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Math Help - one more undetermined coeffecient

  1. #1
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    one more undetermined coeffecient

    the problem is y(quad prime)+2y(double prime) + y = (x-1)^2

    I keep tackling this problem and coming up with nothing anyone help me here?

    its suppose to come out in the form of y = c1cos(x)+c2 sin(x) +c3 xcos(x)+c4sin(x) +x^2-2x-3
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  2. #2
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    From m^{4}+2m^{2}+1=0, we find m_{1}=m_{3}=i and m_{2}=m_{4}=-i

    Then y_{c}=C_{1}cos(x)+C_{2}sin(x)+C_{3}xcos(x)+C_{4}xs  in(x)

    y_{p}=Ax^{2}+Bx+C

    Sub into the DE as usual and get:

    A=1, \;\ B=-2, \;\ 4A+C=1

    Then A=1, \;\ B=-2, \;\ C=-3

    y_{p}=x^{2}-2x-3

    Therefore,

    y=C_{1}cos(x)+C_{2}sin(x)+C_{3}xcos(x)+C_{4}xsin(x  )+x^{2}-2x-3

    Your troubles are probably just coming from the algebra involved after you sub back into the DE?. That's the best place to go astray.
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  3. #3
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    Thanks agian man, im just getting back to college and my algerba is hazy at best. I ussally start second guessing some steps i take then i forget what step it was by the time i get there.

    Also can you show me the factoring steps on that ? fourth power polynomials is something that is slipping my mind.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by neven87 View Post
    Thanks agian man, im just getting back to college and my algerba is hazy at best. I ussally start second guessing some steps i take then i forget what step it was by the time i get there.

    Also can you show me the factoring steps on that ? fourth power polynomials is something that is slipping my mind.
    m^4 + 2m^2 + 1 = 0
    This is a "biquadratic" equation. Let a = m^2.

    Then
    a^2 + 2a + 1 = 0

    (a + 1)^2 = 0

    a + 1 = 0

    Thus
    m^2 = -1

    Thus
    m = \pm i

    Both of these roots are doubled since the discriminant of the a equation is 0.

    -Dan
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