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Homework Help: Differential Equation Help!

  1. May 22, 2005 #1
    Find constants A,B and C such that the function:
    [tex]y = Ax^2 + Bx + C[/tex]

    satisfies the differential equation:
    [tex]y'' + y' - 2y = x^2[/tex]

    [tex]\frac{d}{dx} (y) = \frac{d}{dx} (Ax^2 + Bx + C) = 2Ax + B[/tex]
    [tex]y' = 2Ax + B[/tex]

    [tex]\frac{d}{dx} (y') = \frac{d}{dx} (2Ax + B) = 2A[/tex]
    [tex]y'' = 2A[/tex]

    [tex]2A + 2Ax + B - 2y = x^2[/tex]

    I have been assigned a problem that is not yet covered for another 7 chapters.

    I do not understand the question...

    Any suggestions?

  2. jcsd
  3. May 22, 2005 #2


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    Dearly Missed

    Why didn't you continue with the substitution:
  4. May 22, 2005 #3
    arildno is right. Here is a hint.

    [tex]2A+2Ax+B-2Ax^{2}-2Bx-2C=x^{2}[/tex] can be cleaned up to

    [tex]-2Ax^2+(2A-2B)x+2(A-C)+B = x^2[/tex]

    Do you see any terms [itex]x[/itex] terms on the RHS of the equation? What does this tells you about [itex]2A-2B[/itex]? More over what should [itex]-2A[/itex] equal to, so it can satisfy the RHS of the equation? Apply the same idea for the terms [itex]2(A-C)+B[/itex]
  5. May 22, 2005 #4


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    You should regard what Arildno wrote as what it should really be,viz. an identity

    [tex] (2A-2C+B)+(2A-2B)x-2Ax^{2}\equiv x^{2} [/tex]

  6. Jun 13, 2005 #5
    [tex]A = -\frac{1}{2}[/tex]

    [tex]B = -\frac{1}{2}[/tex]

    [tex]C = -\frac{3}{4}[/tex]
    Last edited: Jun 13, 2005
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