Show that for $\eta > 0$, $\alpha_\eta=-\infty$ and for the Cauchy problem $z'=x-z^2,z(0)=\eta$ the saturated solution is defined up to $+\infty$.

Question

We consider the differential equation $y'=x+y^2$ and for any $\eta \geq 0$ we denote by $\phi_\eta:(\alpha_\eta,\beta_\eta)\rightarrow\mathbb{R}$ the unique saturated solution of the Cauchy Problem $y'=x+y^2$, $y(0)=\eta$. I am required to do the change of variable $z(x)=y(-x)$ and I succeded here, obtaining $y'(-x)=y^2(-x)-x$ and so $z'(x)=x-z^2(x)$, then I was required to represent the $0-isocline$ and it was also simple as I obtained a parabolla and also I was required to write the equation of the tangent in the point $(0,\eta)$ which was also pretty simple as I obtained $\phi'_\eta(0)=\eta^2$ and the equation of the tangent is $y=\eta+\eta^{2}x$.

Now, I am required to find $\phi^{(5)}_0(0)$( that $5$ indicates the number of differentiations), which puzzles me as I don't know the form of the solution and if it can be found out and also I am told to show that $\frac{1}{\eta}\geq\beta_\eta$

What I did not prove after the first edit : and that for $\eta > 0$, $\alpha_\eta=-\infty$ and for the Cauchy problem $z'=x-z^2,z(0)=\eta$ the saturated solution is defined up to $+\infty$.

At this last ones I am lost. Any help, please?

Edit: I have passed beyond finding $\phi^{(5)}_0(0)$ and I think I did show that $\frac{1}{\eta}\geq\beta_\eta$, now it remains to prove the last questions.

Answer 1

For $z>\sqrt{x}$ the solution is falling, so that once the solution is below $\sqrt{x}+1$, it stays below that curve.

Setting $z=\sqrt{x}+v$, then $$ v'=-v^2-2v\sqrt{x}-\frac1{2\sqrt{x}} $$ will be positive for $v=-1$ if $x>1$. So if one can ensure that $z(1)>0$, the solution stays above $\sqrt{x}-1$. As the solution is then bounded inside this channel, it can be extended infinitely.

Perhaps one can use the Taylor polynomial of the first part to fill the gap in this argument.

See also Properties of the solutions to $x'=t-x^2$, where it is simply observed that on the line $z=0$ the right side is positive, so no solution can cross from positive to negative values. Also that $b(x)=\sqrt{x}-\frac1{\sqrt{x}}$ is a lower bound for all solutions with $\eta=z(0)>0$.