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Math Help - laplace transforms to solve IVPs

  1. #1
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    laplace transforms to solve IVPs

    Use Laplace Transforms to solve the following Initial Value Problems:

    (i)

    \dot{y_1} = y_{2}
    \dot{y_2} = -y_{1} + e^t

    with y_{1}(0) = 0, y_{2}(0) = 0

    (ii)

    \dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases} with y(0) = 0, \dot{y}(0) = 0 and \ddot{y}(0) = 0
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by wik_chick88 View Post
    Use Laplace Transforms to solve the following Initial Value Problems:

    (i)

    \dot{y_1} = y_{2}
    \dot{y_2} = -y_{1} + e^t

    with y_{1}(0) = 0, y_{2}(0) = 0
    There are two ways (at least) of doing this the first involves differentiating one of the equation and then substituting the other into the new equation:

    \ddot{y}_1=\dot{y}_2=-y_{1} + e^t

    Then solving the new equation for \dot y:

    \ddot{y}_1=-y_{1} + e^t

    Laplace transforming:

    s^2Y_1(s)-s y_1(0)-\dot{y}_1(0)=Y_1(s)-\frac{1}{s-1}, \ \ \ \ s>1

    Applying the initial conditions y_1(0)=0, \dot{y}_1=0 this becomes:

    Y_1(s)=-\frac{1}{(s-1)^2(s+1)}

    which you invert by the use of partial fractions and a table of inverse Laplace transforms.

    The second method involves writing this as a vector ODE:

    \dot{ {\bold{y}} }={\bold{A}}{\bold{y}} + (0,e^t)^T

    where:

    {\bold{A}}=\left[\begin{array}{rr}0&1\\-1&0\end{array}\right]

    and: {\bold{y}}=(y_1,y_2)^T

    Now take the Laplace transform:

    s{ {\bold{Y}} }(s)-{\bold{y}}(0)={\bold{A}}{\bold{Y}}(s) + (0,-1/(s-1))^T

    Then (as : {\bold{y}}(0)=(0,0)^T)

    {\bold{Y}}(s)=({\bold{A}}-s{\bold{I}})^{-1}(0,1/(s-1))^T

    Now with some matrix algebra you can find Y_1(s) and Y_2(s) and proceed as before.

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by wik_chick88 View Post

    (ii)

    \dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases} with y(0) = 0, \dot{y}(0) = 0 and \ddot{y}(0) = 0
    Here you have:

    \dddot{y} =u(t-3) t

    where u(.) is the Heavyside unit step function, and:

    \mathcal{L}\dddot{y}=s^3Y(s)-s^2 \ddot{y}(0)-s\dot{y}(0)-y(0).

    The rest is routine.

    CB
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  4. #4
    MHF Contributor chisigma's Avatar
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    ... use Laplace Transforms to solve the following Initial Value Problems:


    \dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases} with y(0) = 0, \dot{y}(0) = 0 and \ddot{y}(0) = 0

    If we intend to apply the LT for resolving a DE, then we have to impose that t \ge 0. So the proposed third order differrential equation must be written as…

    \dddot{y} = \begin{cases} t & \ 0 \le t < 3 \\ 0 & \ t \geq 3 \end{cases} = t\cdot u(t) - (3+t) \cdot u(t-3)


    ... where u(*) is the Heavyside unit step function...

    Kind regards

    \chi \sigma
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by chisigma View Post
    If we intend to apply the LT for resolving a DE, then we have to impose that t \ge 0. So the proposed third order differrential equation must be written as…
    No we don't, the default LT in use is the one sided LT and so what the function does at negative arguments is irrelevant to the forward transform.

    But the inverse transform is only valid for t\ge 0, since any function that agree on the non-negative half real axis have the same (one-sided) LT.

    So the solution obtained using the LT to a IVP is only valid for non-negative times.

    CB
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    There are two ways (at least) of doing this the first involves differentiating one of the equation and then substituting the other into the new equation:

    \ddot{y}_1=\dot{y}_2=-y_{1} + e^t

    Then solving the new equation for \dot y:

    \ddot{y}_1=-y_{1} + e^t

    Laplace transforming:

    s^2Y_1(s)-s y_1(0)-\dot{y}_1(0)=Y_1(s)-\frac{1}{s-1}, \ \ \ \ s>1

    Applying the initial conditions y_1(0)=0, \dot{y}_1=0 this becomes:

    Y_1(s)=-\frac{1}{(s-1)^2(s+1)}

    which you invert by the use of partial fractions and a table of inverse Laplace transforms.
    hi thanks so much! i do have a question though...
    when you laplace transformed it, why did the -y_{1} + e^t get laplace transformed to Y_{1}(s) - \frac{1}{s-1} wouldn't it be -Y_{1}(s) + \frac{1}{s-1}??
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