Question

How do you find the particular solution to `(dr)/(ds)=e^(r-2s)` that satisfies r(0)=0?

Answer 1

`r=ln((2e^(2s))/(1+e^(2s)))`

Explanation:

We will want to separate the variables first, which means that we can treat `(dr)/(ds)` like division. Move all the terms with `s` to one side and all those with `r` to the other side of the equation.

`(dr)/(ds)=e^r/e^(2s)`

Rearranging now (separating variables):

`(dr)/e^r=(ds)/e^(2s)`

Rewriting and integrating:

`inte^-rdr=inte^(-2s)ds`

Integrating both sides:

`-int-e^-rdr=-1/2int-2e^(-2s)ds`

`-e^-r=-1/2e^(-2s)+C`

We can find the constant of integration now using `r(0)=0`:

`-e^0=-1/2e^0+C`

Since `e^0=1`:

`-1=-1/2+C`

`C=-1/2`

So:

`-e^-r=-1/2e^(-2s)-1/2`

Now solving for `r`:

`e^(-r)=1/2e^(-2s)+1/2=(e^(-2s)+1)/2`

`-r=ln((e^(-2s)+1)/2)`

`r=ln(2/(e^(-2s)+1))=ln((2e^(2s))/(1+e^(2s)))`
graph{ln((2e^(2x))/(1+e^(2x))) [-10, 10, -5, 5]}