Please solve this equation for r. I have the following equation:
$$A^2t^2\space dt^2=-c^2dt^2+\frac{dr^2}{1-Kr^2}+r^2(d\theta^2+\sin^2 \theta\space d\phi^2)$$
The angular separation isn't used for the measurements we're taking, so $d\theta$ and $d\phi$ are zero.  So the first simplification is easy:
$$A^2t^2\space dt^2=-c^2dt^2+\frac{dr^2}{1-Kr^2}$$
Can someone walk me through the solution to this equation for r?  I'm pretty sure it will have the general form of
$$r=\int_{t_1}^{t_0}(c+A t)\space dt$$
But I'm not exactly sure how the curvature, (${1-Kr^2}$), will factor into the solution.

$$(A^2t^2+c^2) dt^2=\frac{dr^2}{1-kr^2}$$
The ODE is separable :
$$\pm\int \sqrt{A^2t^2+c^2}dt=\pm\int\frac{dr}{\sqrt{1-kr^2}}+\text{constant}$$
$$\frac12 t \sqrt{A^2t^2+c^2}+\frac{c^2}{2A}\ln\left(A^2t+A\sqrt{A^2t^2+c^2}\right)=\pm\frac{1}{\sqrt{k}}\sin^{-1}(r\sqrt{k})+C$$
$$\frac12 t \sqrt{A^2t^2+c^2}+\frac{c^2}{2A}\ln\left(At+\sqrt{A^2t^2+c^2}\right)=\pm\frac{1}{\sqrt{k}}\sin^{-1}(r\sqrt{k})+C'$$
$$r=\pm\sin\left(\frac{\sqrt{k}}{2} t \sqrt{At^2+c^2}+\frac{c^2\sqrt{k}}{2A}\ln\left(At+\sqrt{A^2t^2+c^2}\right)-C'\right)$$
