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Second Order Differential Nonhomogeneous Equation

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Use the method of undetermined coefficients to find one solution of
    http://img85.imageshack.us/img85/6844/4ab921ad6ba6851cc91401c.png [Broken]
    Note that the method finds a specific solution, not the general one.

    2. Relevant equations
    Y = Yc + Yp
    Yc = C1e^(r1t)+C2e^(rt) when roots are not the same.

    3. The attempt at a solution
    For the homogeneous part I used the quadratic equation to get the roots -1 and -4.
    Y = Yc + Yp
    Yc = C1e^-t +C2e^-4t

    I tried Yp = (At^2+Bt+C)*De^(4t) but it doesn't seem correct after solving a bunch of derivatives and plugging it back into y'' + 3y' -4y and solving for constants.

    How to determine the Y-particular for this problem?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 9, 2011 #2
    Hii. Method of undetermined coefficients is sometimes lengthy but it sure does give the correct particular integral. As u hav already written the rough Yp,what u hav to do now is just replace it for y in the original equation and compare the coefficients of (e^4t)t^2,(e^4t)t and e^4t on both sides of the equation and u will get the expression for Yp. There are however other useful and short methods for finding Yp. One is the D Operator method and the other is the variation of parameters method. These are too long for me to describe elaborately here,but u must check these out in the net or any other reference books.Any standard engineering mathematics book would contain these topics.Personally I recommend using the D Operator method as it is very useful when exploited efficiently.
     
  4. Dec 9, 2011 #3
    And welcome to Physics Forums......
     
  5. Dec 9, 2011 #4

    Dick

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    Can you show what you did and where it went wrong? The numbers aren't pretty but it certainly seems to be working fine for me. You don't need the D parameter. Set it equal to 1. Then I'll get you started. A=(-1/24).
     
    Last edited by a moderator: May 5, 2017
  6. Dec 9, 2011 #5
    And one more thing, the roots of your auxiliary equation are 1 and -4 respectively.
     
  7. Dec 9, 2011 #6

    Dick

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    That would be true. Nice catch. But you don't need them to find the particular solution.
     
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