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Second order ODE initial value problem

  1. Aug 2, 2011 #1
    So the question is y" - y' - 6y = e^-x + 12x, y(0)=1,y'(0)=-2

    First I found the general solution which came out to be, Ae^3x + Be^-2x

    I then Substituted y=ae^-x + bx + c
    y'=-ae^-x + b
    y"=ae^-x
    Then I just compared the coefficients to get a=-1/4, B=-2 and C=-1/6

    So I am getting y = Ae^3x + Be^-2x -(e^-x)/4 - 2x - 1/6
    Im not sure if this is right, I have done the rest but I get some funny answers for A and B so I was wondering if someone could verify if this answer is right. Thanks
     
  2. jcsd
  3. Aug 2, 2011 #2
    Your C value isn't correct; try plugging everything back into the equation and see if you get a different C.
     
  4. Aug 2, 2011 #3
    I just tried it again. Got the same answer.
     
  5. Aug 2, 2011 #4
    Could you show what you did?
     
  6. Aug 2, 2011 #5
    I did ae^-x - (-ae^-x+b) - 6(ae^-x+bx+c) = e^-x + 12x
    Then I got e^-x(-4a-1) - b(1+6c) = x(12+6b)

    After that I did -4a = 1 so a = -1/4
    6c=-1 so c = -1/6
    6b=-12 so b = -2

    Could you tell me where I am going wrong?
     
  7. Aug 2, 2011 #6

    ehild

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    the red one. It is -(b+6c)

    ehild
     
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