I was reading Planetary Motion (page 117) in Barry Spain's Tensor calculus, and stupidly enough, I didn't understand this. The equations are:
$$\frac{d^2\psi}{d\sigma^2} + \frac{2}{r}\frac{dr}{d\sigma}\frac{d\psi}{d\sigma} = 0,$$
$$\frac{d^2t}{d\sigma^2} + \frac{2m}{c^2r}\left(1-\frac{2m}{c^2r}\right)^{-1}\frac{dr}{d\sigma}\frac{dt}{d\sigma} = 0,$$
And the next statement reads
we integrate the above to get
$$r^2\frac{d\psi}{d\sigma} = h, \quad \left(1-\frac{2m}{c^2r}\right) \frac{dt}{d\sigma}= k$$
respectively, for constants $k$ and $h$.
I believe it is a simple question, as no steps are given, but I am unable to get it. I tried everything I could, substitutions et al, but to no avail. So please entertain this silly question. Thank you very much...
EDIT
Please show the integration, I dont want to confirm the validity of the answers by working backwards, but I want to establish them.
EDIT (2)
I want to see the actual integration. If I am right, then the second term($2r\frac{dr}{d\sigma}\frac{d\psi}{d\sigma}$) of the first equation yields the desired result which is $r^2\frac{d\psi}{d\sigma}$ when integrated wrt $d\sigma$. It means that the integral of $r^2\frac{d^2\psi}{d\sigma^2}$ is $0$...I think the integral simplifies to $\frac{d}{d\sigma}\int{r^2d\psi}$, and to get the required answer, it should be zero or a constant.....But HOW??? PLease help....