# Very Awesome Nonlinear ODE

1. Jul 9, 2011

### TylerH

$$y^2=y' \Rightarrow y=\frac{y'}{y} \Rightarrow \int y dx = ln \left( y \right) \Rightarrow y=e^{\int y dx}=e^{\int e^{\int y dx} dx}=e^{\int e^{\int e^{\int y dx} dx} dx}=\cdots$$

Is that correct?

2. Jul 9, 2011

### hunt_mat

Why not just say:
$$\frac{y'}{y^{2}}=1\Rightarrow -\frac{1}{y}=x+C$$

3. Jul 9, 2011

### HallsofIvy

Staff Emeritus
Which is the same as y= -1/(x+C) so that $\int ydx= -ln(x+C)+ C_2$ so that
$e^{\int ydx}=C_2 \frac{1}{x+ C}= \frac{-1}{x+ C}= y$
so, yes, your chain of exponentials is correct- but a very complicated way of writing a very simple function.

Last edited by a moderator: Jul 9, 2011
4. Jul 9, 2011

### pmsrw3

This reminds me of a problem in a math contest I once saw. Solve:

$$x^{x^{x^{x^\ldots}}} = 2$$

The answer is $\sqrt{2}$.

5. Jul 9, 2011

### TylerH

Yeah, I found the real solution. But, it's the fact that something so contrived is equal to something so simple that makes it awesome.

6. Jul 9, 2011

### TylerH

That was the first thing that came to mind. Btw, you [STRIKE]would[/STRIKE] could solve that "infinite power tower" using (4) here: http://mathworld.wolfram.com/LambertW-Function.html.

Last edited: Jul 10, 2011
7. Jul 10, 2011

### HallsofIvy

Staff Emeritus
If
$$x^{x^{x^{x^\ldots}}}= 2$$
then
$$x^{\left(x^{x^{x^\ldots}}\right)}= x^2=2$$

More generally, if
$$x^{x^{x^{x^\ldots}}}= a> 0$$
then $x=\sqrt{a}$.

8. Jul 10, 2011

### mheslep

almost:
$x=\sqrt[a]{a}$

9. Jul 10, 2011

### pmsrw3

Actually, it should be $a^{1/a}$, no?

But this is true only if there is a solution. When I substitute $\sqrt{2}$ for x and iterate it, it does indeed converge to 2. But when I try it for cube root of 3, it doesn't converge to 3. It converges, but to 2.47805. What's more, when I ask Mathematica for $-\frac{\text{ProductLog}[-\text{Log}[z]]}{\text{Log}[z]}$, which is supposed to give the infinite power tower of z (see http://mathworld.wolfram.com/PowerTower.html), it does in fact come to 2.47805. MathWorld says it converges only up to $e^{1/e}$. The interesting thing is, $2.47805^{1/2.47805}$ equals the cube root of 3.