Second-order nonlinear ordinary differential equation

[lewis198]

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.

 

[tiny-tim]

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

Find x(t).

Hi lewis198!

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

 

[lanedance]

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]

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]

I'll just find t(x) then t-1(x)

 

[tiny-tim]

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

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]

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]

similar outcome is to susbtitute to get a seprable equation then intgerate twice (ash means derivative w.r.t. t)

ash?

 

[lanedance]

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)
$$u = x' = \frac{dx}{dt}$$
then
$$x'' = u' = \frac{du}{dt} =\frac{du}{dx} \frac{dx}{dt} = \frac{du}{dx}u = \frac{1}{x^2}$$