Find the transformations that transform $$X = -y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$$ to $\bar{X}$ = $\frac{\partial}{\partial s}$ That is X(s)=1 and X(t)=0
I know we have to solve differential equations for dependent and independent invariants then express s and t in terms of x and y, but I'm having trouble because of the solution to the differential equation involving s. Here's what I've done so far:
$-y \frac{\partial s}{\partial x} + x \frac{\partial s}{\partial y} =1$ and $-y \frac{\partial t}{\partial x} + x \frac{\partial t}{\partial y} =0$ therefore we solve
$\frac{dx}{-y} = \frac{dy}{x} (= \frac{ds}{1})$ and $\frac{dx}{-y} = \frac{dy}{x} (= \frac{dt}{0})$
Solving $\frac{dx}{-y} = \frac{dy}{x}$ we get xdx=-ydy and integrating gives us $\frac{1}{2} x^2 + \frac{1}{2} y^2= b_1$ since $\frac{1}{2}$ is a constatnt we can write the independent invariant as $b_1 = x^2 + y^2$ Now we solve $\frac{dx}{-y} = \frac{ds}{1}$ and $\frac{dy}{x} = \frac{ds}{1}$ We are now working with three variables so we look at $\frac{dy}{ds} = x$ and $\frac{dx}{ds} = -y$. Taking the second derivative of the first one and then substituting in the second we have $\frac{d^2y}{ds^2} = \frac{dx}{ds} =-y$ Therefore $\frac{d^2y}{ds^2} +y =0$ which a differential equation of the form $(D^2+1)y=0$ which has non real roots so the solution is of the form $y=Acos(s) + Bsin(s)$. This is where I'm stuck I know I have to find a dependent invariant $k_1 = f(b_1)$ then solve for s but I'm not sure how to find a dependent invariant from the solution to the differential equation.
Any thoughts would be greatly appreciated.
Once you got $x^2+y^2=C$, you are tempting to substitute it into $$\frac{dy}{x}=\pm\frac{dy}{\sqrt{C^2-y^2}}=ds$$ and thus $$s=\pm\arcsin\frac yC$$