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Secondorder nonlinear ordinary differential equation 
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#1
Nov2009, 01:16 AM

P: 96

1. The problem statement, all variables and given/known data
Given the Secondorder 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/((C2GM/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. 


#2
Nov2009, 03:00 AM

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P: 26,157

Standard trick: multiply both sides by x'(t) 


#3
Nov2009, 03:07 AM

HW Helper
P: 3,307

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) 


#4
Nov2109, 11:32 AM

P: 96

Secondorder nonlinear ordinary differential equation
Hi lanedance, I tried that method before and got
(1/2)*(u^2)=1/(x^2) dt=dx/((C2GM/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. 


#5
Nov2509, 10:01 AM

P: 96

I'll just find t(x) then t1(x)



#6
Nov2509, 11:38 AM

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No, the RHS is wrong and you've left out the constant of integration. 


#7
Nov2509, 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 nonlinear 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) 


#8
Nov2509, 09:30 PM

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#9
Nov2509, 10:10 PM

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P: 3,307

 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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