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Second-order nonlinear ordinary differential equation

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lewis198
#1
Nov20-09, 01:16 AM
P: 96
1. The problem statement, all variables and given/known data

Given the Second-order nonlinear ordinary differential equation

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

Find x(t).


2. Relevant equations

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


3. 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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tiny-tim
#2
Nov20-09, 03:00 AM
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Quote Quote by lewis198 View Post
x''(t)=1/(x(t)^2)

Find x(t).
Hi lewis198!

Standard trick: multiply both sides by x'(t)
lanedance
#3
Nov20-09, 03:07 AM
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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)

lewis198
#4
Nov21-09, 11:32 AM
P: 96
Second-order nonlinear ordinary differential equation

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.
lewis198
#5
Nov25-09, 10:01 AM
P: 96
I'll just find t(x) then t-1(x)
tiny-tim
#6
Nov25-09, 11:38 AM
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Quote Quote by lewis198 View Post
x''(t)=1/(x(t)^2)
Quote Quote by lewis198 View Post
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 )

No, the RHS is wrong and you've left out the constant of integration.
lanedance
#7
Nov25-09, 06:43 PM
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P: 3,307
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)
Mark44
#8
Nov25-09, 09:30 PM
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Quote Quote by lanedance View Post
similar outcome is to susbtitute to get a seprable equation then intgerate twice (ash means derivative w.r.t. t)
ash?
lanedance
#9
Nov25-09, 10:10 PM
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Quote Quote by Mark44 View Post
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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