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Cauchy Euler, non-homogeneous, weird condition

  1. Jul 13, 2012 #1
    xy''+y'=-x
    y(1)=0, y(0) bounded (so the natural log, 1/x etc. terms drop out)
    homogeneous, cauchy euler: y=a+bx
    variation of parameters, and using the conditions gives y=1-x, I think (i tried this previously and I think this is what I got, I didnt write it down). Very different from what I have on my solution handout.
    First off, does it matter when to apply the bounded condition? (as in, do we drop the blnx term first? the problem here is variation of parameters doesnt work then),
    Is there another way to solve a non-homogeneous cauchy euler besides using variation of parameters?
     
  2. jcsd
  3. Jul 14, 2012 #2

    tiny-tim

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    hi ericm1234! :smile:
    in this particular case, isn't the LHS obviously a perfect derivative? :wink:
     
  4. Jul 14, 2012 #3
    Man..thanks Tim I see that now. Very easy once that is spotted.
    I tried over and over to solve this as a cauchy euler, then using variation of parameters; would that approach be invalid? I couldnt get the answer.
     
  5. Jul 14, 2012 #4
    I get the homogeneous solution y=a+blnx
    variation of parameters gives u1= x^3/3-x^3/9 and u2=-x^3/3, y=a+blnx+u1+u2lnx
    Different from what I get (the correct answer) the other way.
    Help.
     
  6. Jul 14, 2012 #5
    I should add, again: y(1)=0, and y(0) bounded, are the conditions.
    Correct answer should be 1/4(1-x^2), so again why isn't my cauchy euler way working
     
  7. Jul 14, 2012 #6

    HallsofIvy

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    If you want someone to point out what you did wrong, then you will have to show what you did! I did this problem using "variation of parameters" and I certainly got nothing involving "[itex]x^3[/itex]".
     
    Last edited: Jul 14, 2012
  8. Jul 14, 2012 #7
    I see what I did wrong; my method of variation of parameters requires first dividing by the lead coefficient before going to solve for the particular solution. All good now.
    Thanks.
     
    Last edited: Jul 14, 2012
  9. Jul 14, 2012 #8
    Cauchy euler non-homogeneous problem

    1. The problem statement, all variables and given/known data
    Must solve by cauchy euler, with variation of parameters:
    Original problem: xy''+y'=-x, y(1)=0, y(0) bounded (which I guess means all 1/x, lnx terms etc drop out)




    3. The attempt at a solution

    homogeneous: xy''+y'=0 make substitution y=x^r, so homogeneous soln is y=a+blnx (call it yc)
    variation of parameters:
    u1'=a determinant with entries (from left to right, top to bottom) 0, lnx, -x, 1/x then divide that by the wronskian= 1/x
    u2'= a determinant with entries (from left to right, top to bottom) 1, 0, 0 -x, then divide by wronskian= 1/x
    u1= x^3/3*lnx-x^3/9 u2=-x^3/3
    yp=u1*1+u2*lnx
    y=yc+yp

    But this is wrong; what am I doing wrong?
     
  10. Jul 14, 2012 #9

    HallsofIvy

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    Re: Cauchy euler non-homogeneous problem

    Okay, the characteristic equation is [itex]r^2= 0[/itex] which has r= 0 as a double root. That gives the general solution to the associated homogeneous equation, y= a+ b ln(x).

    Using "variation of parameters", we look for a solution of the form y(x)= u1(x)ln(x)+u2(x). Then y'= u1'ln(x)+ u1/x+ u2'. Because there are, in fact, an infinite number of such solutions, we narrow the search and simplify by requiring that u1'ln(x)+ u2'= 0. Then y'= u1/x so that [itex]y''= u'/x- u/x^2[/itex].

    With that, [itex]xy''+ y'= x(u1'/x- u1/x^2)+ u1/x= u1'= -x[/itex]. Integrating that [itex]u1= -x^2/2[/itex]. [itex]u1'ln(x)+ u2'= -xln(x)+ u2'= 0[/itex] gives us [itex]u2'= xln(x)[/itex]. That is NOT going to give your u1 and u2.
     
  11. Jul 14, 2012 #10
    Re: Cauchy euler non-homogeneous problem

    Thanks.
    I learned a different method (or different way of viewing it anyway) for variation of parameters, and actually it required diving out the lead coefficient before beginning to solve for the particular solution. Everything's good now.
     
    Last edited: Jul 14, 2012
  12. Jul 15, 2012 #11

    HallsofIvy

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    ericm1234, you also posted this under "homework" so I am going to merge the two threads there.

    Double posting is a violation of the forum rules and can result in penalties and a possible ban from the forum.
     
  13. Jul 15, 2012 #12
    I reposted it there as it seems more of a "coursework" relevant post.
     
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