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For the D.E. $$y'=x^2+y^2$$ show that the solution with $y(0) = 0$ has a vertical asymptote at some point $x_0$. Try to find upper and lower bounds for $x_0$:

$$y'=x^2+y^2$$ $$x\in \left [ a,b \right ]$$ $$b> a> 0$$ $$a^2+y^2\leq x^2+y^2\leq b^2+y^2$$ $$a^2+y^2\leq y'\leq b^2+y^2$$ $$y'\geq a^2+y^2$$ $$\frac{y}{a^2+y^2}\geq 1$$ $$\int \frac{dy}{a^2+y^2}\geq \int dx=x+c$$ $$\frac{1}{a}\arctan \frac{y}{a}\geq x+c$$ $$\arctan \frac{y}{a}\geq a(x+c)$$ $$\frac{y}{a}\geq\tan a(x+c)$$ $$y\geq a\tan a(x+c)$$ $$a(x+c)\simeq \frac{\pi}{2}$$

But where to from here?

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1. $x_0$ exists

First note that $y'''(x)$ is increasing$^{[1]}$. It is also easy to see that $y'(0)=y''(0)=0$ but $y'''(0)=2$$^{[2]}$, so by Taylor's theorem$^{[3]}$, $$ y(x)=\frac{x^3}{6}y'''(c)\ge \frac{x^3}{3},\qquad (*) $$ for all $x>0$ such that $y$ is defined. Choose one such $x=\epsilon>0$. Then if $x>\epsilon$, we get $$ y'(x)\ge \epsilon^2+y(x)^2, $$ which, since $y(\epsilon)>0$, implies $y(x)\to\infty$ as $x\to x_0<\infty$ for some $x_0>\epsilon$.

Edits:

$[1]$: Since $y'(x)=x^2+y(x)^2\ge 0$, $y$ is increasing. Since $y\ge 0$ and $x\ge 0$, we have $y''(x)=2x+2y(x)y'(x)\ge 0$, so $y'$ is also increasing. In a similar way, we deduce that $y'''(x)\ge 0$ and $y^{(4)}(x)\ge 0$.

$[2]$: Since $y(0)=0$, we have $y'(0)=0$. Therefore, $y''(0)=2x+2y(x)y'(x)|_{x=0}=0$. On the other hand, $y'''(0)=2+2y'(x)^2+2y(x)y''(x)|_{x=0}=2$.

$[3]$: First note that $y$ is smooth. Indeed, since $y$ is continuous and $y'(x)=x^2+y(x)^2$, we see that $y'(x)$ is continuous. Since $y''(x)=2x+2y(x)y'(x)$ and the right hand side is continuous, so is $y''$. In a similar way, all derivatives of $y$ are continuous. Since $y$ is smooth, Taylor's theorem can be applied: $$ y(x)=y(0)+xy'(0)+\frac{1}{2}x^2y''(0)+\frac{1}{6}x^3 y'''(c),\qquad x>0, $$ where $c\in(0,x)$. But the first three terms are zero by [2], so (*) holds.

2. Lower bound:

Since a finite $x_0>0$ exists, we get $$ y'(x)\le x_0^2+y(x)^2, $$ which, since $y(0)=0$, implies $$ y(x)\le x_0 \tan (x_0\,x). $$ If it were true that $x_0^2<\pi/2$, then $y(x_0)<\infty$, so $x_0\ge\sqrt{\pi/2}=:z$.

3. Upper bound

For $x>z$, where $z$ is the lower bound, we have $$ y'(x)\ge z^2+y(x)^2, $$ which implies $$ y(x)\ge z\,\tan z(x+c), $$ where $$ c=-z+\frac{1}{z}\arctan\frac{y(z)}{z}\ge-z+\frac{1}{z}\arctan\frac{z^2}{3} $$ by inequality (*). Let $$ \zeta=\frac{\pi}{2z}-c\le \frac{\pi}{2z}+z-\frac{1}{z}\arctan\frac{z^2}{3}\approx 2.12. $$ Then $y(\zeta)$ does not exist, so $x_0<\zeta$. Note that $z\approx 1.25$.

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  • $\begingroup$ Can you flesh out your methods for arriving at (*)? $\endgroup$ – liana1000 Jul 7 '17 at 5:33
  • $\begingroup$ @liana1000 I added more details before 2.. Let me know if I can clarify. $\endgroup$ – user254433 Jul 7 '17 at 6:47
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The usual trick to get a better manageable equation out of this Riccati equation is to substitute $y=-\frac{u'}{u}$ which results in the linear ODE of second order

$$ u''+x^2u=0,\quad u(0)=1,\, u'(0)=0 $$

While this still does not lead to a symbolic solution without involving (very) special functions (Convert $\frac{d^2y}{dx^2}+x^2y=0$ to Bessel equivalent and show that its solution is $\sqrt x(AJ_{1/4}+BJ_{-1/4})$), one can easily find a power series solution $$ u(x)=1-\frac{x^4}{3·4}+\frac{x^8}{3·4·7·8}-\frac{x^{12}}{3·4·7·8·11·12}\pm… $$ This is an alternating series with eventually monotonically falling absolute values of the terms. For $x<\sqrt7$ one gets the bounds by partial sums $$ 1-\frac{x^4}{3·4}\le u(x)\le1-\frac{x^4}{3·4}+\frac{x^8}{3·4·7·8}. $$ The first positive root of $u(x_0)=0$ is the location of the first pole of $y$. From the bounds one gets the root bounds

$$ \sqrt[4\,]{12}\le x_0\le \sqrt[4\,]{16+4(3-\sqrt7)} $$

which numerically gives the interval $$ [1.8612097182041991,\; 2.042882110200651] $$ while the numerator $-u'(x)=\frac{x^3}{3}(1-\frac{x^4}{4·7}\pm…)$ has its first positive root above $\sqrt[4\,]{28}$.

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    $\begingroup$ A symbolic solution would have to use parabolic cylinder functions of a complex argument, that's not very practical. The smallest positive zero $x_0$ of $u(x)$ can be found numerically with Newton iteration. A nice detail: since $u''(x_0)=0$ due to the differential equation, we have cubic convergence in this case. The approximation $x_0=2.003147359426885$ can be obtained in three steps (starting from $1.8$), so the upper bound is rather good. $\endgroup$ – Professor Vector Jul 6 '17 at 11:24
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From numerical solution comes out that $x=2$ and $x=-2$ are vertical asymptotes. Trying to solve as a Bernoulli equation gives a mess and the substitution $w=\frac{1}{y}$ gives problems as $$w=\frac{1}{y};\;w'=-\frac{y'}{y^2}$$ Divide the original equation by $y^2$ $$\frac{y'}{y^2}=\frac{x^2}{y^2}+1\rightarrow -w'=x^2w^2+1$$ $$w'+x^2w^2=-1\rightarrow w(x)=c\;e^{-\frac{x^3}{3}}+\frac{e^{-\frac{x^3}{3}} x \Gamma \left(\frac{1}{3},-\frac{x^3}{3}\right)}{3^{2/3} \sqrt[3]{-x^3}}$$ The problem is now with the initial value, as $w\to\infty$ as $x\to 0$

Anyway the general solution is $$y=\frac{1}{w(x)}=\frac{3 e^{\frac{x^3}{3}}}{3 c+x E_{\frac{2}{3}}\left(-\frac{x^3}{3}\right)}$$ where $E_k(x)$ is the integral exponential function defined by $$E_k(x)=\int_1^{\infty }\frac{e^{-tx}}{t^k}\,dt$$ and has a vertical asymptote for any $c\in\mathbb{R}$

Hope this helps

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    $\begingroup$ This seems to be wrong, it looks pretty much like the general solution of $-w'=x^2w+1,$ instead. $\endgroup$ – Professor Vector Jul 6 '17 at 8:04

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