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

##### 1 Answer

May 24, 2017

# r = Ae^(0.05s) #

#### Explanation:

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

# r = Ae^(0.05s) #