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Hard diff-eq

  1. Oct 31, 2004 #1
    I'm finding this diff equation hard
    [tex] y'' + \ln{y} = yx [/tex]
    How do I solve it?
  2. jcsd
  3. Oct 31, 2004 #2

    James R

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  4. Oct 31, 2004 #3


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    Willing to agree with the previous comment .... I'm not sure how well the classical numerical solution methods bite, but if nothing else you can discretize and do for example a simplistic DM or FE etc. analysis.
  5. Oct 31, 2004 #4
    I don't want it numerically, I want it exact if possible
  6. Oct 31, 2004 #5

    Dr Transport

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    rearrange: [tex] y'' - xy = -\ln(y) [/tex], solve the homogeneous equation for [tex] y [/tex] then use those solutions as an integrating factor, or Green's function to solve the equation.

  7. Nov 1, 2004 #6


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    No, that won't work: this is not a linear equation. The right hand side is ln(y), NOT ln(x)!
  8. Nov 1, 2004 #7


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    We MIGHT make some headway if we seek an asymptotic approximation in the vicinity of x=0 (i.e, we make a supposition that our unknown function y(x) is defined at x=0)
    Let us state an initial condition:
    In addition, we set:

    so that:

    In the vicinity of x=0, we have:

    Hence, close to x=0, we have the differential equation in Y:
    [tex]Y''+\frac{Y}{y_{0}}=y_{0}x-ln(y_{0}), Y(0)=0 (1)[/tex]
    The general solution of the homogenous equation (that is, [tex]Y''+\frac{Y}{y_{0}}=0[/tex]) is:
    A particular solution to (1) is the linear function:
    We therefore set

    Hence, we get the asymptotic solution, to first order:

    I would like to emphasize that this is only a first order approximation, valid in the limit [tex]x\to0[/tex]
    Last edited: Nov 1, 2004
  9. Dec 4, 2004 #8
    I've thought about this a lot ...

    (1) y'' + ln(y) = yx
    (2) ln(y'' + ln(y)) = ln(yx) = ln(y)+ln(x)
    (3) ln(y'' + ln(y)) = ln(y) + ln(x)
    (4) ln(y'' + ln(y)) - ln(y) = ln(x)
    (5) ln(y''/ln(y) + ln(y)/ln(y)) = ln(x)
    (6) ln(y''/ln(y) + 1) = ln(x)
    (7) y''/ln(y) + 1 = x
    (8) y'' + ln(y) = ln(y)*x

    Since the left side of (8) is equal to the left hand side of (1) this must give that ln(y) * x = y * x => ln(y) = y => y = e^y.

    However I don't think there exist a solution to y = e^y ... if it exists it must a complex integer so y is a constant function. Using math software this gives y = 0.3181315052047641 - 1.3372357014306895i
  10. Dec 4, 2004 #9


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    Step 5) is wrong hedlund
    From 4), we have:
    As I'm sure you agree with..
  11. Dec 4, 2004 #10
    Yeah it was plain stupid ... I thought of it as ln(ln(y)) :/
  12. Dec 4, 2004 #11


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    To able to find exact solutions to (non-linear) diff.eqs are more often than not just a happy chance, NEVER the rule.
    In fact, you'll meet just about every exact solution of diff.eqs which has been /can be found through your studies, they are extremely few in number.
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