# Maximal solution and interval of differential equation $y'=y\ln(1+y)$

Sorry for my bad english.

Let the differential equation defined by $\begin{cases} y'=y\ln(1+y) \\ y(0) = a > -1\end{cases}$.

We want to find the maximal solution defined on the maximal interval $J$. We can notice that $u_1=0$ is a maximal solution.

Let an other solution $(u_2,J)$ distinct of $u_1$. Then $\forall x \in J, u_2(x)>0$ OR $-1<u_2(x)<0$. In the second case, I found that $J=\mathbb{R}$ because $u_2$ is bounded.

In the first case, $\inf(J) = -\infty$ because $u_2$ is incrasing and bounded from below.

But how to study $\sup(J)$ ? Someone could help me ? Thank you in advance !

• Note that $u_1\equiv 0$ is a solution if and only if $a=0$. – Chee Han Apr 24 '17 at 23:34

## 1 Answer

If $y(x)>0$ for all $x\in J$ then $y(x)\ln(1+y(x))\le (1+y(x))\ln(1+y(x))$ and so $$\frac{1}{(1+y(x))\ln(1+y(x))}y'(x)\le 1.$$ Since $a>0$, integrating both sides you get
$$\ln(\ln(1+y(x)))-\ln(\ln(1+a))\le x,$$ which gives $$1+y(x)\le e^{\ln(1+a)e^{x}}.$$In particular, if $J$ is bounded from above, then so is $y$ but then, since it is increasing, there exists $$\lim_{x\to\sup J}y(x)=\ell<\infty$$ which is a contradiction since we could extend $y$ by considering the initial value problem $y(\sup J)=\ell$. Thus $J=\mathbb {R}.$