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A problem about laplace

  1. Jul 28, 2010 #1
    hi guys

    i cannot solve this.

    say that y'' + y = g(t) derive the formula y(t) = c1*cost + c2*sint + convolution of sint and g(t)
     
  2. jcsd
  3. Jul 28, 2010 #2
    Either solve it via variation of parameters or since you mentioned laplace, take the laplace transform of both sides:

    [tex]\mathcal{L}\left\{y''+y=f(x)\right\}[/tex]

    and letting:

    [tex]\mathcal{L}\left\{y\right\}=\widetilde{y}[/tex]

    solve for [itex]\widetilde{y}[/itex], invert, and use the convolution theorem to express the solution in terms of a convolution.

    Look in any DE text book and this problem will be solved both ways.
     
    Last edited: Jul 28, 2010
  4. Jul 28, 2010 #3
    Use the method of variation of arbitrary constants. The solution to the homogeneous equation is:

    [tex]
    y_{0}(t) = C_{1} \, \cos{t} + C_{2} \, \sin{t}
    [/tex]

    Then, assume [itex]C_{i} \rightarrow C_{i}(t), i = 1, 2[/itex] where these functions satisfy the following conditions:

    [tex]
    \left[\begin{array}{cc}
    \cos{t} & \sin{t} \\

    -\sin{t} & \cos{t}
    \end{array}\right] \cdot \left[\begin{array}{c}
    C'_{1}(t) \\

    C'_{2}(t)
    \end{array}\right] = \left[\begin{array}{c}
    0 \\
    g(t)
    \end{array}\right]
    [/tex]

    The solution for [itex]C'_{i}(t)[/itex] is:

    [tex]
    \left[\begin{array}{c}
    C'_{1}(t) \\

    C'_{2}(t)
    \end{array}\right] = \left[\begin{array}{cc}
    \cos{t} & -\sin{t} \\

    \sin{t} & \cos{t}
    \end{array}\right] \cdot \left[\begin{array}{c}
    0 \\
    g(t)
    \end{array}\right] = \left[\begin{array}{c}
    -\sin{t} \, g(t) \\

    \cos{t} \, g(t)
    \end{array}\right]
    [/tex]

    One integration gives the following:

    [tex]
    C_{1}(t) = C_{1} - \int_{t_{0}}^{t}{g(t') \, \sin{t'} \, dt'}
    [/tex]

    [tex]
    C_{2}(t) = C_{2} + \int_{t_{0}}^{t}{g(t') \, \cos{t'} \, dt'}
    [/tex]

    Substituting this into the expression for the general solution, one gets:

    [tex]
    y(t) = C_{1} \, \cos{t} + C_{2} \, \sin{t} + \int_{t_{0}}^{t}{g(t') \, \left(\sin{t} \, \cos{t'} - \cos{t} \, \sin{t'}\right) \, dt' \right)}
    [/tex]

    where, the integrating constants [itex]C_{1/2}[/itex], are determined from the initial conditions:

    [tex]
    y(t_{0}) = C_{1}
    [/tex]

    [tex]
    y'(t_{0}) = C_{2}
    [/tex]

    Using the addition theorem for the sine function:

    [tex]
    \sin{(t - t')} = \sin{t} \, \cos{t'} - \cos{t} \, \sin{t'}
    [/tex]

    we see that the above integral can be written as:

    [tex]
    \int_{t_{0}}^{t}{g(t') \, \sin{(t - t')} \, dt'}
    [/tex]

    Take [itex]t_{0} = 0[/itex] and compare with the definition for convolution, you will get your desired result.
     
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