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Varation of Parameters fun diff EQ question, where do i go next?

  1. Feb 17, 2006 #1
    Varation of Parameters fun!! diff EQ question, where do i go next?

    This is my first attempt at doing Variation of parameters, didn't go to bad, things cancled out pretty well but now i'm almost done but i'm stuck!
    The problem says: Find the solution of y''+15y'+56y = 54*e^(-5t), with y(0) = 8, and y'(0) = 2, y = ?

    Here is my work:


    If you don't follow me or see an error right off the bat, please let me know! Thanks! :smile:
  2. jcsd
  3. Feb 18, 2006 #2


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    Now integrate u1= -54e3t and u2'= 54e2t to find u1 and u2 of course! Once you know those put them into "u1e-8t+ u2e-7t to get a "specific solution" and add that to the general solution of the homogenous equation.
  4. Feb 18, 2006 #3
    ahh thanks Ivey, i'm alittle confused on what u mean when u said add it to the homogenous equation, the orginal equation wasn't homogenous was set equal to 54e^(-5t), are you saying add it to the r^2+15r+56 = 0? Thanks here is what i got now:
  5. Feb 18, 2006 #4


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    Well, that's pretty much the whole idea! If [itex] y(t)= C_1e^{-8t}+ C_2e^{-7t}[/itex] is the general solution to y"+ 15y'+ 56y= 0 and [itex]y(t)= 9e^{-5t}[/itex] satisfies y"+ 15y'+ 56y= 54e-5t, then
    The general solution to y"+ 15y'+56y= 54e-5t is
    [itex]y(t)= C_1e^{-8t}+ C_2e^{-7t}+ 9e^{-5t}[/itex].
  6. Feb 18, 2006 #5
    Thanks again Ivey!
    I don't know if its right because webhw is still down, yay!
    but this is what I got:
  7. Feb 18, 2006 #6
    I personally think it's easier to solve it like this: First, as usual, you solve the homogeneous equation which is basically setting the right hand side (RHS) to zero, i.e. [itex]y"+ 15y'+56y=0[/itex]. This will give you [itex]y_{h}(t)=C_1e^{-8t}+ C_2e^{-7t}[/itex]. Next you solve the "special" or "particular" equation by making an ansatz (educated guess) depending on what expression you have on the RHS. In this case I would choose the ansatz [itex]y_{p}(t)=Ae^{-5t}[/itex]. Differentiate this and insert into the original equation gives you a value on A. The total solution is simply [itex]y=y_{h}+y_{p}[/itex]. And then use your initial conditions to get the values of [itex]C_1[/itex] and [itex]C_2[/itex].

    Sound complicated, but that's how I used to solve these things. :smile:

    And yes, I too got [itex]C_1=-40[/itex] and [itex]C_2=39[/itex].
  8. Feb 19, 2006 #7
    i like ur method alot better! But how did you make an educational guess of [itex]y_{p}(t)=Ae^{-5t}[/itex] ? I understand why u would guess an e^(t) but how did u konw it should be e^(-5t)? Thanks!
  9. Feb 19, 2006 #8
    You can find out more about that method if you look up 'method of undetermined coefficients.' I find that reduction of order is generally faster because you don't need to memorise specific forms for certain situations.
  10. Feb 19, 2006 #9
    I made the ansatz [itex]y_{p}(t)=Ae^{-5t}[/itex] because the the RHS of the original equation had the exponential term [itex]e^{-5t}[/itex]. More generally, I could have chosen the ansatz [itex]y_{p}(t)=Ae^{-5t}+Be^{+5t}[/itex] but the second term is redundant.
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