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Help for solving a 2nd order non-linear ODE

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data
    [tex]yy''-y'^2 = y^2lny[/tex]

    3. The attempt at a solution
    well, since the equation is of the form [itex]f(y,y',y'')=0[/itex] I turn it into the form [itex]f(y,p,p dp/dy)=0[/itex].
    After those substitutions are made, we'll have the following equation:
    [tex]yp (\frac{dp}{dy})-p^2-y^2 lny=0[/tex]
    which is a Bernoulli equation that can be solved easily. the solution of this ODE is:
    [tex]p^2=y^2lny+cy^2[/tex]
    Here's where I'm stuck, because If I substitute y'=p again I'll have an ODE that I don't know how to solve it, I guess the general solution will be parametric, but I don't know how to proceed from this step.
     
    Last edited: Nov 16, 2011
  2. jcsd
  3. Nov 16, 2011 #2

    Stephen Tashi

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    One thought is:
    [tex] D( \frac{y'}{y}) = ln(y) [/tex]
    [tex] \frac{y'}{y} = \int ln(y) [/tex]
     
  4. Nov 16, 2011 #3

    dextercioby

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    Have you tried a substitution y(x)=eq(x) ?
     
  5. Nov 16, 2011 #4
    Would you explain more please?

    I don't know how that would work. Would you explain more?
     
  6. Nov 16, 2011 #5

    Mark44

    Staff: Mentor

    Stephen has these backwards.

    D(ln(y)) = y'/y
    so ln(y) = ∫(y'/y)
     
  7. Nov 16, 2011 #6
    Yes, I understood that, but how does that help?

    I think this is what he means:
    [itex]y'^2 = y^2 lny + cy^2[/itex]→ [itex]y^{-2}y'^2-lny=C[/itex]
    → [itex]\frac{y'^2}{y^2} - lny = C[/itex]→ [itex](\frac{y'}{y})^2 - (\int{\frac{y'}{y}dy})=C[/itex]
    Hence, the equation has turned into the form [itex]u'^2 - u = C[/itex]. Has anything changed? Can I solve this one?
     
  8. Nov 16, 2011 #7

    Stephen Tashi

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    That's not my thought. My thought is that the original equation is equivalent to the equation [itex] D(\frac{y'}{y}) = ln(y) [/itex]

    But employing what you said:

    [tex] D(\frac{y'}{y}) = \int \frac{y'}{y} [/tex]
     
    Last edited: Nov 16, 2011
  9. Nov 16, 2011 #8
    Ahh, now I see what you mean. That's a great way of doing this ODE once you know that D(y'/y)=(yy''-y'^2)/y^2. Is there anyway to proceed with my own method?
     
  10. Nov 17, 2011 #9

    Stephen Tashi

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    I don't see how to use your method. As understand the situation, the solution of the Bernoulli equation that you employ assumes that [itex] p [/itex] is a known function. But [itex] p [/itex] isn't known.
     
  11. Nov 17, 2011 #10

    dextercioby

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    It would get you rid of the natural logarithm and perhaps the transformed ODE would be integrable...
     
  12. Nov 17, 2011 #11
    Your solution is wrong at this step!

    The correct solution is:

    [tex]
    p y^2 (C + \ln^2{y}) = 1, \ p = y'
    [/tex]

    If you split the variables, the integral over y is expressible in terms of elementary functions.
     
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