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...

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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.**