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    http://www.mediafire.com/view/?grhasyd8ilzemsw


    I'm not completely sure what I'm being asked to do in the first part.

    This is what I've done :

    I let y = e^(mx) and the substitute into the given equation to get a quadratic:

    m2-bm+4 Since b<0 it must be negative

    I then complete the square to obtain two solutions :

    Click here

    The problem that I now face is how do I express the general solution. Depending on the value of b^2 in the square root I may have something of the form:

    y=Acoswx + Bsinwx

    or

    y=Aex+Bex
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    The critical case is when the quadratic has a repeated root. For which value negative of b does this occur? You should also know that when the root is repeated (call it r), the solution to the second-order linear homogenous ODE is (A+Bx)exp(rx).
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    (Original post by Ari Ben Canaan)
    http://www.mediafire.com/view/?grhasyd8ilzemsw


    I'm not completely sure what I'm being asked to do in the first part.

    This is what I've done :

    I let y = e^(mx) and the substitute into the given equation to get a quadratic:

    m2-bm+4 Since b<0 it must be negative

    I then complete the square to obtain two solutions :

    Click here

    The problem that I now face is how do I express the general solution. Depending on the value of b^2 in the square root I may have something of the form:

    y=Acoswx + Bsinwx

    or

    y=Aex+Bex
    "critical case" means double root.

    So you want the quadratic

    m^2 + bm + 4=0

    to have real, equal roots.

    If

    \lambda

    is the value for which the quadratic holds, then it is the repeated root so

    y = (A+Bx)e^{\lambda x}

    For the second part, try

    \lambda \sin x + \mu \cos x

    for your particular solution.

    You could (and I would) use complex variables to make the algebra less messy.
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    (Original post by Lord of the Flies)
    The critical case is when the quadratic has a repeated root. For which value negative of b does this occur? You should also know that when the root is repeated (call it r), the solution to the second-order linear homogenous ODE is (A+Bx)exp(rx).
    Ah, I see. I never knew that referred to a repeated root case. So when b^2 = 4ac b^2 = 16 Hence, my answer to the quadratic simplifies to :

    (16)^0.5 / 2 = ±2 However, we are told b<0 hence the general solution is :

    y = (A+Bx)e^2x

    Is that correct ?

    Thank you everyone ! +rep to all.
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    (Original post by Ari Ben Canaan)
    Ah, I see. I never knew that referred to a repeated root case. So when b^2 = 4ac b^2 = 16 Hence, my answer to the quadratic simplifies to :

    (16)^0.5 / 2 = ±2 However, we are told b<0 hence the general solution is :

    y = (A+Bx)e^2x

    Is that correct ?

    Thank you everyone ! +rep to all.
    Yes
 
 
 
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