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Differential Equation, Bernoulli

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    3y^2*y'+y^3= e^-x



    3. The attempt at a solution

    I am using Bernoulli's equation to substitute V in and I keep coming out with
    Y^3= e^x(-e^-x+c)

    my V = y^3
    Y = v^(1/3)
    and dy/dv=(1/3)v^(-2/3)

    I peeked to see if I was correct, the right answer is supposed to be Y^3 = e^-x (x+c)

    Where did I go wrong?
     
  2. jcsd
  3. Sep 15, 2009 #2

    Dick

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    v=y^3 is certainly a good substitution to use, and v=e^(-x)*(x+c) is certainly the correct answer. What I don't understand is how you used Bernoulli's equation to get a wrong answer. Can you explain?
     
  4. Sep 15, 2009 #3
    Ok I will try and explain what I did:

    After getting those substitutions I applied them all to the original equation and got out:

    (1/3)v(-2/3)* Dv/Dx + (1/3)v(1/3) = (1/(3ex))*v(-2/3)

    Multiplied by the inverse of dy/dv to get:

    dv/dx + v = 1/(ex)

    This is where I got confused, the function I am supposed to put into Bernoulli's formula usually lies in front of V, so I figured I would use 1, so I did:

    eintegral(1, dx) = ex

    Multiplied both sides by this and integrated to get the answer that I got. I hope I better explained what I did. Do you see where I went wrong, because I still do not.

    EDIT: When I multiplied both sides by ex I seemed to have spaced and not multiplied the RHS... Thus giving me 1 on the right hand side which integrates to be x+C, thus giving me the correct answer... Having me go back through it really helped me figure out what I did wrong. Thanks for helping.
     
    Last edited: Sep 15, 2009
  5. Sep 15, 2009 #4
    P.S. I really love this forum, everyone is so helpful!
     
  6. Sep 15, 2009 #5

    Dick

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    Ok, so now you've got e^x*v'+e^x*v=1, right? That's (e^x*v)'=1. Integrate both sides then solve for v.
     
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