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 $$v(x,y,y') = \frac{x^{k-1}}{y'}$$