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!