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This problem is hindering my overall work

  1. Nov 8, 2006 #1
    This problem is hindering my overall work !!!

    Need to find particular solution.
    y"+6y'+5y=-9te^(5t)

    What would be my guess work: I tried y_p=Ate^(5t)+Bte^(5t)
    Any thought on the guess?
    Thanks,
    Dmitriy.
     
  2. jcsd
  3. Nov 9, 2006 #2

    dextercioby

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    If you like lenghty calculations, you might use the method of variation of constants. Assume for example that the particular solution to the given ODE is of the form

    [tex] y_{p}=C(t)e^{-5t} [/tex]

    You need to find the exact expression of C(t). For that plug in the ODE and then get another nonhomogenous second ODE whose particular solution you need to find

    I get something like

    [tex] C''(t)-4C'(t)=-9te^{10t} [/tex]

    for which i use the ansatz

    [tex] C(t)=D(t)e^{4t} [/tex]

    and finally getting the ODE

    [tex]D''(t)=-9te^{6t} [/tex]

    which can be integrated twice to get a solution depending on 2 constants of integration which you'll determine by plugging in the initial ODE.

    Daniel.
     
    Last edited: Nov 9, 2006
  4. Nov 9, 2006 #3

    dextercioby

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    It's too complicated what i did above (perhaps even incorrect).

    Try the ansatz

    [tex] y_{p}=C(t)e^{5t} [/tex]

    THe ODE for C(t) is

    [tex] C''(t)+16C'(t)+60C(t)=-9t [/tex]

    whose particular solution is

    [tex]C_{p}(t)=\frac{1}{25}-\frac{3}{20}t [/tex]

    and whose general solution, after using the ansatz leads to the same general solution of the initial ODE, so nothing new. Ergo, the particular solution to the initial ODE is

    [tex] y_{p}(t)=\left(\frac{1}{25}-\frac{3}{20}t\right)e^{5t} [/tex]

    Daniel.
     
  5. Nov 9, 2006 #4
    dexterciby: This is works perfect.
    however, it certainly interfere with my previous knowledge about non homogeneous EQ's. one of the solutions for the characteristic equation is e^(-5t), wouldn't we have to multiply everything by another "t"?
    Please explain?
     
    Last edited: Nov 9, 2006
  6. Nov 10, 2006 #5

    dextercioby

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    No, i absorbed any linear "t" dependence in the unknown function C(t).

    Daniel.
     
  7. Nov 10, 2006 #6

    HallsofIvy

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    Since e5t is not a solution to the associated homogeneous equation (e-5t is), a solution to the entire equation will be of the form yp(t)= (At+ B)e5t. Plug that into the equation and solve for A and B.
     
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