Modeling epidemics - solving differential equation

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  • #1
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Homework Statement


Solve for y: ##\frac {dy}{dx} = \frac {1+y^6}{xy^5}## , where y(1) = 1.
Answer ## y = \sqrt[6] {2x-1}##

Homework Equations




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!
 
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Answers and Replies

  • #2
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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##.
 
  • #3
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Yes I definitely messed up on the ##\ln |1| = 0## but I don't understand where this ##\frac {1}{6}## is coming from.
 
  • #4
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Yes I definitely messed up on the ##\ln |1| = 0## but I don't understand where this ##\frac {1}{6}## is coming from.
If ##u=1+y^6## then ##\frac{du}{dy}=6y^5##. Where did the factor go to?
 
  • #5
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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?
 
  • #6
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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.
 
  • #7
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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##
 
  • #8
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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##
Yes. Or likewise ##\ln|x| + C = \ln \sqrt[6]{1+y^6}## and ##C = \frac{1}{6} \ln (1+1^6) - \ln 1##.
 
  • #9
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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.
 
  • #10
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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.
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.
 
  • #11
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Now I got my answer I'm just curious where the highlighted 2 came from.
 
  • #12
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Now I got my answer I'm just curious where the highlighted 2 came from.
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.
 
  • #13
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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.
 
  • #14
Ray Vickson
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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.
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.
 
  • #15
epenguin
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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.

Unless the problem was miscopied at some stage and the x should have been in the numerator.
 
  • #16
Ray Vickson
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Unless the problem was miscopied at some stage and the x should have been in the numerator.

In that case the DE would be ##y' = x(1+y^6)/y^5, \; y(1) = 1##, whose solution is
$$y(x) = \left( 2 e^{3x^2-3}-1\right)^{1/6}.$$
 
  • #17
epenguin
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Unless the problem was miscopied at some stage and the x should have been in the numerator.

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

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

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