Say we have a stationary particle located at $(0,0)$ with a positive unit charge, and a stationary particle located at $(a,0)$ with charge a negative unit charge. I want to find the field lines associated with the force field generated by the two charges.
Suppose we have a charge with positive unit charge located at $(x_0,y_0)$. The force acting on this charge can be described with the following (disgusting) vector: $$\left(\frac{x_0}{\left(x_0^2+y_0^2\right)^{3/2}}+\frac{a-x_0}{\left((x_0-a)^2+y_0^2\right)^{3/2}},\frac{y_0}{\left(x_0^2+y_0^2\right)^{3/2}}-\frac{y_0}{\left((x_0-a)^2+y_0^2\right)^{3/2}}\right)$$ This yields the following (disgusting) differetial-equation for the field lines associated with the force field: $$y'\left(\frac{x}{\left(x^2+y^2\right)^{3/2}}+\frac{a-x}{\left((x-a)^2+y^2\right)^{3/2}}\right)=\frac{y}{\left(x^2+y^2\right)^{3/2}}-\frac{y}{\left((x-a)^2+y^2\right)^{3/2}}$$ I can sort of simplify (but not really): $$y'\left(x\left((x-a)^2+y^2\right)^{3/2}+(a-x)\left(x^2+y^2\right)^{3/2}\right)=y\left(\left((x-a)^2+y^2\right)^{3/2}-\left(x^2+y^2\right)^{3/2}\right)$$ Unfortunately I have no idea how to solve this thing. Any help would be very much appreciated.
I have posted the exact same question on physics.stackexchange
Update: doing the same thing in polar coordinates yields the following differential equation (I think). $$\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}\left(\frac{\cos\theta}{r^2}+\frac{1}{\left(r^2\sin^2\theta+\left(a-r\cos\theta\right)^2\right)\sqrt{\left(\left(\frac{r\sin\theta}{a-r\cos\theta}\right)^2+1\right)}}\right)=\frac{\sin\theta}{r^2}-\frac{r\sin\theta}{\left(a-r\cos\theta\right)\left(r^2\sin^2\theta+\left(a-r\cos\theta\right)^2\right)\sqrt{\left(\left(\frac{r\sin\theta}{a-r\cos\theta}\right)^2+1\right)}}$$ I think the above only works for points whose $x$-coordinate is less than $a$.