I have the differential equation

$$y''y+n(y')^2=0\tag 1$$

I haven't found any references to equations of this type by searching. Does anyone have a suggestion as to how to proceed?

It comes about from the following progression:

$$y=(y')^n-\frac 1n\tag 2\\ y'=ny''(y')^{n-1}\implies1=ny''(y')^{n-2}\\ 0=n(n-2)(y'')^2(y')^{n-3}+ny'''(y')^{n-2}\\ 0=(n-2)(y'')^2+y'''y'$$

Taking $y$ in place of $y'$ and $n$ in place of $n-2$ I get the form shown in $(1)$. Are there any good ways to solve any of the other intermediate forms, or even the initial form shown in $(2)$?

Further constraints:

$y(0)\approx 1.27$ is an otherwise-unknown constant that I am attempting to build an expression for.

The initial problem is the recurrence relation with $a_1\approx 1.27$ the unknown constant and

$$a_{n+1}=(a_n)^{n+1}-\frac 1{n+1}$$

with the additional constraint $\lim_{n\to\infty}a_n=1$.

  • $\begingroup$ Hm, it's not evident that the problem is correctly posed. This might very well be the case where the $\lim=1$ for all initial values (or an interval of them). $\endgroup$ – TZakrevskiy Nov 27 '13 at 16:16
  • $\begingroup$ I think $a_1\approx 1.27$ is the only solution that is monotone-decreasing for this sequence. $\endgroup$ – abiessu Nov 27 '13 at 16:29
  • $\begingroup$ This somehow reminds me of the Mandelbrot set. $\endgroup$ – TZakrevskiy Nov 27 '13 at 16:31
  • $\begingroup$ Good point! I didn't think of that... The only differences are the limitation to real numbers and the increasing exponent between iterations... $\endgroup$ – abiessu Nov 27 '13 at 16:34

Have you thought about doing these:

$$y'=u\to y''=u\frac{du}{dy}$$ and so $$y''y+n(y')^2=0\to u(u'(y)y+nu)=0$$

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  • $\begingroup$ Perhaps you could expand this a bit further? I'm not following what the next step might be... $\endgroup$ – abiessu Nov 27 '13 at 16:01
  • $\begingroup$ It gives us $y'=0$ or we have $\frac{du}{u}=\frac{-n}{y}$. This is a separable ODE. $\endgroup$ – mrs Nov 27 '13 at 16:04
  • $\begingroup$ Thank you, that makes more sense now, thus giving the above solution $\ln y'=-n\ln y+c$. And then perhaps one more step to evaluate? $\endgroup$ – abiessu Nov 27 '13 at 16:24
  • $\begingroup$ @abiessu: Yes. ${}$ $\endgroup$ – mrs Nov 27 '13 at 16:26
  • $\begingroup$ @B.S.: You made it easy to understand +1 $\endgroup$ – Amzoti Nov 28 '13 at 7:01

hint: as long as $yy' \ne 0$ you can divide through by it, giving: $$\frac{y''}{y'} + n\frac{y'}{y} = 0 $$ giving $ln\; y' = -n \;ln \; y + c$

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  • $\begingroup$ Does the next step go like this: $\frac {y'}{y^n}=e^c$ and thus $-\frac 1{y^{n-1}}=e^{c_1}x+c_2$? $\endgroup$ – abiessu Nov 27 '13 at 16:26

For Cauchy problem with initial data in zero you can multiply everything by $(n+1)y^{n-1}(t)$ because, as you say, the initial data is not zero. You get $$(y^{n+1})''=0,$$ therefore, $y^{n+1}(t)=a+bt$. Now you solve for $a$, $b$, which should give

$$y(t) = \left(y^{n+1}(0)+t(n+1)y^{n/(n+1)}(0)y'(0)\right)^{1/(n+1)}.$$

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  • $\begingroup$ The difficulty with this is that $y(0)$ is an undefined constant ($\approx 1.27$) that I am attempting to solve for... $\endgroup$ – abiessu Nov 27 '13 at 16:02
  • $\begingroup$ So, if you want to find $y(0)$, then what data do you have initially? Values in other points? Estimations of behaviour? $\endgroup$ – TZakrevskiy Nov 27 '13 at 16:12
  • $\begingroup$ I edited the question to include the sequence where I got the problem; the limit of my sequence is $\lim a_n=1$ for recurrence $a_{n+1}=(a_n)^{n+1}-\frac 1{n+1}$. $\endgroup$ – abiessu Nov 27 '13 at 16:14

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