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

1 Answer
Dec 4, 2016

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]}