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Going from dy/dx + P(x)y = Q(x)y^n to du/dx + (1-n)P(x) u = (1-n)Q(x)

  1. Jul 13, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    There's this problem I'm presented with that has a part before asking the question which states a fact and, I want to be able to derive that fact.

    I want to go from

    dy/dx + P(x) y = Q(x) y^n

    to

    du/dx + (1 - n) P(x) u = (1 - n) Q(x).

    2. Relevant equations
    Bernoulli differential equation process, u = y^(1 - n)

    3. The attempt at a solution
    u = y^(1 – n), du/dx = (1 – n) y^(-n) dy/dx, u^(n – 1) = y, y^n = u^(n^2 – n)

    dy/dx + P(x) y = Q(x) y^n

    du/dx 1/(n – 1) y^n + P(x) u^(n – 1) = Q(x) u^(n^2 – n)

    du/dx y^n + (1 – n) P(x) u^(n – 1) = (1 – n) Q(x) u^(n^2 – n)

    du/dx u^(n^2 – n) + (1 – n) P(x) u^(n – 1) = Q(x) u^(n^2 – n)

    du/dx + (1 – n) P(x) u^(-n^2 + 2n – 1) = (1 – n) Q(x)

    but, instead, I should be getting

    du/dx + (1 – n) P(x) u = (1 – n) Q(x).

    What am I doing wrong?

    Any input would be greatly appreciated!
     
    Last edited: Jul 13, 2013
  2. jcsd
  3. Jul 13, 2013 #2

    CAF123

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    Gold Member

    Hi s3a,


    The error is there.
     
  4. Jul 13, 2013 #3

    s3a

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    Thanks for pointing out which part was wrong but, via "reverse engineering", I think it should be u^(n^2 - n + 1) but, looking at u = y^(1 - n), I get u^(1/(1 - n)) = y.

    What do I do now?
     
  5. Jul 13, 2013 #4

    CAF123

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    Gold Member

    ##y = u^{\frac{1}{1-n}}## is correct. Now you have ##\frac{dy}{dx},\,\,y\,\,\text{and}\,\,y^n## all in terms of ##u## and ##\frac{du}{dx}##, so sub them into your original equation and simplify.
     
    Last edited: Jul 13, 2013
  6. Jul 13, 2013 #5

    s3a

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    Sorry for still not getting it.

    Here's my latest work.:

    dy/dx + P(x) y = Q(x) y^n

    du/dx y^n / (1 – n) + P(x) u^(1/(1 – n)) = Q(x) u^(n^2 – n)

    du/dx u^(n^2 – n) + (1 – n) P(x) ^(1/(1 – n)) = (1 – n) Q(x) u^(n^2 – n)

    du/dx + (1 – n) P(x) u^[(n^2 – n)/(1 – n)] = (1 – n) Q(x)

    du/dx + (1 – n) P(x) u^[-n(1 – n)/(1 – n)] = (1 – n) Q(x)

    du/dx + (1 – n) P(x) u^(-n) = (1 – n) Q(x)
     
  7. Jul 13, 2013 #6

    CAF123

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    Gold Member

    You appear to still have the expression for y in terms of u incorrect.
    If ##u = y^{1-n}## then ##u^{1/(1-n)} = \left(y^{1-n}\right)^{1/(1-n)} = y## as you mentioned in your last post.

    How did you get ##y = u^{n^2 -n}##?
     
  8. Jul 14, 2013 #7

    s3a

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    Actually, I solved it, now!

    Thanks! :)
     
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