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Find a particular solution?

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Find a particular solution of y"-5y'+6y=-e^(x)[(4+6x-x^2)cosx-(2-4x+3x^2)sinx].

    2. Relevant equations
    None.

    3. The attempt at a solution
    yp=e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
    y'p=e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
    y"p=e^(x)[(Ax^2+Bx+C)(-cosx+sinx)+(2Ax+B)(-sinx-cosx)+(2Ax+B)(-sinx-cosx)+2A(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
    subbing:
    y"p-5y'p+6yp=4Ax^2e^xcosx+4Bxe^xcosx+4Ce^xcosx+2Ax^2e^xsinx+2Bxe^xsinx+2Ce^xsinx+2Axe^xsinx+Be^xsinx-10Axe^xcosx-5Be^xcosx+2Ae^xcosx-2Ae^xsinx=-e^x[(4+6x-x^2)cosx-(2-4x+3x^2)sinx]
    So 4Ax^2e^xcosx=x^2e^xcosx
    and therefore A=1/4.
    But the answer to this problem is yp=e^(x)(x^2*cosx+2sinx). I don't think the coefficients I got are right. So what's the right initial guess for this problem?
     
  2. jcsd
  3. Mar 6, 2016 #2

    vela

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    How did you come up with your guess?
     
  4. Mar 6, 2016 #3
    Just by making my best guess.
     
  5. Mar 6, 2016 #4

    vela

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    What made you think it was the best as opposed to some other form? I'm asking because writing down the form of particular solution is a cookbook process, yet you managed to come up with something different.
     
  6. Mar 6, 2016 #5
    Since there's -e^x in the very front, so I got e^x as well in forming my guess yp. And then there's (4+6x-x^2) in front of cosx and same thing in front of sinx. So I got yp=e^(x)[(Ax^2+Bx+C)(cosx-sinx)].
     
  7. Mar 6, 2016 #6

    vela

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    The sine doesn't have the same factor multiplying it as cosine in the forcing function, or are you saying the cosine and the sine should both have a quadratic multiplying it? If the latter, why would you expect it to be the exact same quadratic?
     
  8. Mar 6, 2016 #7
    So this is a forcing function?
     
  9. Mar 6, 2016 #8

    vela

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    The forcing function is the stuff that appears on the righthand side of the differential equation.
     
  10. Mar 6, 2016 #9
    So what's the way to find the guess?
     
  11. Mar 6, 2016 #10

    Ray Vickson

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    Have you checked whether the supposed "answer" actually solves the DE?
     
    Last edited: Mar 7, 2016
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