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Homework Help: Second order non-linear differential equation involving log

  1. Dec 13, 2013 #1
    EDIT: my problem is solved, thank you to those who helped

    1. The problem statement, all variables and given/known data


    [itex]x y^{\prime \prime} = y^{\prime} \log (\frac{y^{\prime}}{x})[/itex]

    Note: This is the first part of an undergraduate applications course in differential equations. We were taught to solve second order non-linear equations by doing tricks such as substituting [itex]u = y^{\prime}[/itex], factoring, and then dealing with the two cases that result.

    2. Relevant equations


    3. The attempt at a solution

    I tried [itex]u := y^{\prime}[/itex], so [itex]u^{\prime} = y^{\prime \prime}[/itex], and so substitution gives

    [itex]x u^{\prime} = u \log(\frac{u}{x})[/itex]

    but I can't seem to be able to do anything with that.

    I also I noticed that

    [itex] y^{\prime \prime} = y^{\prime} x^{-1} \log(y^{\prime} x^{-1}) [/itex]

    which I thought was interesting. But I couldn't really do anything with it.

    Thank you.
    Last edited: Dec 13, 2013
  2. jcsd
  3. Dec 13, 2013 #2


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    How about log (y'/x) = log (y') - log (x)?
  4. Dec 13, 2013 #3


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    You want to try and find a variable change that will make it separable. Since you have u/x in the log, substituting xv=u might be a good thing to try.
  5. Dec 13, 2013 #4
    You have [itex]x u^{\prime} = u \log(\frac{u}{x})[/itex]. This can be rewritten as
    [tex]\frac{d \ln u}{d\ln x}=\ln(u)-\ln(x)[/tex]
    Let ln(u) = w and ln(x) = z
  6. Dec 13, 2013 #5
    thanks for the help.

    OK, that was productive. Here's what I did:

    [itex]v := \frac{y^{\prime}}{x} \Longrightarrow v^{\prime} = \frac{x y^{\prime \prime} - y^{\prime}}{x^2} \Longrightarrow x^2 v^{\prime} + y^{\prime} = x y^{\prime \prime}[/itex]


    [itex]x^2 v^{\prime} + y^{\prime} = y^{\prime} \log(v) \Longrightarrow x v^{\prime} + v = v \log(v)[/itex]
    [itex]\Longrightarrow \frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow (v - v \log(v)) dx + x dv = 0[/itex]

    [itex]M := v - v \log(v)[/itex]
    [itex]N := x[/itex]

    [itex]M_v = 1 - \log(v) - 1 = -\log(v) \neq 1 = N_x[/itex]

    so the equation is not exact. But

    [itex]\displaystyle{\frac{N_x - M_v}{M} = \frac{1 + log(v)}{v(1 - \log(v))} \neq f(v)}[/itex]
    [itex]\displaystyle{\frac{M_v - N_x}{N} = -\frac{\log(v) + 1}{x} \neq f(x)}[/itex]
    [itex]\displaystyle{\frac{N_x - M_v}{x M - v N} = -\frac{1 + \log(v)}{x v \log(v)} \neq f(xv)}[/itex]

    so I am stuck (one of those works when we are given a separable equation in my course).
  7. Dec 13, 2013 #6


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    xv'+v=vlog(v) is already separable. Just separate it and go from there.
  8. Dec 13, 2013 #7
    thanks for the help.

    I don't know how to evaluate

    [itex]\frac{d \log(u)}{d \log(x)}[/itex]

    I understand that I would be differentiating [itex]\log(u)[/itex] as a function of [itex]\log(x)[/itex], but I don't see how that is a function of [itex]\log(x)[/itex].
  9. Dec 13, 2013 #8
    true. OK, then

    [itex]\displaystyle{\frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow \int{\frac{1}{v} \cdot \frac{1}{\log(v) - 1}} dv = \int{\frac{1}{x}}} dx[/itex]
    [itex]\displaystyle{\Longrightarrow \log(\log(v) - 1) = \log(x) + C \Longrightarrow \log(v) - 1 = C x \Longrightarrow v = \frac{y^{\prime}}{x} = e^{C x + 1}}[/itex]
    [itex]\displaystyle{\Longrightarrow y = \int{x e^{C x + 1}} dx = x \frac{1}{C} e^{C x + 1} - \frac{1}{C} \int{e^{C x + 1}} dx = \frac{1}{C^2} e^{C x + 1} (C x - 1) + D}[/itex]

    where I just re-used [itex]C[/itex] for [itex]e^C[/itex].

    I haven't verified it yet by substituting
    Last edited: Dec 13, 2013
  10. Dec 13, 2013 #9


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    Looks fine for v, but I think you dropped a C when you were integrating by parts.
  11. Dec 13, 2013 #10
    If u is a function of x, then it is a function of ln(x).
  12. Dec 13, 2013 #11
    awesome. Also, I also think you are correct about the missing [itex]C[/itex], thank you--I shall update that post. Thank you for your help!
  13. Dec 13, 2013 #12


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    Very welcome! If you want some extra practice you should try doing it Chestermiller's way as well, it's very nice if you don't mind keeping track of a few extra substitution variables.
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