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Math Help - A simple DE...for some.

  1. #1
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    A simple DE...for some.

    If y=e^{ax}cos(bx) and y'''+2y'+3y=0 what is a and b?
    That is the problem to solve and I don't even know what to do.
    Can anyone help me out a little and give me a suggestion or two for the solution of this one?
    Last edited by a4swe; August 31st 2006 at 04:14 AM. Reason: I missed the cos(bx)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by a4swe
    If y=e^{ax} and y'''+2y'+3y=0 what is a and b?
    That is the problem to solve and I don't even know what to do.
    Can anyone help me out a little and give me a suggestion or two for the solution of this one?
    You don't appear to have a "b".

    But never mind. What you do is calculate the derivatives from your given
    form for y(x), and substitute them into the DE. That will give you an
    algebraic equation (after dividing the common factor e^{ax} which will occur
    on the left hand side of the equation), this will allow you to determine what
    a (and b?) have to be.

    RonL
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  3. #3
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    Excuse me, that should be

    <br />
y=e^{ax}cos(bx)<br />
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by a4swe
    Excuse me, that should be

    <br />
y=e^{ax}cos(bx)<br />
    The basic idea is to keep differentiating, and at each stage substitute for
    complicated expression simpler ones base on what you already know.

    Finally you should arrive at an equation of the required form, then you can
    equate coefficients between the given DE and what you have found from
    repeated differentiateion to get some algebraic equations for a and b.

    (You will need to check the algebra in what follows)

    If:

    <br />
y=e^{ax}\cos(bx)<br />

    Then:

    <br />
y'=ae^{ax}\cos(bx)- be^{ax}\sin{bx} =a y - be^{ax}\sin{bx}<br />
,

    and so:

    <br />
y''=a y' - bae^{ax}\sin{bx} -b^2e^{ax}\cos(bx) <br />
=ay'-b^2y- bae^{ax}\sin{bx}=ay'-b^2y+ay'-a^2y<br />
,

    hence:

    <br />
y''=2ay'-(a^2+b^2)y<br />

    so:

    <br />
y'''=2ay''-(a^2+b^2)y'=4a^2y'-2a(a^2+b^2)y-(a^2+b^2)y<br />

    which you should now rearrange into the form y'''+Ay'+By=0 then equate
    coefficients with the given form to get the algebraic equations for a and
    b.

    RonL
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  5. #5
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    Why didn't I think about this?
    Thank you very much, this helps a lot.
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  6. #6
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    Quote Originally Posted by a4swe
    If y=e^{ax}cos(bx) and y'''+2y'+3y=0 what is a and b?
    That is the problem to solve and I don't even know what to do.
    Can anyone help me out a little and give me a suggestion or two for the solution of this one?
    Here is another way.
    Since this is a homogenous linear differencial equation of order three with constand coefficients look at the charachachteriistic polynomial.
    ---
    k^3+2k+3=0
    Trivially, k=-1 is a solution.
    Synthetic division,
    k^3+2k+3\div k+1=k^2-k+3
    The solution of,
    k^2-k+3=0 are,
    k=\frac{1}{2}\pm i\frac{\sqrt{11}}{2}
    Therefore the general solution of this differential equation is,
    C_1e^{-x}+C_2e^{1/2 x}\sin \left( \frac{\sqrt{11}}{2} x\right)+C_3e^{1/2 x}\cos \left( \frac{\sqrt{11}}{2} x\right)
    The particular solution you have is when C_1=C_3=0,C_2=1
    Thus,
    e^{1/2 x}\sin \left( \frac{\sqrt{11}}{2} x \right)=e^a \sin (bx)
    Thus,
    a=1/2 \mbox{ and }b=\frac{\sqrt{11}}{2}
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