How to you find the general solution of #(dr)/(ds)=0.05r#?

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Answer 1 · 2017-05-24T08:06:26

# r = Ae^(0.05s) #

We have:

# (dr)/(ds)=0.05r#

Which is a First Order linear separable DE. We can simply separate the variables to get

# int \ 1/r \ dr = int \ 0.05 \ ds #

Then integrating gives:

# \ \ ln | r | = 0.05s + C #
# :. | r | = e^(0.05s + C) #
# :. | r | = e^(0.05s) *e^C #

And as #e^x >0 AA x in RR#, and putting #A=e^C# we can write the solution as:

# r = Ae^(0.05s) #