Solve complex 2nd order differential equation

[tetris11]

Homework Statement

[tex]t''[x] = \frac{2}{t} t'[x]^{2}[/tex]

The Attempt at a Solution

This is not your basic 2ODE, since I can't separate the components into y'', y' and y.

Help?

I've so far tried:
[tex]\frac{d^{2}t}{dx^{2}}=\frac{2}{t}(\frac{dt}{dx})^{2}[/tex]

[tex]\frac{dt}{dx}=\frac{2}{t}(\frac{dt}{dx})^{2}dx[/tex]

[tex]dt=\frac{2}{t}(\frac{dt^{2}}{dx})[/tex]

[tex]\frac{1}{2}dx=\frac{1}{t}dt[/tex]

[tex]\frac{1}{2}x+k=ln[t]+c[/tex]

[tex]t = e^{k-c}e^{\frac{1}{2}x[/tex]

somehow this doest seem right...

 

[fzero]

How about writing the equation as

[tex]\frac{t''}{t'} = 2 \frac{t'}{t}?[/tex]

 

[tetris11]

So I then get:

ln(t') +c = 2ln(t) +k ?

2ln(t) - ln(t') -R = [tex]ln(\frac{t^{2}}{t'}) = ln(R)[/tex] (my maths is pretty rusty)

 

[fzero]

So I then get:

ln(t') +c = 2ln(t) +k ?

2ln(t) - ln(t') -R = [tex]ln(\frac{t^{2}}{t'}) = R[/tex] (my maths is pretty rusty)

Yep, it might be easier to write this in the form

[tex]ln(\frac{t^{2}}{t'}) = \ln C,[/tex]

since the next step is to manipulate this into an easy first-order equation for t.

 

[tetris11]

[tex] ln(\frac{t^{2}}{t'}) = ln(R)[/tex]

[tex] t^{2} = R\frac{dt}{dx}[/tex]

[tex]\int\frac{1}{t^{2}} dt =\int -R dx [/tex]

[tex]\frac{-1}{t} +k = -Rx +c [/tex]

[tex] t = \frac{A}{Rx+j} [/tex]

thanks dude!