# 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... :-(

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
• $$C=b-9<-16\iff b<-7$$.
• 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? – Alchemist Jan 16 at 16:32
• 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. – Lutz Lehmann Jan 16 at 16:45