution $y$ becomes singular where $z=\frac1y$ has a root. As $(x+1)$ has no root on $[0,4]$, it remains to consider the other side. If $C>0$ then there is no problem. For negative $C$ to avoid roots on the interval one needs $C<-16$. So either

*

*$C=b-9>0\iff b>9$ or

*$C=b-9<-16\iff b<-7$.

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