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Linear differential equaton

  1. Nov 13, 2005 #1
    for the following question:
    y``+2y`-35y=12e^(5x)+37sin5x

    my problem:
    yh=c1e^(-7x)+c2e^(5x)
    suppose yp=c3xe^(5x)+c4sinx+c5cos5x
    then yp`=c3e^(5x)+5c3xe^(5x)+5c4cos5x-5c5sin5x
    so yp``=8c3e^(5x)+(-60c4-10c5)sin5x+(-60c5+10c4)
    which means that 8c3=12 => c3=2/3
    also from (-60c4-10c5)=37 and (10c4-60c5)=0 implies that
    c4=-3/5 and c5=-1/10
    so y=c1e^(-7x)+c2e^(5x)+(3/2)xe^(5x) +(3/2)sin5x-(1/10)cos5x
    but the correct answer should be
    y=c1e^(-7x)+c2e^(5x)+xe^(5x) +(3/2)sin5x-(1/10)cos5x

    i've checked this many times, but i don't know where my calculations went wrong... :P
     
  2. jcsd
  3. Nov 13, 2005 #2

    HallsofIvy

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    Science Advisor

    Since your coefficents are correct for sine and cosine, just recheck the exponential.
    Let y= c3xe5x

    Then y'= c3e5x+ 5c3xe5x just as you say.

    y"= 5c3e5x+ (5c3e5x+ 25c3xe5x)
    = 103e5x+ 25c3xe5x

    I don't see where you got the "8".

    Now y"= 103e5x+ 25c3xe5x
    +2y'= 2c3e5x+10c3xe5x
    -35y= -35c3xe5x

    So we must have 12c3= 12.
     
  4. Nov 13, 2005 #3
    thank you very much!!! i'll redo my calculations...
     
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