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Solve y'y'''=y''

  1. Aug 8, 2008 #1
    How to use power series to solve this non-linear differential equation?
     
  2. jcsd
  3. Aug 8, 2008 #2

    HallsofIvy

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    First, because this is a third order equation, its general solution will involve 3 undetermined constants. So, assume y(0)= A, y'(0)= B, y"(0)= C.

    From y'y"'= y", y"'= y"/y' and so y"'(0)= C/B. Now, differentiating y"'= y"/y' again, yiv= (y'y"'- y"2)/y'2 so yiv(0)= (B(C/B)- C^2)/B2= (C- C2)/B2.

    So far, we have y(x)= y(0)+ y'(0)x+ (1/2)y"(0)x2+ (1/3!)y"'(0)x3+ (1/4!)yiv(0)x4+ ...= A+ Bx+ (1/2)Cx2+ (1/6)(C/B)x3+ (1/24){(C- C2)/B2}x^4+ ...

    Continue like that to get higher terms.
     
  4. Aug 11, 2008 #3
    Re-arrange your equation as

    [tex]y^{\prime \prime \prime} = \frac{y^{\prime \prime}}{y^{\prime}}[/tex]

    Now integrate with respect to x to get

    [tex]y^{\prime \prime} = \kappa + \ln{y^{\prime}}[/tex]

    where [tex]\kappa[/tex] is a constant of integration. Now re-arrange and integrate to get

    [tex]\int{\frac{d y^{\prime}}{\kappa + \ln{y^{\prime}}} = x + \epsilon[/tex]

    where [tex]\epsilon[/tex] is another constant.

    I checked the Integrator (Wolfram site), and it gave the integral as:

    [tex]\int{\frac{dw}{a + \ln{w}}} = e^{-a}Ei(a + \ln{w})[/tex]

    where [tex]Ei[/tex] is the Exponential Integral.

    Hence we can apply this to our integral to get

    [tex]e^{- \kappa} Ei(\kappa + \ln{y^{\prime}) = x + \epsilon [/tex]

    which we can re-arrange as

    [tex]y^{\prime} = exp(Ei^{-1}(e^{\kappa}(x + \epsilon)) - \kappa)[/tex]

    We can tidy this up a bit by making

    [tex]e^{\kappa} = \alpha[/tex]

    and

    [tex]\epsilon \alpha = \beta [/tex]

    and then integrate to get

    [tex]y(x) = \frac{1}{\beta}\int{exp(Ei^{-1}(\alpha x + \beta)) dx}[/tex]

    and then you'll have to try some numerical techniques to obtain values for y(x) (I have to idea how to express the inverse of "Ei".)
     
    Last edited: Aug 11, 2008
  5. Aug 11, 2008 #4
    Sorry that should read:

    [tex]y(x) = \frac{1}{\alpha}\int{exp(Ei^{-1}(\alpha x + \beta)) dx}[/tex]
     
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