I am interested in solving a differential equation from the two-body problem. This is not homework, and I just thought of this on my own during physics class one day. However, after many attempts, I still cannot find a way to go at this. I know Wikipedia does something with the center of mass of the system but that is not a solution that satisfies me or one that I understand. That's why I'm asking in Math.SE and not Physics, which would just tell me to look at the com. So, if the following is solvable (general solution), please, let me know how.
For two masses $m_1$ and $m_2$, and constant $G$, we know that $$ m_1\frac{d^2\vec r_1}{dt^2}=G\frac{m_1 m_2 (\vec r_1-\vec r_2)}{|\vec r_1-\vec r_2|^3} $$And that $$ m_1\frac{d^2\vec r_1}{dt^2}=-m_2\frac{d^2\vec r_2}{dt^2} $$ My approach:
Let $\vec v_n$ correspond to $\frac{d\vec r_n}{dt}$ $$ \int_0^t\int_0^tm_1\frac{d^2\vec r_1}{dt^2}dtdt=\int_0^t\int_0^t-m_2\frac{d^2\vec r_2}{dt^2}dtdt $$ $$ \int_0^tm_1(\vec v_1(t)-\vec v_1(0))dt=\int_0^t-m_2(\vec v_2(t)-\vec v_2(0))dt $$ $$ m_1(\vec r_1(t)-r_1(0)-t\vec v_1(0))=-m_2(\vec r_2(t)-r_2(0)-t\vec v_2(0)) $$ $$ \frac{-m_1}{m_2}(\vec r_1(t)-\vec r_1(0)-t\vec v_1(0))+\vec r_2(0)+t\vec v_2(0)=\vec r_2(t) $$ Now that $\vec r_2(t)$ can be expressed with $\vec r_1(t)$ and constants, I replace it in the DE. For brevity consider that for a function $f$, $\vec r_2 = f(\vec r_1)$ $$ m_1\frac{d^2\vec r_1}{dt^2}=G\frac{m_1 m_2 (\vec r_1-f(\vec r_1))}{|\vec r_1-f(\vec r_1)|^3} $$ Now, convert to $x_1$ and $y_1$ $$ m_1\frac{d^2x_1}{dt^2}=G\frac{m_1 m_2 (x_1-f(x_1))}{D} $$ $$ m_1\frac{d^2y_1}{dt^2}=G\frac{m_1 m_2 (y_1-f(y_1))}{D} $$ D is the denominator: $$ D=|\vec r_1-f(\vec r_1)|^3=((x_1+\frac{m_1}{m_2}x_1 +c_1 )^2+(y_1+\frac{m_1}{m_2}y_1 +c_2 )^2)^{3/2} $$ Where $c_1$ and $c_2$ in the above are constants dependent on the initial conditions.
I can't even imagine beginning to take the Laplace transform of those equations, let alone using it to solve for $x_1$ and $y_1$.
Any suggestions?
