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Math Help - 2nd order of differential equations

  1. #16
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    laplace transform can be converted using this formula right?

    <br />
{e}^{at}\leftrightarrow \frac{1}{(s+a)}<br />
    that is where I got <br />
{e}^{-10s}= \frac{1}{s+10}<br />
    Last edited by jenjen; March 19th 2006 at 02:28 PM.
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  2. #17
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    Quote Originally Posted by jenjen
    laplace transform can be converted using this formula right?
    <br />
{e}^{at}\leftrightarrow \frac{1}{(s+a)}<br />

    that is where I got {e}^{-10s}= \frac{1}{s+10}
    You mean,
    \mathcal{L}\{e^{at}\}=\frac{1}{s-a}
    You mistake is,
    \frac{1}{s+a}
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  3. #18
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    yeah that is correct. THANK YOU. so then how do i suppose to approach this problem altogether? do i have to use partial expansion?
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  4. #19
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    Quote Originally Posted by ThePerfectHacker
    You mean,
    \mathcal{L}\{e^{at}\}=\frac{1}{s-a}
    You mistake is,
    \frac{1}{s+a}
    Except we are in the s domain, we want \mathcal{L}^{-1} e^{-10s}.

    RonL
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  5. #20
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    so if that is true, then do we use partial fraction expansion to solve the problem? I am still confused about why we don't use partial fractions. because since
    {e}^{at}\leftrightarrow \frac{1}{(s-a)}

    {e}^{-10s}= \frac{1}{s+10}

    \left( \frac {1}{s+10} \right) \left( \frac {2}{s^2+4} \right)

    \frac {2}{(s+10)(s^2+4)}

    \frac {A}{s+10}+ \frac {B}{s^2+4}

    from here, I am stuck. Is this the right method?
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  6. #21
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    Quote Originally Posted by jenjen
    so if that is true, then do we use partial fraction expansion to solve the problem? I am still confused about why we don't use partial fractions. because since
    {e}^{at}\leftrightarrow \frac{1}{(s-a)}

    {e}^{-10s}= \frac{1}{s+10}

    \left( \frac {1}{s+10} \right) \left( \frac {2}{s^2+4} \right)

    \frac {2}{(s+10)(s^2+4)}

    \frac {A}{s+10}+ \frac {B}{s^2+4}

    from here, I am stuck. Is this the right method?
    If you want to decompse
    \frac{2}{(s+10)(s^2+4)}
    Then you need to write it as, (because of the quadradic )
    \frac{A}{s+10}+\frac{Bs+C}{s^2+4}
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  7. #22
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    can you help me solve that?? I tried many times but I can't seem to get it to work.

    I also have another problem I couln't do.

    problem: the absent-minded Dr. R claims that {y_1}=cosh(t), {y_2}=sinh(t) form a linearly independent set of solutions for the differential equation {y"-y=0}. Is he right? If so: how can we conclude that we can solve the IVP, {y"-y=0}, {y(5)}={7}, {y'(5)}={10} with a function of the form {y}={c_1y_1+c_2y_2} where {c_1} and {c_2} are constants.
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  8. #23
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    Quote Originally Posted by jenjen
    can you help me solve that?? I tried many times but I can't seem to get it to work.

    I also have another problem I couln't do.

    problem: the absent-minded Dr. R claims that {y_1}=cosh(t), {y_2}=sinh(t) form a linearly independent set of solutions for the differential equation {y"-y=0}. Is he right? If so: how can we conclude that we can solve the IVP, {y"-y=0}, {y(5)}={7}, {y'(5)}={10} with a function of the form {y}={c_1y_1+c_2y_2} where {c_1} and {c_2} are constants.
    Remember that,
    \cosh x=\frac{e^x+e^{-x}}{2}
    \sinh x=\frac{e^x-e^{-x}}{2}
    They only differ by signs joining exponents. We need to show that y''-y=0 for \frac{e^x\pm e^{-x}}{2}
    Thus,
    y=\frac{e^x\pm e^{-x}}{2}
    y'=\frac{e^x\mp e^{-x}}{2}
    y''=\frac{e^x\pm e^{-x}}{2}
    In other words y''=y thus, y''-y=0
    Notice that \cosh x\not =\sinh x for all x for some k. Thus, they are linearly independent.

    Thus, the general solution to this differencial equation is,
    y=C_1\cosh x+C_2\sinh x

    Please next time start a new thread, if it is off topic.
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  9. #24
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    how come the y(5)=7 and y'(5)=10 isn't included in the answer?
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  10. #25
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    Quote Originally Posted by jenjen
    so if that is true, then do we use partial fraction expansion to solve the problem? I am still confused about why we don't use partial fractions. because since
    {e}^{at}\leftrightarrow \frac{1}{(s-a)}
    This post is an orphan, because you have not quoted anything form the
    previous post we are guessing what your question is at this point.


    RonL
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  11. #26
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    Quote Originally Posted by ThePerfectHacker
    You mean,
    \mathcal{L}\{e^{at}\}=\frac{1}{s-a}
    You mistake is,
    \frac{1}{s+a}
    No his mistake is that he wants:

    \mathcal{L}^{-1} e^{as}

    which I have already dealt with in an earlier post in this thread.

    RonL
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  12. #27
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    well, thank you so much Captainblack for all the help. They were very helpful. I am going to school to take my final now.


    Jen
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