Ordinary Differential Equation, Second OrderDate: 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/ |
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