Second-order nonlinear ordinary differential equation

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Homework Help Overview

The discussion revolves around a second-order nonlinear ordinary differential equation given by x''(t) = 1/(x(t)^2). Participants are attempting to find the function x(t) and are exploring various methods to approach the problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants have attempted using Laplace transforms and linear methods, but found them unhelpful. There are discussions about finding t(x) and the challenges of inverting this to obtain x(t). Some suggest multiplying both sides of the equation by x'(t) as a potential approach. Others mention substituting variables to create a separable equation and integrating twice. Questions arise about the elegance of these methods and the implications of boundary conditions on the solutions.

Discussion Status

The discussion is ongoing, with various methods being explored. Some participants express uncertainty about the effectiveness of their approaches, while others highlight the complexities introduced by the nonlinearity of the equation. There is no explicit consensus on a single method, but multiple lines of reasoning are being examined.

Contextual Notes

Participants note that the behavior of the solution may become complicated near x=0 and tends to a straight line for large values of x. There are also mentions of constants of integration that may affect the solutions, indicating that boundary conditions could play a significant role in simplifying the problem.

lewis198
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Homework Statement



Given the Second-order nonlinear ordinary differential equation

x''(t)=1/(x(t)^2)

Find x(t).

Homework Equations



I tried use Laplace transforms, and solving it using linear methods but that is not useful.

The Attempt at a Solution



I tried to find t(x) and got to dt=dx/((C-2GM/x)^0.5) or something like that.

I guess you could find t(x) then find [inverse t(x)] = x(t)
But I would like to know how to solve it properly really.
 
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lewis198 said:
x''(t)=1/(x(t)^2)

Find x(t).

Hi lewis198! :smile:

Standard trick: multiply both sides by x'(t) :wink:
 
similar outcome is to susbtitute to get a seprable equation then intgerate twice (ash means derivative w.r.t. t)
u = x'
then
x" = u' = (du/dx)(dx/dt) = (du/dx)
 
Hi lanedance, I tried that method before and got

(1/2)*(u^2)=1/(x^2)

dt=dx/((C-2GM/x)^0.5).

this will therefore give me an integral t(x).

But I need x(t). It will be quite messy doing the inverse won't it?

I'm not sure where to go from multiplying LHS and RHS by x'(t).

Is there a more elegant way to get x(t)? For example if I had

ax''+bx'+cx=f(x)

I could get y=A*e^(mt)+B*e(mt)

But since in my characteristic equation b=0 and c=0 my m quadratic equation is void.
 
I'll just find t(x) then t-1(x)
 
lewis198 said:
x''(t)=1/(x(t)^2)

lewis198 said:
Hi lanedance, I tried that method before and got

(1/2)*(u^2)=1/(x^2)

(try using the X2 tag just above the Reply box :wink:)

No, the RHS is wrong and you've left out the constant of integration. :wink:
 
yeah I'm not too sure, don't know if Tim has any other ideas, but simple generic general solutions don't always apply to non-linear de's - this one is gets a little crazy near x=0, and tends to a straight line for x>>1

So the general solution may not be able to be solved simply for t^(-1). That said if you have the right boundary conditions, this one could simplify a bit... (in particular if you could set the 1st constant of integration to zero)
 
Last edited:
lanedance said:
similar outcome is to susbtitute to get a seprable equation then intgerate twice (ash means derivative w.r.t. t)

ash?
 
Mark44 said:
ash?
there's a fire... few mistakes, so i clarified below
--------------------------------------------------------------------------------

similar outcome is to susbtitute to get a separable equation then integrate twice (dash means derivative w.r.t. t)
[tex]u = x' = \frac{dx}{dt}[/tex]
then
[tex]x'' = u' = \frac{du}{dt} =\frac{du}{dx} \frac{dx}{dt} = \frac{du}{dx}u = \frac{1}{x^2}[/tex]
 

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