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.