1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Second order diff eq with two variables
  1. Second Order Diff Eq (Replies: 7)

Loading...