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Homework Help: I can't solve this diff eqtn

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Find a particular soln

    [tex]y'' -5y' +6y = te^t[/tex]


    The characteristic eqtn is [tex]r^2 - 5r + 6 = (r - 2)(r - 3)[/tex]

    r = 2, 3

    According to my book

    [PLAIN]http://img202.imageshack.us/img202/4539/unledxe.jpg [Broken]

    So I thought

    [tex]y = t(At + A_0)e^{2t} + t(Bt + B_0)e^{3t} [/tex]


    But not working...

    what exatly is a simple root?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 30, 2011 #2
    Hi flyingpig! :smile:

    You seemed to use r=2 and r=3, which is incorrect.
    What is the correct r? You only need to look at the right-hand side of

    [tex]y^{\prime\prime}-5y^\prime+6y=te^t[/tex]
     
  4. Jun 30, 2011 #3

    Mark44

    Staff: Mentor

    To elaborate on what micromass said, r = 2 and r = 3 are roots of the characteristic equation, meaning that e2t and e3t are solutions of the homogeneous equation y'' - 5y' + 6y = 0. All solutions of this equation have the form yc = c1e2t + c2e3t, where yc is the complementary solution (the solution to the homogeneous problem).

    What you're asked for is a particular solution of the equation y'' - 5y' + 6y = tet. This solution will involve et, but won't involve e2t or e3t.
     
  5. Jun 30, 2011 #4

    hunt_mat

    User Avatar
    Homework Helper

    I would try a particular solution of the form:
    [tex]
    y_{p}=ate^{t}+be^{t}
    [/tex]
     
  6. Jul 1, 2011 #5
    But that's what (14) says
     
  7. Jul 1, 2011 #6

    hunt_mat

    User Avatar
    Homework Helper

    and did you try it?
     
  8. Jul 1, 2011 #7
    [tex]y = t(At + A_0)e^{t}[/tex]?
     
  9. Jul 1, 2011 #8

    ehild

    User Avatar
    Homework Helper

    No. It is not the same function hunt_mat suggested. Try that.


    ehild
     
  10. Jul 1, 2011 #9

    Mark44

    Staff: Mentor

    You're not reading the problem correctly, particularly item (i).
    "s = 0 if r is not a root of the associated auxiliary equation."
     
  11. Jul 2, 2011 #10
    Yeah I don't understand what dose that mean? r is a root, i have two??
     
  12. Jul 2, 2011 #11
    r has nothing to do with the characteristic equation. r is determined by the right-hand side of the equation, that is: [itex]te^t[/itex].
    Thus t=1 here.

    Now, to determine s, the characteristic equation does come into play. You'll need to check whether r is a root of the characteristic equation.
     
  13. Jul 3, 2011 #12
    That clears up ALOT.

    Just one other thing. Is the table one of those things yo jsut have to remember?
     
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