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

1 Answer
mason m · Shwetank Mauria
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]}

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