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Modeling epidemics - solving differential equation

  1. Apr 22, 2017 #1
    1. The problem statement, all variables and given/known data
    Solve for y: ##\frac {dy}{dx} = \frac {1+y^6}{xy^5}## , where y(1) = 1.
    Answer ## y = \sqrt[6] {2x-1}##
    2. Relevant equations


    3. The attempt at a solution
    ##\frac {dy}{dx} = \frac {1+y^6}{xy^5}##
    ##\frac{dy (y^5)}{1+y^6} = dx \frac {1}{x}##
    u= 1+y6
    ##\frac {du}{y^5}=dx##
    ##\int \frac{1}{u}du = \int \frac {1}{x}dx##
    ##\ln|1+y^6| = \ln|x| + C##
    The natural logs cancel out.
    Substituting in my (1,1)
    ## 1+1^6 = 0+C##
    ##C=2##
    This is where I'm a bit lost. I'm not sure where I messed up but I don't know how to get the 2x-1. Any help would be greatly appreciated!
     
    Last edited: Apr 22, 2017
  2. jcsd
  3. Apr 22, 2017 #2

    fresh_42

    Staff: Mentor

    You lost me at the substitution. Shouldn't there be a ##\frac{1}{6}## somewhere? I also got a different result than the one you expect as an answer. Did you differentiate it? Plus ##\ln |1| = 0##, not ##1##.
     
  4. Apr 22, 2017 #3
    Yes I definitely messed up on the ##\ln |1| = 0## but I don't understand where this ##\frac {1}{6}## is coming from.
     
  5. Apr 22, 2017 #4

    fresh_42

    Staff: Mentor

    If ##u=1+y^6## then ##\frac{du}{dy}=6y^5##. Where did the factor go to?
     
  6. Apr 22, 2017 #5
    Oh I see. Would the integral look like this then
    ##\ln |6(1+y^6)|##? Or do I put the ##\frac{1}{6}## on the outside of the natural log?
     
  7. Apr 22, 2017 #6

    fresh_42

    Staff: Mentor

    As I couldn't follow your calculation beyond the substitution I did it step by step:
    ##\frac{dy}{dx}=\frac{u}{xy^5}## and ##y^5=\frac{1}{6}\frac{du}{dy}##. Now the ##dy## canceled out and leaves ##\frac{dx}{x}=\frac{1}{6}\frac{du}{u}##. The factor in front of the logarithm becomes an exponent in the logarithm and thus the ##6-##th root. Next I replaced ##u## and calculated the constant ##C##. The rest was some algebra. What's always a good idea in such cases, is to differentiate the result again and check whether the original equation pops up.
     
  8. Apr 22, 2017 #7
    Sorry, I'm still a little confused. After integrating should it look like this then?
    ##\frac {1}{6} \int \frac {1}{u}##
    ##= \frac {1}{6} \ln |1+y^6| = \ln|x| + C##
     
  9. Apr 22, 2017 #8

    fresh_42

    Staff: Mentor

    Yes. Or likewise ##\ln|x| + C = \ln \sqrt[6]{1+y^6}## and ##C = \frac{1}{6} \ln (1+1^6) - \ln 1##.
     
  10. Apr 22, 2017 #9
    I'm still not understanding how to get the answer. If ##\frac {1}{6} \ln(2)-0 = C## then C ≈ 0.1155... or at least that is what my calculator says.
     
  11. Apr 22, 2017 #10

    fresh_42

    Staff: Mentor

    You don't need to calculate it. Just drop the ##\ln## as you did in your solution. It's probably more convenient to first multiply by ##6## and get ##\ln |1+y^6| = \ln 2|x|^6## and then drop the ##\ln## as you did before.
     
  12. Apr 22, 2017 #11
    Now I got my answer I'm just curious where the highlighted 2 came from.
     
  13. Apr 22, 2017 #12

    fresh_42

    Staff: Mentor

    The ##2## is from ##\ln 2 + \ln |x|^6 = \ln (2x^6)## or what did you mean? And it isn't your answer, it is ##y = \sqrt[6]{2x^6-1}##, not ##y = \sqrt[6]{2x-1}##. At least if I made no mistake, but I also differentiated it again.
     
  14. Apr 22, 2017 #13
    The answer in my course is 2x-1 but it could be wrong, it is easily the worst course I've done. Thank you for all your help though.
     
  15. Apr 23, 2017 #14

    Ray Vickson

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    Science Advisor
    Homework Helper

    The given solution ##y(x) = \sqrt[6]{2x-1}## is incorrect: if you substitute it into the DE you will see that it does not work.
     
  16. Apr 23, 2017 #15

    epenguin

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    Gold Member

    Unless the problem was miscopied at some stage and the x should have been in the numerator.
     
  17. Apr 24, 2017 #16

    Ray Vickson

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    In that case the DE would be ##y' = x(1+y^6)/y^5, \; y(1) = 1##, whose solution is
    $$y(x) = \left( 2 e^{3x^2-3}-1\right)^{1/6}.$$
     
  18. Apr 24, 2017 #17

    epenguin

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    OK, OK. :oldbiggrin:
    Just Unless the problem was miscopied at some stage....
     
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