I Derivation of planetary orbit equation with Lagrangian

it is because ##\frac{\partial L}{\partial \varphi}=0## and due to the Lagrange equation: ##\frac{d}{dt}\frac{\partial L}{\partial \dot\varphi}=\frac{\partial L}{\partial \varphi}## one gets ##\frac{d}{dt}\frac{\partial L}{\partial \dot\varphi}=0##

 

The most general solution of the differential equation
$$\frac{\mathrm{d}^2 w}{\mathrm{d} \varphi^2}=-w$$
is given by
$$w(\varphi)=A_1 \cos \varphi+A_2 \sin \varphi,$$
where ##A_1## and ##A_2## are constants of integration. This follows, because for a homogeneous 2nd-order linear differential equation the solution is the linear combination of two arbitrary linearly independent solutions, and cos and sin fulfill these conditions.

Now you can write this general solution also in another form,
$$w(\varphi)=A \cos(\varphi+\delta),$$
where ##A## and ##\delta## are constants. To see this just use the addition theorem for cos
$$w(\varphi)=A \cos \delta \cos \varphi - A \sin \delta \sin \varphi.$$
Comparing with the general solution above you have
$$A_1=A \cos \delta, \quad A_2=-A \sin \delta,$$
and you always find ##A>0## and ##\delta \in (-\pi,\pi]## to fulfill these equations. First we have
$$A=\sqrt{A_1^2+A_2^2}, \quad \delta =-\mathrm{sign} A_2 \arccos \left (\frac{A_1}{\sqrt{A_1^2+A_2^2}} \right).$$
So indeed the given solution is the most general one for the differential equation for ##w##.

 

No, both equations are equivalent, as demonstrated in the previous posting. You can always map from the pair ##(A_1,A_2)## of integration constants to the pair ##(A,\delta)##. It just depends on what you consider more convenient. In this case, it's easier to see that the orbit is a conic section (ellipse, parabola, hyperbola), using the ##(A,\delta)## notation.