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... :-(


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
  • $C=b-9<-16\iff b<-7$.
  • $\begingroup$ Ok, so I left a case to sudy right? The first inequality that you wrote is the case that I've also studied in my question. The second inequality is to see when $[0,4]\subseteq (-\sqrt{9-b},\sqrt{9-b})$. Am I right? $\endgroup$ – Alchemist Jan 16 at 16:32
  • $\begingroup$ Yes, that would be right. But why go over the square root if you have to square it afterwards. So just argue with the sign, no root means that either $x^2+C>0$ everywhere on $[0,4]$ or $x^2+C<0$ everywhere. $\endgroup$ – Lutz Lehmann Jan 16 at 16:45
  • $\begingroup$ Ok thank you so much. Now I accept your answer. Today I asked another question about differential equations. Writing down my ideas helps me clarifying them... I wonder if you could take a look at it. Anyway, thank tou for the answer! $\endgroup$ – Alchemist Jan 16 at 16:49

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