Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 22

This differential equation can be solved by separating the variables.
(dr)/(ds) = e^(r - 2s)
Dividing by e^r and multiplying by ds results in the variables r and s on the different sides of the equation:
(dr)/e^r = e^(-2s)ds
This is equivalent to
e^(-r) dr = e^(-2s)ds
Now we can take the integral of the both sides of the equation:
-e^(-r) = 1/(-2)e^(-2s) + C , where C is an arbitrary constant.
From here, e^(-r) = 1/2e^(-2s) - C
and -r = ln(1/2e^(-2s) - C)
or r = -ln(1/2e^(-2s) - C)
Since the initial condition is r(0) = 0, we can find the constant C:
r(0) = -ln(1/2e^(-2*0) - C) = -ln(1/2 - C) = 0
This means 1/2 - C = 1
and C = -1/2
Plugging C in in the equation for r(s) above, we can get the particular solution:
r = -ln((e^(-2s) + 1)/2) . This is algebraically equivalent to
r = ln(2/(e^(-2s) + 1)) . This is the answer.