Ordinary Differential Equation, Second Order

Date: 10/26/2002 at 10:17:38
From: Stefan Hau-Riege
Subject: Ordinary differential equation, second order

Hi,

I've been looking for solutions to the differential equation

   y'' = x^2 / y^2

I only came up with the solution y(x) = x^(4/3). However, I am looking 
for solutions with y(0) > 0 and y'(0) = 0.  Any ideas?

Thanks,
-Stefan

Date: 12/10/2002 at 17:04:39
From: Doctor Nitrogen
Subject: Re: Ordinary differential equation, second order

Hi, Stefan:

I think you might have to prove the following conjecture in order to 
solve the D.E., assuming, of course, y is a continuous, twice  
differentiable function; that is, you must prove the conjecture below 
for the problem

   y'' = x^2/y^2, y(0) > 0, y'(0) = 0:

Conjecture: IF K(y), where K(y) is a continuous, twice differentiable 
implicit function of y and where y = y(x), is a continuous, twice 
differentiable function of x such that

   [A] K'(y(0)) = 0,

   [B] K''(y(x)) = (2y)(y')^2,

then the solution to

     y'' = x^2/y^2, y(0) > 0, y'(0) = 0

is   y^3/3 = x^4/12 + K(y(x)).

If this Conjecture is true, then here is how you could solve your 
D.E.:

[1] Let   dy/dx = (1/3)(x^3/y^2) + C_1(y(x)),

where C_1 is some function of y(x), and assume a solution

   y/3 = x^4/(12y^2) + C(y(x)), 

or, rearranged,

[2]  y^3/3 = x^4/12 + y^2C(y(x))

[3]  = x^4/12 + K(y(x)), 

where  K(y(x)) = y^2C(y(x)).

If the conjecture is right and [A] above is true, then differentiating 
[2]-[3] implicitly twice w.r.t. x:

[4]  (y^2)y' = x^3/3 + (dK/dy)(dy/dx)

[5]  2y(y')^2 + (y^2)(y'') = x^2 + d/dx[(dK/dy)(dy/dx)].

Now when x = 0, Equation [4] becomes 0, as is required, since 

y'(0) = 0, and if 

   d/dx[(dK/dy)(dy/dx)] = 2y(y')^2,

then you can cancel it out on both sides of [5] to obtain

[6]  (y^2)(y'') = x^2, which reduces to

[7]  y'' = x^2/y^2.

All this of course depends on the conjecture stated above actually 
being correct.

All I can say about equations [2]-[3] is that they were not easy for 
me to come by. 

I hope this was some help. You are welcome to return to The Math 
Forum/Doctor Math whenever you have any math-related questions.

- Doctor Nitrogen, The Math Forum
  http://mathforum.org/dr.math/