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Math Help - Different Results when Using Laplace or Complamentary Functions/Particular Integrals

  1. #1
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    Post Different Results when Using Laplace or Complamentary Functions/Particular Integrals

    Hi,

    y''(t) + 3y'(t) + y(t) = 2x(t)

    Where x(t) = u(t), and the system is initially at rest. We are interested only when t>=0, so x(t) = 1 always.

    I can solve it by Laplace transforms:

    Original eqn:
    y''(t) + 3y'(t) + y(t) = 2

    Taking Laplace of both sides (and assuming system was at rest):
    Y(s)[s^2 + 3s + 2] = 2/s
    Y(s) = [2/s]/[s^2 + 3s + 2]
    Y(s) = A/s + B/(s + 1) + C/(s + 2)

    Finally: y(t) = 1 - 2exp[-t] + ext[-2t]

    --------

    The problem is when I try solving it by equating coefficients. I get a different answer:

    Original eqn:
    y''(t) + 3y'(t) + y(t) = 2

    To get complementary function, y_comp(t):
    y''(t) + 3y'(t) + y(t) = 0
    Auxiliary eqn: s^2 + 3s + 1 = 0; s = -1, -2
    y_comp(t) = Aexp[-t] + Bexp[-2t]

    To get particular integral, y_pt(t):
    Forcing function: x(t) = 1
    Gives y_pt(t) = 1

    General solution:
    y(t) = y_comp(t) + y_pt(t) = Aexp[-t] + Bexp[-2t] + 1

    Solving for the constants A and B (problem is here):
    y'(t) = -Aexp[-t] -2Bexp[-2t]
    y''(t) = Aexp[-t] + 4Bexp[-2t]
    Given the system was at rest, I get:
    0 = -A - 2B (1)
    0 = A + 4B (2)

    Solving these gives A=B=0. But this means that y(t) = 1, which is different to the Laplace solution of y(t) = 1 - 2exp[-t] + ext[-2t].

    Thanks for any help.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by algorithm View Post
    Hi,

    y''(t) + 3y'(t) + y(t) = 2x(t)

    Where x(t) = u(t), and the system is initially at rest. We are interested only when t>=0, so x(t) = 1 always.

    I can solve it by Laplace transforms:

    Original eqn:
    y''(t) + 3y'(t) + y(t) = 2

    Taking Laplace of both sides (and assuming system was at rest):
    Y(s)[s^2 + 3s + 2] = 2/s
    Y(s) = [2/s]/[s^2 + 3s + 2]
    Y(s) = A/s + B/(s + 1) + C/(s + 2)

    Finally: y(t) = 1 - 2exp[-t] + ext[-2t]

    --------

    The problem is when I try solving it by equating coefficients. I get a different answer:

    Original eqn:
    y''(t) + 3y'(t) + y(t) = 2

    To get complementary function, y_comp(t):
    y''(t) + 3y'(t) + y(t) = 0
    Auxiliary eqn: s^2 + 3s + 1 = 0; s = -1, -2
    y_comp(t) = Aexp[-t] + Bexp[-2t]

    To get particular integral, y_pt(t):
    Forcing function: x(t) = 1
    Gives y_pt(t) = 1

    General solution:
    y(t) = y_comp(t) + y_pt(t) = Aexp[-t] + Bexp[-2t] + 1

    Solving for the constants A and B (problem is here):
    y'(t) = -Aexp[-t] -2Bexp[-2t]
    y''(t) = Aexp[-t] + 4Bexp[-2t]
    Given the system was at rest, I get:
    0 = -A - 2B (1)
    0 = A + 4B (2)

    Solving these gives A=B=0. But this means that y(t) = 1, which is different to the Laplace solution of y(t) = 1 - 2exp[-t] + ext[-2t].

    Thanks for any help.
    When you apply initial conditions, there isn't one for the second derivative of the function!

    Your coefficient equations should be:

    A+B+1=0 (for y(0)=0)

    -A-2B=0 (for y^{\prime}(0)=0)

    Can you take it from here?
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  3. #3
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    Joined
    Nov 2008
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    Hi,

    Thanks for the reply. So when we assume a 2nd order system is initially at rest, the initial conditions are only y(0) = y'(0) = 0?
    Can you please explain why?

    Thanks
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  4. #4
    MHF Contributor chisigma's Avatar
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    near Piacenza (Italy)
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    The second order linear DE is...

    y^{''} (t) + 3\cdot y^{'} (t) + 2\cdot y(t) = 2\cdot \mathcal{U} (t) , y(0)=y^{'}(0)=0 (1)

    ... being \mathcal{U} (*) the 'Haeviside Step Function'. In terms of LT the (1) becomes...

    s^{2}\cdot Y(s) - s\cdot y(0) - y^{'} (0) +3\cdot s\cdot Y(s) -3\cdot y(0) + 2\cdot Y(s) = \frac{2}{s} (2)

    ... and from it...

    Y(s)= \frac{(s-3)\cdot y(0) + y^{'} (0)}{s^{2} + 3s +2} + \frac{2}{s\cdot (s^{2} + 3s + 2)} (3)

    Because is y(0)=y^{'}(0)=0 the only contribution is the 'forced response' ...

    Y(s) = \frac{2}{s\cdot (s^{2} + 3s + 2)} = \frac{1}{s} - \frac{2}{s+1} + \frac{1}{s+2} (4)

    ... so that is...

    y(t)=\mathcal{L}^{-1} \{Y(s)\} = (1-2\cdot e^{-t} + e^{-2t})\cdot \mathcal{U} (t) (5)

    Kind regards

    \chi \sigma
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