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Homework Help: Second order diff eq with two variables

  1. Mar 8, 2007 #1
    Here's the problem:
    x^2y''-3xy'-12y=0

    with initial conditions y(1)=0 and y'(1)=7

    I'm supposed to solve for y in the form y=c1y1+c2y2

    y1 = x^6 by inspection

    Now to solve for y2

    y2=y1v

    v can be solved for by the equation
    y1v''+(2y1'+py1)v'=0
    where p is the function in front of y' when there is no function in front of y'':
    x^2y''-3xy'-12y=0 divide by x^2
    y''-3/xy'-12/x^2y=0
    so p=-3/x

    x^6v''+(2*6x^5+(-3/x)x^6)v'=0
    x^6v''+(12x^5-3x^5)v'=0
    x^6v''=-9x^5v'
    so v'=e^(-x^2/18)
    and v= -x/9(e^(-x^2/18)+c

    so y2=x^6(-x/9(e^(-x^2/18)))
    y2=-x^(7)/9*e^(-x^2/18)

    Now y=c1y1+c2y2 and my initial conditions are y(1)=0 and y'(1)=7

    c1*x^6+c2*-x^(7)/9*e^((-x^2)/18)
    y'=c1y1'+c2y2'
    = c1*6*x^5+
    c2*((-7/9*x^6*e^((-x^2)/18)+((-x^7)/9*(-x/9)*e^((-x^2)/18)

    Subbing in 1 for x for both equations I get
    c1+c2*-1/9*e^(-1/18)
    and
    6c1+c2*(-7/9*e^(-1/18)+(-1/9*-1/9*e^(-1/18))
    =
    6c1+c2*(-63/81*e^(-1/18)-1/81*e^(-1/18)
    =
    6c1-64/81*e^(-1/18)c2

    I solved for c1 in the first equation:
    c1=c2*1/9*e^(-1/18)
    and plugged that into the second equation
    6*c2*1/9*e^(-1/18)-64/81*e^(-1/18)*c2=7
    c2*e^(-1/18)*(54/81-64/81)=7
    c2*e^(-1/18)*-10/81=7
    c2=-81/10*7*e^(1/18)

    so c1=(-81/10*7*e^(1/18))*1/9*e^(-1/18)
    c1=-567/10*e^(1/18)*1/9*e^(-1/18)
    =-567/90

    Now I plug those back in the equation c1y1+c2y2=y and that's not the right answer. There's a good chance I messed up in there somehow, although I've double checked everything. Does anyone see what I did wrong, or better yet, an easier way to solve this problem. This way is just ridiculous.

    Thanks a lot.
     
  2. jcsd
  3. Mar 8, 2007 #2

    Dick

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    If the first solution is a power law (x^6) it's a good guess that there may be another power law solution. Put y=x^n and solve for n.
     
  4. Mar 8, 2007 #3
    I think the two different solutions have to be independent of each other - or not be multiples of each other. I'll give it a shot and see what happens though. Thanks.
     
  5. Mar 8, 2007 #4
    wow, yeah, that's all I had to do. Thanks a lot.
     
  6. Mar 8, 2007 #5

    Dick

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    x^n and x^m where n and m are different are independent.
     
  7. Mar 9, 2007 #6

    HallsofIvy

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    In other words, with u= v', u'= (-9/x)u
    du/u= -9dx/x so ln(u)= -9 ln x+ C, u= v'= Cx-9
    v= Cx-8 and so vx6= x-2
    NOT what you give:

     
  8. Mar 9, 2007 #7

    D H

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    You were too quick on the draw with "y1 = x^6 by inspection".
    An easy way to solve this is to assume each [itex]y_i[/itex] is of the form [itex]y_i=x^n[/itex]. Then

    [tex]x^2y_i^{\prime\prime}-3xy_i^{\prime}-12y_i =
    n(n-1)x^2x^{n-2} -3nxx^{n-1}-12x^n = (n(n-1)-3n-12)x^n[/tex]

    The characteristic polynomial

    [tex]n(n-1)-3n-12 = n^2-4n-12[/tex]

    has zeros [itex]n=6[/itex] and [itex]n=-2[/itex]. Thus

    [tex]y=c_1x^6 + c_2x^{-2}[/tex]

    Applying the given conditions [itex]y(1)=0[/itex], [itex]y'(1)=7[/itex] to the above leads to a unique solution for [itex]c_1[/itex] and [itex]c_2[/itex].
     
  9. Mar 9, 2007 #8

    HallsofIvy

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    Yes, that's the simplest way to do it, but I suspect he was told to find a second solution by reducing the order- and that "x6 is a solution by inspection" was in the book!
     
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