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Differential Equation dont understand t>0 how to apply that to equation

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data

    y' + (2/t)y = cos(t) / t^2 initial cond: y(pi) = 0 .... t>0

    2. Relevant equations
    integrating factor is t^2

    so.. integ (yt^2)' = integ cos (t)

    = yt^2 = -sin (t) + c

    y = ( -sin (t) /t^2 ) + c

    ..this brings up my first question.. when i divided through by t^2 to isolate y, am i correct in saying that I dont need to also divide c by t^2, because a constant divided by a constant is still just another constant.. right?

    so .. with that assumption...

    0 = ( - sin (pi) / pi^2 )+ c

    0 = ( - 0 / pi^2 ) + c
    0 = c

    this where im stuck, because i realize IC said that t>0.. but what does it mean to apply that IC to the equation?? Does this lead to an actual value for c somehow? thanks
    edit: if it says that t = pi .. that is, y(pi) = 0 .. then why would it also say t>0 ?? pi is clearly bigger than zero, so why the added condition?




    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 11, 2010 #2

    Mark44

    Staff: Mentor

    Mistake in the line above.
    No, you are not correct. You need to divide c by t^2 -- t^2 is not a constant.
    Because the diff. equation is undefined for t = 0.
     
  4. Apr 11, 2010 #3
    thanks for reply.

    yt^2 = sin(t) + c

    y = sin(t) / t^2 + c/t^2

    IC: y(pi) = 0

    0 = 0 / pi^2 + c/pi^2

    c = 0(pi^2)

    c = 0

    i understand why it is undefined at zero but Im not grasping how to apply that info to solve the equation. thanks for any more help
     
  5. Apr 11, 2010 #4

    Mark44

    Staff: Mentor

    You don't need to apply the information that t > 0. And you almost there with the solution.

    You have y = sin(t) / t^2 + c/t^2 and c = 0, so y = ?
     
  6. Apr 11, 2010 #5
    if t = 0 then i can cancel last term

    y = sin(t) / t^2

    that is the solution to the problem?
     
  7. Apr 11, 2010 #6

    Mark44

    Staff: Mentor

    No, t can't be zero.

    You have y(t) = sin(t) / t^2 + C/t^2 and y(pi) = 0, so
    0 = y(pi) = sin(pi)/t^2 + C/t^2
    ==> C/t^2 = 0 ==> C = 0

    So y(t) = sin(t) / t^2, and we still have the restriction that t > 0.
     
  8. Apr 12, 2010 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You mean "if c= 0". And you should include "t> 0" in your solution:

    y= sin(t)/t^2, t> 0.

    Since the differential equation does not hold at 0, not only can t not be 0 but you cannot continue the solution to negative values of t.
     
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