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

Dec 4, 2016

$r = \ln \left(\frac{2 {e}^{2 s}}{1 + {e}^{2 s}}\right)$

#### Explanation:

We will want to separate the variables first, which means that we can treat $\frac{\mathrm{dr}}{\mathrm{ds}}$ like division. Move all the terms with $s$ to one side and all those with $r$ to the other side of the equation.

$\frac{\mathrm{dr}}{\mathrm{ds}} = {e}^{r} / {e}^{2 s}$

Rearranging now (separating variables):

$\frac{\mathrm{dr}}{e} ^ r = \frac{\mathrm{ds}}{e} ^ \left(2 s\right)$

Rewriting and integrating:

$\int {e}^{-} r \mathrm{dr} = \int {e}^{- 2 s} \mathrm{ds}$

Integrating both sides:

$- \int - {e}^{-} r \mathrm{dr} = - \frac{1}{2} \int - 2 {e}^{- 2 s} \mathrm{ds}$

$- {e}^{-} r = - \frac{1}{2} {e}^{- 2 s} + C$

We can find the constant of integration now using $r \left(0\right) = 0$:

$- {e}^{0} = - \frac{1}{2} {e}^{0} + C$

Since ${e}^{0} = 1$:

$- 1 = - \frac{1}{2} + C$

$C = - \frac{1}{2}$

So:

$- {e}^{-} r = - \frac{1}{2} {e}^{- 2 s} - \frac{1}{2}$

Now solving for $r$:

${e}^{- r} = \frac{1}{2} {e}^{- 2 s} + \frac{1}{2} = \frac{{e}^{- 2 s} + 1}{2}$

$- r = \ln \left(\frac{{e}^{- 2 s} + 1}{2}\right)$

$r = \ln \left(\frac{2}{{e}^{- 2 s} + 1}\right) = \ln \left(\frac{2 {e}^{2 s}}{1 + {e}^{2 s}}\right)$
graph{ln((2e^(2x))/(1+e^(2x))) [-10, 10, -5, 5]}