Solve complex 2nd order differential equation

[tetris11]

Homework Statement

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

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:
\frac{d^{2}t}{dx^{2}}=\frac{2}{t}(\frac{dt}{dx})^{2}

\frac{dt}{dx}=\frac{2}{t}(\frac{dt}{dx})^{2}dx

dt=\frac{2}{t}(\frac{dt^{2}}{dx})

\frac{1}{2}dx=\frac{1}{t}dt

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

t = e^{k-c}e^{\frac{1}{2}x

somehow this doest seem right...

 

[fzero]

How about writing the equation as

\frac{t''}{t'} = 2 \frac{t'}{t}?

 

[tetris11]

So I then get:

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

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

 

[fzero]

So I then get:

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

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

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

ln(\frac{t^{2}}{t'}) = \ln C,

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

 

[tetris11]

ln(\frac{t^{2}}{t'}) = ln(R)

t^{2} = R\frac{dt}{dx}

\int\frac{1}{t^{2}} dt =\int -R dx

\frac{-1}{t} +k = -Rx +c

t = \frac{A}{Rx+j}

thanks dude!