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Math Help - 2nd order of differential equations

  1. #1
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    2nd order of differential equations

    Hi, I am studying for my final on Monday, 20th and I came across 4 questions on the 2nd order of differential equations that I couldn’t do. Please help me.

    Problem #1: For what values of c do solutions y"+cy'+y = 0 have infinitely many zeros and tend to zero as t goes to infinity (here we are considering y as a function of t).

    Problem #2: Solve the IVP: y"-y = t, y(0) = 0, y'(0) = 0

    Problem 3: For what values of w do solutions of y"+9y = cos (wt) become unbounded as t goes to infinity. For what values of w are solutions periodic with a period of 3?

    Problem #4: find the inverse Laplace Transform of the function <br />
e^{-10_s}\frac{2}{s^{2}+4}<br />

    thanks very much

    Jen
    Last edited by jenjen; March 18th 2006 at 11:32 PM.
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  2. #2
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    Quote Originally Posted by jenjen
    Hi, I am studying for my final on Monday, 20th and I came across 3 questions on the 2nd order of differential equations that I couldn’t do. Please help me.

    Problem #1: For what values of c do solutions y”+cy’+y = 0 have infinitely many zeros and tend to zero as t goes to infinity (here we are considering y as a function of t).
    The solutions of:

    <br />
y''+cy'+y=0<br />

    are all of the form:

    <br />
Ae^{k_1t}+Be^{k_2t}<br />

    where A and B are arbitrary constants and
    e^{k_it},\ i=1,2 are also solutions of the DE.

    Now lets try y(t)=e^{kt} as a trial solution of the equation:

    <br />
k^2y(t)+cky(t)+y(t)=0<br />

    So:

    <br />
k^2+ck+1=0<br />

    and hence:

    <br />
k=\frac{-c\pm\sqrt{c^2-4}}{2}<br />

    Now if the k's are real the solutions have at most a finite number
    of zeros, so for there to be an infinite number of zeros of y
    we need the k's to be complex so then we must have c<2 (Note
    this does not guarantee that the solution has any zeros for real t,
    but solutions which are real will have an infinite number of zeros).

    Also for y(t) to die away as t \to \infty we need
    that \Re k<0, so c>0. This is because:

    <br />
y(t)=e^{(a+ib)t}<br />

    is oscillatory with increasing amplitude if a>0, of constant amplitude
    if a=0 and of decaying amplitude if a<0.

    Hence for y(t) to have an infinite number of zeros and to die
    away as t \to \infty we need 0<c<2

    RonL
    Last edited by CaptainBlack; March 18th 2006 at 11:09 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack
    where A and B are arbitrary constants and
    e^{k_it},\ i=1,2 are also solutions of the DE.
    why did you choose 1, 2 for i? but not other numbers?
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  4. #4
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    Quote Originally Posted by jenjen
    Problem #2: Solve the IVP: y”-y = t, y(0) = 0, y’(0) = 0
    The general solution of

    <br />
y''-y=t<br />
,

    may be written as h(t)+p(t), where h(t) is the
    general solution of y''-y=0, and p(t) is a particular
    solution (that is any solution of) of y''-y=t.

    Now we know that the general solution of y''-y=0 is:

    <br />
h(t)=Ae^t+Be^{-t}<br />

    Also it is clear that p(t)=-t is a particular solution of y''-y=t,
    so the general solution is:

    <br />
y(t)=Ae^t+Be^{-t} -t<br />
.

    Now we apply the initial conditions and these give:

    <br />
A+B=0<br />
<br />
A-B-1=0<br />
,

    so A=1/2 and B=-1/2. Hence:

    <br />
y(t)=\frac{1}{2}e^t-\frac{1}{2}e^{-t} -t<br />
,

    is the solution to the IVP.

    RonL
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  5. #5
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    Quote Originally Posted by jenjen
    why did you choose 1, 2 for i? but not other numbers?
    because there are only two solutions to the quadratic that you get by
    assuming a exponential solution and k_1 and k_2
    are these two solutions, thats all there is to it.

    The subscript is just an index indicating which solution for k
    we are talking about.

    RonL
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    thanks to you CaptainBlack you are really helpful but I'm still studying hard for my final and I'm still confused on that one and none of my friends are able to do that so...

    Problem #4: find the inverse Laplace Transform of the function <br />
e^{-10_s}\frac{2}{s^{2}+4}<br />
    Last edited by jenjen; March 19th 2006 at 11:17 AM.
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  7. #7
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    Quote Originally Posted by jenjen
    thanks to you CaptainBlack you are really helpful but I'm still studying hard for my final and I'm still confused on that one and none of my friends are able to do that so...

    Problem #4: find the inverse Laplace Transform of the function <br />
e^{-10_s}\frac{2}{s^{2}+4}<br />
    There is some confusion here, I'm not clear on what the e^{-10_s}
    is supposed to denote, I guess it is supposed to be e^{-10s},
    is that right?

    RonL
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    Quote Originally Posted by CaptainBlack
    There is some confusion here, I'm not clear on what the e^{-10_s}
    is supposed to denote, I guess it is supposed to be e^{-10s},
    is that right?

    RonL

    yeah you are right. sorry.
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  9. #9
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    Quote Originally Posted by jenjen
    Problem #4: find the inverse Laplace Transform of the function <br />
e^{-10s}\frac{2}{s^{2}+4}<br />
    First thing we need is the convolution theorem for Laplace transforms:

    <br />
\mathcal{L}^{-1} f(s)g(s) = \int_0^t F(u)G(t-u)\ du<br />

    Then we need the following:

    <br />
\mathcal{L}^{-1} e^{-as}=\delta (t-a)<br />
,

    and:

    <br />
\mathcal{L}^{-1} \frac{1}{s^2+a^2}=\frac{\sin(at)}{a}<br />
.

    So putting this all together we have:

    <br />
\mathcal{L}^{-1} e^{-10s}\frac{2}{s^2+4}=\int_0^t \delta (u-10) \sin(2(t-u)) \ du<br />
,

    which from the definition of the \delta functional gives:

    <br />
\mathcal{L}^{-1} e^{-10s}\frac{2}{s^2+4}=\sin(2(t-10))\ \  t>10<br />
    <br />
=0\ \ t<10<br />

    This also follows directly form the translation theorem.

    RonL
    Last edited by CaptainBlack; March 19th 2006 at 12:47 PM.
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    Quote Originally Posted by CaptainBlack

    <br />
\mathcal{L}^{-1} \frac{1}{s^2+a^2}=\frac{\sin(at)}{a}<br />
    RonL
    why sin(at) and not cos(at) and where does the denominator (a) come from? my book formula states cos(at) without the denominator
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  11. #11
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    nevermind, you are right!! sorry.
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  12. #12
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    but I don't understand why you didn't have to convert <br />
\mathcal{L}^{-1} e^{-10s}<br />
to 1/(s+10)
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  13. #13
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    if we converted that to 1/(s+10), can we use partial fraction expansion to solve the problem? or we don't need partial fraction in this situation?
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  14. #14
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    Quote Originally Posted by jenjen
    why sin(at) and not cos(at) and where does the denominator (a) come from? my book formula states cos(at) without the denominator
    and my book has:

    <br />
\mathcal{L}^{-1} \frac{s}{s^2+a^2}=\cos(at)<br />

    RonL
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  15. #15
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    Quote Originally Posted by jenjen
    but I don't understand why you didn't have to convert <br />
\mathcal{L}^{-1} e^{-10s}<br />
to 1/(s+10)
    Not sure I understand this.

    RonL
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