Answer
$$r=\ln{\left(\frac{1}{\frac{1}{2}e^{-2s}+\frac{1}{2}}\right)}$$
Work Step by Step
$\frac{dr}{ds}=e^{r-2s}=e^re^{-2s}$
$\frac{1}{e^r}dr=e^{-2s}ds$
$\int e^{-r}dr=\int e^{-2s}ds$
$-e^{-r}=-\frac{1}{2}e^{-2s}+C$
$e^{-r}=\frac{1}{2}e^{-2s}+C$
$-r=\ln{\left(\frac{1}{2}e^{-2s}+C\right)}$
$r=-\ln{\left(\frac{1}{2}e^{-2s}+C\right)}=\ln{\left(\frac{1}{\frac{1}{2}e^{-2s}+C}\right)}$
Now applying our initial condition we get
$0=\ln{\left(\frac{1}{\frac{1}{2}+C}\right)}$
$\frac{1}{2}+C=1$
$C=\frac{1}{2}$
and so our particular solution becomes
$$r=\ln{\left(\frac{1}{\frac{1}{2}e^{-2s}+\frac{1}{2}}\right)}$$
