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Nonlinear second order differential equation

  1. Nov 11, 2012 #1
    I am having a problem finding the solution for this eq:

    y''(x)+(2/x)y'(x)+(w^2)y(x)=0

    I couldnt find examples in the text book that goes on a similar line, and have been searching the internet as well, but no use. I am thinking of using substitution v=y' but not sure how to do that in the presence of y??

    any help would be much appreciated!

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 11, 2012 #2
    Is w a constant? You could always multiple by x and you would have a Cauchy-Euler type.
     
    Last edited: Nov 11, 2012
  4. Nov 11, 2012 #3
    Yes w is a constant! thank you VERY much Dustinsfl! I am now reading about the Cauchy-euler equations and how to solve them!

    very appreciated!!! :)
     
  5. Nov 11, 2012 #4
    ahh, just checked the cauchy-euler equation, and from what i understood it does not apply in this case! as the x is in the denominator and only in the second term!
     
  6. Nov 11, 2012 #5
    Multiple by x.
     
  7. Nov 11, 2012 #6
    I did, but substituting for trial solutions does not deal to cancelling x, for example y=x^m, you will have x left,,
     
  8. Nov 11, 2012 #7
    May be you can try reduction of order.
     
  9. Nov 11, 2012 #8
    tried that, no implicit solution,,,
     
  10. Nov 12, 2012 #9

    HallsofIvy

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    I should point out that this is NOT a "non-linear" equation. It is a linear equation with variable coefficients. The standard method for such equations is to use a "power series"

    I would start by multiplying both sides by x, as Dustinsfl suggested, but that does NOT give a "Cauchy Euler equation", we get, rather, [itex]xy''+ 2y'+ \omega^2 xy= 0[/itex]. That is not a "Cauchy-Euler equation" because the coefficient of y'' is x, not x2. We could get that by multipling both sides by x2 but then the coefficient of y is [itex]x^2[/itex] also.

    Because that "1/x" is undefined at 0, and, after multipying by x the coefficient of y'' is 0 at x= 0, x= 0 is a "regular singular point" and should use "Frobenius' method" rather than a regular power seties. That is, we look for a solution of the form [itex]y= \sum_{n= 0}^\infty a_nx^{n+ c}[/itex] where "c" is not necessarily a positive integer (not necessariy an integer).

    Differentiating term by term, [itex]y'= \sum_{n=0}^\infty(c+n)a_nx^{n+c- 1}[/itex] and [itex]y''= \sum_{n=0}^\infty(n+c)(n+ c- 1)a_nx^{n+ c- 2}[/itex]
    Putting those into the equation, we have
    [itex]\sum_{n=0}^\infty (n+c)(n+c-1)a_nx^{n+ c- 1}+ \sum_{n=0}^\infty 2(n+ c)x^{n+ c- 1}+ \sum_{n=0}^\infty \omega^2 a_n x^{n+c+ 1}[/itex].

    The lowest power will occur when n= 0 which would give power [itex]x^{c- 1}[/itex] in the first and second terms, [itex]x^{c+1}[/itex] for the last term. That is, the lowest power wil be [itex]x^{c- 1}[/itex] and its coefficient will be [itex]c(c-1)a_0+ 2ca_0= (c(c- 1)+ 2c)a_0= (c^2+ c)a_0= 0[/itex]. . We could choose values of c so that the first, "i" terms were 0 but, in order to be specific we we choose so that the very first term, [itex]a_0[/itex] is NOT 0. In order for that to be true we must have [itex]c^2+ c= 0[/itex]. This is called the "indicial equation". Obviously c= 0 and c= -1 satisfy this.

    Putting c= 0 and then c= -1 into the equations for the coefficients leads to recursive equations for [itex]a_n[/itex]. Does any of that sound familiar to you?
     
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