How do you find the particular solution to (dr)/(ds)=e^(r-2s) that satisfies r(0)=0?
1 Answer
Dec 4, 2016
Explanation:
We will want to separate the variables first, which means that we can treat
(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
-e^0=-1/2e^0+C
Since
-1=-1/2+C
C=-1/2
So:
-e^-r=-1/2e^(-2s)-1/2
Now solving for
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]}