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Separable Differential Equation (Perfect Derivative)

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data


    [tex]-y + xy' = 0 [/tex] and y(2)=5


    3. The attempt at a solution

    This first part trips me up. I am supposed to find the perfect derivative, which is [tex](xy') = 0 [/tex] ? Is this legal, or does the -y not allow for that?

    If that is correct, then I know that I integrate that, which gives me [tex] xy = c [/tex]

    Then, to solve for c, I just plug in 2 and 5.
     
  2. jcsd
  3. Oct 7, 2009 #2

    Dick

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    No, it's not exact. And your solution assuming that is wrong. But it is separable. Try that.
     
  4. Oct 7, 2009 #3

    LCKurtz

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    It's also linear. Have you studied integrating factors?
     
  5. Oct 8, 2009 #4
    Oh, I understand! Just multiply by e^-x, so the problem looks like this:

    [tex]-ye^-^x + xe^-^xy' = 0[/tex]

    Therefore:

    [tex]xe^-^xy = c[/tex]

    Correct?

    I'm having trouble with the next problem also. The next problem is similar, but with a [tex]x^3[/tex] added to it. I can't think of any other way to do it other than by multiplying everything by e^-x and doing integration by parts to [tex]x^3e^-^x[/tex].Any tips for an easier way to do it?
     
    Last edited: Oct 8, 2009
  6. Oct 8, 2009 #5

    LCKurtz

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    No, your solution isn't correct. Your original equation

    [tex] xy' - y = 0[/tex]

    is not in the correct form to find the integrating factor. The coefficient of y' needs to be 1 so you need to divide by x before calculating the integrating factor. Also, as a side note, linear differential equations are typically written with the highest derivatives first much like polynomials, not that it matters that much.
     
  7. Oct 8, 2009 #6
    Oh.. you're right.

    It's not separable though, is it? When I try and separate it I cannot get the Ys on one side and the Xs on one side. Unless [tex]\frac{dy}{y} = \frac{dx}{x}[/tex] is the equation correctly separated.
     
  8. Oct 8, 2009 #7
    remember that is [tex]\frac{1}{x}dx=\frac{1}{y}dy[/tex] You can integrate this.

    As for the other method, do you know how to find an integrating factor for second order linear equations? It's generally not just multiply through by [tex]e^{-x}[/tex]
     
    Last edited: Oct 8, 2009
  9. Oct 8, 2009 #8
    Right. But is that the equation correctly separated?
     
  10. Oct 8, 2009 #9
    Yes but I believe the problem is asking you to find an integrating factor.
     
  11. Oct 8, 2009 #10
    Okay, so if I divide by X, I get [tex]y' - \frac{y}{x} = 0 [/tex]

    So the integrating factor would be -x? That can't be right because that just gives me what I started out with.
     
  12. Oct 8, 2009 #11

    LCKurtz

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    For y' + P(x)y = Q(x) the integrating factor is

    [tex]R(x) = e^{\int P(x) dx}[/tex]
     
  13. Oct 8, 2009 #12
    [tex]e^\int^-^1^/^x[/tex] = [tex]e^-^l^n^x[/tex] = -x. So this means I would multiply the whole problem by -x which gives me what I start out with.

    Or am I doing something wrong?
     
  14. Oct 8, 2009 #13

    LCKurtz

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    Yes you are doing something wrong:

    [tex]e^{- \ln(x)} = e^{\ln(x^{-1})}= x^{-1}[/tex]
     
  15. Oct 8, 2009 #14
    Ahhh! Finally, it makes sense! Thanks!
     
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