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Solving a first order differential equation

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    We have the equation
    ## (\frac{dr}{ds})^2+(\frac{l}{r})^2=1 ##
    and want to solve to get ## r=\sqrt{l^2+(s-s_0)^2}##

    2. Relevant equations


    3. The attempt at a solution
    I have worked backwards, plugging in the solution to prove that it is correct, but the closest I have gotten to actually finding the solution without using r is: ##\frac{dr}{ds}=\frac{\sqrt{r^2-l^2}}{r}##

    Can anyone help with where to go from here?
     
  2. jcsd
  3. Oct 19, 2015 #2

    Geofleur

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    Science Advisor
    Gold Member

    What you have is a good start. Can you rewrite the equation just a little more so that only ## r ## and ## dr ## show up on one side and only ## ds ## shows up on the other?
     
  4. Oct 19, 2015 #3

    BvU

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    Science Advisor
    Homework Helper
    Gold Member

    Hello Stephen,

    I take it you have seen the solution satisfies the differential equation ?

    And doesn't the solution remind you of good old Pythagoras ?
     
  5. Oct 19, 2015 #4

    Mark44

    Staff: Mentor

    Some differential equation problems take the form of "show that this equation is a solution of the differential equation ..." Other differential equation problems ask you to solve a given DE, and don't provide the solution. Your problem appears to be the latter type.

    To start, note that ## (\frac{dr}{ds})^2+(\frac{l}{r})^2=1 ## can be rewritten as ## \frac{dr}{ds} = \pm \sqrt{1 - (\frac{l}{r})^2} ##
     
  6. Oct 23, 2015 #5
    This is the part of the problem I'm having trouble with
     
  7. Oct 23, 2015 #6

    Mark44

    Staff: Mentor

    What is the trouble you're having?
    Starting from ## \frac{dr}{ds} = \pm \sqrt{1 - (\frac{l}{r})^2} ##, separate the variables by dividing both sides by ##\sqrt{1 - (\frac{l}{r})^2} ##, and multiplying both sides by ds. You will need to handle the + and - cases with an equation for each.
     
  8. Oct 23, 2015 #7
    Oh yeah, I have it now. I was being stupid.
     
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