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Bernoulli's DE

  1. Oct 9, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I must solve [itex]xy'+y+x^4y^4e^x=0[/itex].


    2. Relevant equations
    Bernoulli's.


    3. The attempt at a solution
    I divided the original DE by x to get [itex]y'+y \left ( \frac{1}{x} \right )=-x^3e^xy^4[/itex].
    Now let [itex]z=y^{-3} \Rightarrow z'=-3y^{-3}y'[/itex].
    I then multiplied the DE by [itex]-3y^{-4}[/itex] to reduces the DE to [itex]z'-\frac{3}{x}z=3x^3e^x[/itex]. This is a first order linear DE so I should be able to solve it via the integrating factor method, however this doesn't work out for me.
    The integrating factor is [itex]e^{\int -3 /x dx}=x^{-3}[/itex]. So that the general solution of this DE (the z's one) should be [itex]z=-e^xx^3+Cx^3[/itex].
    So that [itex]z'=-e^xx^3-3e^xx^2+3Cx^2[/itex].
    But then when I replace z and z' into [itex]z'-\frac{3}{x}z[/itex] I get that it's worth [itex]-e^xx^3[/itex] rather than [itex]3x^3e^x[/itex]. So it seems that I have a "-3" missing factor. I've rechecked all the algebra like 3 times, including now by typing this post and I still don't see where my mistake lies. I'm almost 100% sure it's in the integrating factor method but I really don't see it. I've even reopened Boas' mathematical methods book for the integrating factor method and I feel I've done it right.
    Thanks for all help!
     
  2. jcsd
  3. Oct 10, 2011 #2

    Hootenanny

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    I would have another look at this bit:
    and in particular, the computation of the integrating factor :wink:
     
  4. Oct 10, 2011 #3

    fluidistic

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    I appreciate your help. However I'm afraid I don't see any error for the integrating factor.
     
  5. Oct 10, 2011 #4

    Hootenanny

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    It's not the IF itself, but rather what you do with it. After multiplication by the IF, your ODE becomes

    [tex]\frac{d}{dx}\left(\frac{z}{x^3}\right) = 3e^x[/tex]

    Do you see where your factor of three is missing now?
     
  6. Oct 10, 2011 #5

    fluidistic

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    Oh nice, yes now, thank you very very very much. :biggrin:
    This works. Continuing on, I reach as final answer [itex]y(x)=(3e^xx^3+cx^3)^{-1/3}=\frac{1}{x \sqrt [3] {3e^x+c}}[/itex].
     
  7. Oct 10, 2011 #6

    Hootenanny

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    My pleasure :smile:
    That is indeed correct (or more accurately, the only real solution for appropriate x).
     
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