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Variation of parameters. unsure why my solution differs from professor's

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data
    what is general solution of 2y'' - 3y' + y = ((t^2) + 1)e^t


    2. Relevant equations

    my particular solution is: (e^t) ((2/3)(t^3) + 6t -4))
    prof particular solution is: ((1/3)(t^3)(e^t)) - 2(t^3)(e^t) + 9(te^t)

    3. The attempt at a solution
    here is how i solved , i hope this is ok to post link tp pdf..its faster than typing this all out..
    https://docs.google.com/open?id=0BwJqUg33PgREQTFaejBCUkRRWlNfRW9maU1oaDNYQQ

    thanks for any help/advice/hints
     
    Last edited by a moderator: Mar 2, 2012
  2. jcsd
  3. Mar 1, 2012 #2

    ehild

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    You can check if your solution is correct by substituting back into the original equation.


    ehild
     
  4. Mar 1, 2012 #3

    ehild

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    As the coefficient of y'' is not 1 in the original equation, you have to divide the right-hand side by 2 when calculating u' and v'.

    I think you did not copy your professor's solution correctly. It has to be
    (1/3 t3-2t2+9t)et.

    ehild
     
    Last edited: Mar 1, 2012
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