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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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    Numerically.
     
  4. Oct 31, 2004 #3

    PerennialII

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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.

    dt
     
  7. Nov 1, 2004 #6

    HallsofIvy

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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

    arildno

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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:
    [tex]y(0)=y_{0}>0[/tex]
    In addition, we set:
    [tex]y'(0)=y'_{0}[/tex]

    Define:
    [tex]y(x)=Y(x)+y_{0}[/tex],
    so that:
    [tex]Y(0)=0,Y'(0)=y_{0}'[/tex]

    In the vicinity of x=0, we have:
    [tex]ln(y)=ln(y_{0}+Y)=ln(y_{0})+ln(1+\frac{Y}{y_{0}})\approx{ln(y_{0})}+\frac{Y}{y_{0}}[/tex]
    [tex]yx=y_{0}x+Yx\approx{y}_{0}x[/tex]

    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:
    [tex]Y_{h}(x)=A\cos(\frac{x}{\sqrt{y_{0}}})+B\sin(\frac{x}{\sqrt{y_{0}}})[/tex]
    A particular solution to (1) is the linear function:
    [tex]Y_{p}=y_{0}^{2}x-y_{0}ln(y_{0})[/tex]
    We therefore set
    [tex]Y(x)=Y_{h}+Y_{p}[/tex]
    [tex]Y(0)=0\to{A}=y_{0}ln(y_{0})[/tex]
    Whereas:
    [tex]Y'(0)=y_{0}'\to{B}=y_{0}'\sqrt{y_{0}}-(y_{0})^{\frac{5}{2}}[/tex]

    Hence, we get the asymptotic solution, to first order:
    [tex]y(x)=y_{0}+y_{0}ln(y_{0})\cos(\frac{x}{\sqrt{y_{0}}})+(y_{0}'\sqrt{y_{0}}-(y_{0})^{\frac{5}{2}})\sin(\frac{x}{\sqrt{y_{0}}})+y_{0}^{2}x-y_{0}ln(y_{0})[/tex]

    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

    arildno

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    Step 5) is wrong hedlund
    From 4), we have:
    ln(y''+ln(y))-ln(y)=ln(y''/y+ln(y)/y)
    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

    arildno

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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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