[net.math] help solve this?

pete@hao.UUCP (Pete Reppert) (01/23/85)

 I've been challenged to solve the following equation 
    y^2 dy/dx - Ay + B(1-x) = 0

    y(x), x on (0,1), A & B constant.
  
    y(1) = 0
 
 If I can get the solution, my friend promises a 'prize'
 of some sort. He said I can use ANY means to solve it
 which means the net is fair game. I'll be happy to give
 the first  one to get it a cut if the prize turns
 out to be something divisible (knowing my friend, it might
 not be something you'd WANT). 

   p.s. - is it true that you couldn't prove there was no solution
	  to such an equation?
  
		 Thanks!

-- 
 Pete Reppert
 HAO/NCAR
 PO BOX 3000
 Boulder, Colorado 80307

gwyn@brl-tgr.ARPA (Doug Gwyn <gwyn>) (01/29/85)

>  I've been challenged to solve the following equation 
[followed by a DE with boundary conditions]

What would constitute a "solution", anyway?  An explicit
y(x) written in terms of well-known functions, or what?

seshacha@uokvax.UUCP (02/06/85)

     I am not a mathematician,however,I would like to recommend the
     following,
                   
     Consider y = Summation of (Am*X^m) m running from 1 through infinity.
     The differential of y would be a similar summation series as above with
     the limits changed.

     Substitute the series exprssions in the original differential equation
     and expand to the terms to certain degree ( till your patience dies out)
     Assuming that the equation has solution,equate the co-efficients of the
     like powers assimilated to zero.Solve for the co-efficients and you have
     your solution.

     Incase you don't get the solution,please don't blame me.

     Lord bless my lunatic prudence!

leimkuhl@uiucdcsp.UUCP (02/10/85)

   This is actually a rather interesting problem.


   If we try the approach suggested above, we get the rather
nasty recurrence relation:

	a1*a0**2 - A*a0 + B = 0
	a2*a0**2 + 2*a0*a1**2 - A*a1 - B = 0

        Sum(i=0,n; a(n-i+1)*Sum(j=0,i; aj*a(i-j)))  -  A*an = 0
                for n=2,3,..

   This is too hard to solve I think.


   

   Let L(y) = y'*y**2 - Ay

   The homogeneous equation L(y)=0 has the relatively simple solution

           y**2 = Ax + K 

   --a family of parabolas.

   When we progress to the seemingly still simple problem L(y)=B, we
   get the solution

           (u**2)/2 - 2*B*u + (B*B)*log u = k + x*A**3,

   where u = Ay + B


   This is already enormously complicated and impossible to solve 
   explicitly for u.


   Now the given problem is to solve L(y) = B - Bx  which obviously
   is not going to have a simple solution (since L(y)=B is just a
   special case).

   In fact, we can rule out simple forms of a solution involving 
   complex exponentials and (what is the same thing, trigonemetric
   functions).  The method of judicious guessing is not likely to
   yield anything of value.

   Some other suggestions?  

-Ben Leimkuhler