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Math Help - second order DEs

  1. #1
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    second order DEs

    i'm having trouble solving these two second order nonhomogenous DEs

    y"+6y'+9y=(e^-3x)/(x^2+1)

    and

    (D^2+a^2)y = cosbt
    where y'(0)=y(0)=0 and a and b are distict positive numbers

    any kinda hints or help would be appreciated thanks
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  2. #2
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    #1:

    Variation of Parameters


    From the auxilaiary equation we have m^{2}+6m+9=0

    C_{1}e^{-3x}+C_{2}xe^{-3x}

    From the Wronskian:

    W=\begin{vmatrix}e^{-3x}&xe^{-3x}\\-3e^{-3x}&(1-3x)e^{-3x}\end{vmatrix}=e^{-6x}

    W_{1}=\begin{vmatrix}0&xe^{-3x}\\\frac{e^{-3x}}{1+x^{2}}&(1-3x)e^{-3x}\end{vmatrix}=\frac{-xe^{-6x}}{x^{2}+1}

    W_{2}=\begin{vmatrix}e^{-3x}&0\\-3e^{-3x}&\frac{e^{-3x}}{1+x^{2}}=\frac{e^{-6x}}{1+x^{2}}

    u'_{1}=\frac{W_{1}}{W}=\frac{-x}{x^{2}+1}

    u'_{2}=\frac{W_{2}}{W}=\frac{1}{x^{2}+1}

    u_{1}=\int\frac{-x}{x^{2}+1}dx=\frac{-ln(x^{2}+1)}{2}

    u_{2}=\int\frac{1}{x^{2}+1}dx=tan^{-1}(x)

    So, we have:

    y=e^{-3x}\left(C_{1}+C_{2}x+xtan^{-1}(x)-\frac{ln(x^{2}+1)}{2}\right)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by action259
    (D^2+a^2)y = cosbt
    where y'(0)=y(0)=0 and a and b are distict positive numbers
    The general solution is the sum of the general solution of the
    homogeneous equation:

    <br />
y''+a^2y=0<br />
,

    and a particular integral of the inhomogeneous equation.

    The solution of the homogeneous equation is routine, for the inhomogeneous
    use a trial solution of the form:

    <br />
y(t)=A \cos(bt) + B \sin(bt)<br />

    Then we find:

    <br />
-A b^2 \cos(bt)-B b^2 \sin(bt)+ A a^2 A \cos(bt)+B a^2 \sin(bt)=\cos(bT)<br />
,

    so B=0, and A=1/(a^2-b^2).

    Fitting the initial conditions is now just routine and left to the reader.

    RonL
    Last edited by CaptainBlack; September 3rd 2006 at 07:01 AM.
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  4. #4
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    Quote Originally Posted by CaptainBlank

    Then we find:

    <br />
-A b^2 \cos(bt)-B b^2 \sin(bt)+a^2 \cos(bt)=\cos(bT)<br />
,
    I belive you made a mistake. You did not substitute after the a^2.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker
    I belive you made a mistake. You did not substitute after the a^2.
    Its a typo, you will note it is implicit in the solution given. Comes from
    typing a TeX version of the answer off of a scrap of paper.

    It will be corrected.

    RonL
    Last edited by CaptainBlack; September 3rd 2006 at 09:39 AM.
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  6. #6
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    Quote Originally Posted by CaptainBlack
    Comes from
    typing a TeX version of the answer off of a scrap of paper.
    I do not do that. I simply type do problem through LaTeX. Sometimes I spend much time because I am unable to complete the solution in the end and delete my post.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    I do not do that. I simply type do problem through LaTeX. Sometimes I spend much time because I am unable to complete the solution in the end and delete my post.
    Yes, but I only get the computer for short periods as the kids are playing
    games on it all day at the moment. So I have to do as much as possible off
    line!

    RonL

    (I keeps saying I will set up a wirless network, but every time I
    get close, someone finds something else to spend the budget on )
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  8. #8
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    Quote Originally Posted by galactus
    #1:

    Variation of Parameters
    I had been contemplating posting a variation of parameters solution, but
    there was so much TeX involved, and it looked so ugly I went off looking
    for another method.

    Looking at the final result it is so ugly this could only have been set as an
    exercises in variation of parameters method.

    RonL

    (Any opinions expressed in this post are the authors and do not necessarily
    reflect those of the management )
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  9. #9
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    Quote Originally Posted by CaptainBlack View Post
    The solution of the homogeneous equation is routine, for the inhomogeneous use a trial solution of the form:

    <br />
y(t)=A \cos(bt) + B \sin(bt)<br />

    Then we find:

    <br />
-A b^2 \cos(bt)-B b^2 \sin(bt)+ A a^2 A \cos(bt)+B a^2 \sin(bt)=\cos(bT)<br />
    How can you find that ? When I put Acos(bt) + Bsin(bt) in the original equation I get 0 = cos(bt), how can you have found A and B with it ?
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