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Dear all, I'm trying to find the general second-order ODE admitting

$$x* = \alpha x$$ $$y* = \alpha^{k} y$$

and reduce it to first-order plus a quadrature.

The general solution I found by using differential invariants is

$$y'' = \frac{1}{x} + H(y, - \frac{y'}{x})$$

where H is an arbitrary function.

But I have NO clue how to reduce it.

Please help me out here. Thank you in advance.

Edit: I made a big mistake here. Since the unit element of the scaling group is not 0, first I should transform into

$$x* = e^{\epsilon} x$$ $$y* = e^{\epsilon \cdot k} y$$ where $$\alpha = e^{\epsilon}$$.

Then the differential invariants are characterized by:

$$\frac{dx}{x}=\frac{dy}{ky}=\frac{d y'}{(k-1)y'} $$.

And the general solution is found as: $$y''=(k-1)\frac{y'}{x}-\frac{y'^{2}}{y}(k-x\frac{y'}{y})H(u,v)$$ where $$u(x,y) = \frac{x^{k}}{y}$$ and $$v(x,y,y') = \frac{x^{k-1}}{y'}$$

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And I dont know why the formulae cannot be displayed in the right way. Sorry for that. :) –  newbie Mar 20 '11 at 5:31
    
If you precede a line by four spaces, it gets displayed verbatim. To get a displayed equation, you can use double dollar signs instead. –  joriki Mar 20 '11 at 6:24
    
I am having some trouble seeing how you got this general solution. Admittedly, it's been a long while since I studied differential invariants. Can you explain how you got there? In particular, I find it hard to believe that the answer does not depend on $k$. –  Sam Lisi Mar 24 '11 at 12:12
    
@Sam Lisi: As I edited above, I made an egregious mistake. Sorry for that :( –  newbie Apr 5 '11 at 1:17
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