2nd order differential equation

1. Sep 20, 2003

alphy

Hi,

When I was younger, a teacher of mine gave me
this problem for training:

This was a test for the admission to a well known
french Engineering School. What was interesting was that
the test contains a question that was unsolved at the
time this examination was given. If someone has the solution,
I'm very interested to know it.

Summary: the goal is to find a solution of
xy''+2y' +x/y=0 defined on [0,1] such that y(1)=e where e is a
given real number. For that purpose one sets g_0=e and
g_{n+1} = e+T(1/g_n) where
T(f)(x) = (1/x-1)\int_0^x t^2f(t)dt + \int_x^1 (t-t^2)f(t)dt

g_{2p} converges to g and g_{2p+1} converges to G. The open
question is to show that g=G ....

Any idea is welcome

2. May 31, 2004

ichiro_w

I don't pretend to understand what this is all about, but make the following query/observations:

1)Is it correct to note that this is a non-linear DE in that the power (-1) of 1/y disqualifies it as linear.

2)It appears to be a second order DE in that y'' is the highest order derivative present. As such, is it not correct that 2 initial conditions are required for solution? The author provides y(1) = e which is one initial condition.

The rest makes little sense to me, no disrespect intended.

ichiro

Last edited: May 31, 2004
3. Jun 1, 2004

HallsofIvy

Staff Emeritus
You are correct that this is a non-linear second order differential equation. In general it is not possible to find a closed form solutionto such a problem and, in these questions, we are not asked to solve the equation but to find some properties of the equation. The last part asks for a graph of [all] solutions that satisfy y(1)= e. There are, of course, an infinite number of such solutions.

4. Jun 1, 2004

ichiro_w

$$xy'' + 2y'+ \frac{x}{y}= 0\quad$$ can be rewritten as $$\quad y'' +2 \frac{y'}{x}+ y^{-1}=0$$
Here is how I think the query reads:
$$xy''+2y'+\frac{x}{y}=0\mbox{ defined on } [0,1] \mbox{ such that } y(1)=e$$
where e is a given real number. For that purpose one sets
$$g_0=e \mbox{ and }g_{n+1}=e+T(\frac{1}{g_n})\mbox{ where }T[f(x)]=(\frac{1}{x-1})\int_0^xt^2f(t)\;\mathrm{dt}$$
$$+\int _x^1(t-t^2)f(t)\;\mathrm{ dt}$$

$$g_{2p}$$ converges to g and $$g_{2p+1}$$ converges to G. The open question is to show that g=G...

So, if one agrees with the way I have written the orginal query and recognizes the elaborate T[ransform?] function for $$g_{n+1}$$ ,
why not illuminate the rest of us as the the thrust of this problem?

Last edited: Jun 11, 2004