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Math Help - Laplace transforms

  1. #1
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    Laplace transforms

    Ok, so I've got the question:

    Use laplace transforms to find x(t) when x and y satisfy the differential equations:

    dx/dt -dy/dt -2x = 0

    and

    dx/dt +dy/dt -2y = t

    With x = 1, y = 2 at t = 0

    So far i've taken laplace transforms of each side and substituted one into the other to get a single equation for Laplace(x)

    and got to the point where

    Laplace(x) = {3*(1/s)*[(s-1)^2/(s-1)^2 + 1]} + [3* 1/(s-1)^2 + 1] -{15/6 *(1/s)*(1/(s-1)^2 + 1)}

    So now i presume that i just inverse the laplace transforms that i have gotten, but how do you invert the parts where there are two transforms multiplied together such as the 1st and 3rd terms in the above?

    is it just a case of : Laplace^-1 {3*(1/s)*[(s-1)^2/(s-1)^2 + 1]} is

    simply the product of the individual transforms once they have been inverted ie

    3 * 1 * exp(t) * cos(t)

    OR is there some other way that i am not aware of?
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  2. #2
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    Quote Originally Posted by macabre View Post
    Ok, so I've got the question:

    Use laplace transforms to find x(t) when x and y satisfy the differential equations:

    dx/dt -dy/dt -2x = 0

    and

    dx/dt +dy/dt -2y = t

    With x = 1, y = 2 at t = 0
    Let, the functional transform be,
    \int_0^{\infty}e^{-st}x(t)dt=F(s)
    \int_0^{\infty}e^{-st}y(t)dt=G(s)
    Then the Laplace Transforms of the derivatives are respectively:
    -x(0)+sF(s)=-1+sF(s)
    -y(0)+sG(s)=-2+sG(s)
    Take Laplace Transform of both sides.
    Thus, we have, from the system of differencial equations,
    -1+sF(s)+2-sG(s)-2F(s)=0
    -1+sF(s)-2+sG(s)-2G(s)=\frac{1}{s}
    Thus,
    F(s)(s-2)=sG(s)-1
    Thus, (note s>2)
    F(s)=\frac{sG(s)-1}{s-2}
    Substitute into second functional equation,
    -3+\frac{s^2G(s)-s}{s-2}+sG(s)-2G(s)=\frac{1}{s}
    Multiply, to clear denominator,
    -3(s-2)+s^2G(s)-s+s(s-2)G(s)-2(s-2)G(s)=\frac{s-2}{s}
    Thus,
    G(s)[s^2+s(s-2)-2(s-2)]-3(s-2)-s=\frac{s-2}{s}
    Thus,
    G(s)[s^2+s^2-2s-2s+4]-3s+4-s=\frac{s-2}{s}
    Thus,
    G(s)[2s^2-4s+4]-4s+4=\frac{s-2}{s}
    Thus,
    G(s)[2s^2-4s+4]=\frac{s-2}{s}+4s-4
    Thus,
    G(s)[2s^2-4s+4]=\frac{s-2+4s^2-4s}{s}
    Thus,
    G(s)=\frac{4s^2-3s-2}{s(2s^2-4s+4)}
    Now I think there is something wrong with your problem because the procedure now is to decompose after factorization but the factorization is irrational, unlikely that they will give that as a homework problem.
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  3. #3
    TD!
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    Quote Originally Posted by ThePerfectHacker View Post
    Thus, we have, from the system of differencial equations,
    -1+sF(s)+2-sG(s)-2F(s)=0
    -1+sF(s)-2+sG(s)-2G(s)=\frac{1}{s}
    The Laplace transform of t is 1/sē, not 1/s. We have:

    \mathcal{L}\left\{ {x' - y' - 2x} \right\} = sX - x_0  - sY + y_0  - 2X = \left( {s - 2} \right)X - sY - 3

    and

    \mathcal{L}\left\{ {x' - y' - 2y - t} \right\} = sX - x_0  - sY + y_0  - 2Y - s^{ - 2}  = sX - \left( {s + 2} \right)Y - s^2  - 3

    So this yield the following system of equations in X and Y:

    <br />
\left\{ \begin{array}{l}<br />
 \left( {s - 2} \right)X - sY = 3 \\ <br />
 sX - \left( {s + 2} \right)Y = s^2  + 3 \\ <br />
 \end{array} \right.<br />

    With solutions:

    <br />
X = \frac{1}{4}s - \frac{3}{2} \; \wedge \; Y = \frac{1}{4}s - \frac{1}{2}s^2  - \frac{3}{2}<br />

    Now x = \mathcal{L}^{-1}(X) \; \wedge \; y = \mathcal{L}^{-1}(Y)

    I doubt you'll have to work with Dirac-impulses though, so I'm assuming you won't have to take those constant terms into account.
    If so, then you get the easy solution x(t) = 1/4 and y(t) = 1/4 - t/2 which indeed satisties the system of DE's; if not: you'll get \delta{(t)} as well.
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  4. #4
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    TD! - the 2nd equation is x' + y' - 2y = t not x'-y' -2x = t

    Also, how have you gone from having s^-2 to s^2 in your working?
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  5. #5
    TD!
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    About the sē, I just forgot to put the minus up there, the solution I gave holds for 1/sē, not sē.
    In the second equation I did have -2y instead of -2x, but I got the sign of y' wrong, apparently.

    Let me fix my mistakes

    \mathcal{L}\left\{ {x' - y' - 2x} \right\} = sX - x_0  - sY + y_0  - 2X = \left( {s - 2} \right)X - sY + 1

    \mathcal{L}\left\{ {x' + y' - 2y - t} \right\} = sX - x_0  + sY - y_0  - 2Y - s^{ - 2}  = sX + \left( {s - 2} \right)Y - s^2  - 3

    This gives the system:

    <br />
\left\{ \begin{array}{l}<br />
 \left( {s - 2} \right)X - sY = -1 \\ <br />
 sX + \left( {s - 2} \right)Y = s^2  + 3 \\ <br />
 \end{array} \right.<br />

    Unfortunately, the solution is a bit more complicated for this one:

    <br />
X = \frac{{2s^2  + 2s + 1}}{{2s\left( {s^2  - 2s + 2} \right)}}<br />
\; \wedge \;<br />
Y = \frac{{4s^3  - 6s^2  + s - 2}}{{2s^2 \left( {s^2  - 2s + 2} \right)}}<br />

    I assume you know how to do a partial fraction decomposition, this gives:

    <br />
\begin{array}{*{20}l}<br />
   X & { = \frac{{3s}}{{4\left( {s^2  - 2s + 2} \right)}} + \frac{3}{{2\left( {s^2  - 2s + 2} \right)}} + \frac{1}{{4s}}}  \\ \\<br />
   {} & { = \frac{3}{4}\frac{{s - 1}}{{\left( {s - 1} \right)^2  + 1}} + \frac{9}{4}\frac{1}{{\left( {s - 1} \right)^2  + 1}} + \frac{1}{{4s}}} \end{array}<br />

    <br />
\begin{array}{*{20}l}<br />
   Y & { = \frac{{9s}}{{4\left( {s^2  - 2s + 2} \right)}} - \frac{3}{{\left( {s^2  - 2s + 2} \right)}} - \frac{1}{{2s^2 }} - \frac{1}{{4s}}}  \\ \\<br />
   {} & { = \frac{9}{4}\frac{{s - 1}}{{\left( {s - 1} \right)^2  + 1}} - \frac{3}{4}\frac{1}{{\left( {s - 1} \right)^2  + 1}} - \frac{1}{{2s^2 }} - \frac{1}{{4s}}}  <br />
\end{array}<br />

    In the last line for X and Y, I rewrote so they're exactly in the standard form as you can look them up in a table.
    To get x(t) and y(t) out of X(s) and Y(s), you now do the inverse Laplace transform as follows:

    <br />
\begin{array}{*{20}l}<br />
   x & {= \mathcal{L}^{ - 1} \left\{ X \right\}}  \\ \\<br />
{} & { = \frac{3}{4}\mathcal{L}^{ - 1} \left\{ {\frac{{s - 1}}{{\left( {s - 1} \right)^2  + 1}}} \right\} + \frac{9}{4}\mathcal{L}^{ - 1} \left\{ {\frac{1}{{\left( {s - 1} \right)^2  + 1}}} \right\} + \frac{1}{4}\mathcal{L}^{ - 1} \left\{ {\frac{1}{s}} \right\}}  \\ \\<br />
   {} & { = \frac{3}{4}e^t \cos t + \frac{9}{4}e^t \sin t + \frac{1}{4}}<br />
\end{array}<br />

    <br />
\begin{array}{*{20}l}<br />
  y &  {= \mathcal{L}^{ - 1} \left\{ Y \right\}}  \\ \\<br />
{} & { = \frac{9}{4}\mathcal{L}^{ - 1} \left\{ {\frac{{s - 1}}{{\left( {s - 1} \right)^2  + 1}}} \right\} - \frac{3}{4}\mathcal{L}^{ - 1} \left\{ {\frac{1}{{\left( {s - 1} \right)^2  + 1}}} \right\} - \frac{1}{2}\mathcal{L}^{ - 1} \left\{ {\frac{1}{{s^2 }}} \right\} - \frac{1}{4}\mathcal{L}^{ - 1} \left\{ {\frac{1}{s}} \right\}}  \\ \\<br />
   {} & { = \frac{9}{4}e^t \cos t - \frac{3}{4}e^t \sin t - \frac{t}{2} - \frac{1}{4}} <br />
\end{array}<br />

    I hope I didn't make any mistakes this time, but I checked the solution with the 'classic method' and it should be ok:

    <br />
\left\{ \begin{array}{l}<br />
 x\left( t \right) = \frac{1}{4}\left( {3e^t \cos t + 9e^t \sin t + 1} \right) \\ \\ <br />
 y\left( t \right) = \frac{1}{4}\left( {9e^t \cos t - 3e^t \sin t - 2t - 1} \right)<br />
 \end{array} \right.<br />
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