Path of particle under gravity

Question

If a particle is subjected to gravity then $$\frac{∂^2 u}{∂\theta^2} +u = \frac{GM}{h^2} $$ where $$ u = \frac{1}{r}$$ and $$h = r^2\dot{\theta}.$$ If you solve this you get $$u = A\sin\theta+B\cos\theta + \frac{GM}{h^2}.$$ But the general solution for this is just in terms of $\cos\theta$ because then you have the eqn for a conic. So why is $A = 0$?

Answer 1

You can also write the general solution to this differential equation (under a fixed set of polar coordinates) as $$u = C_1 \cos(\theta+C_2) + \frac{GM}{h^2}$$ by defining $C_1=\sqrt{A^2+B^2}$ and $C_2=\tan^{-1}\left(\dfrac{A}B\right)$.

However, by redefining $\theta=0$ appropriately, we can have the same solution with $C_2=0$, leaving us with only a cosine term.