Page 2 of 2 FirstFirst 12
Results 16 to 22 of 22

Math Help - Laplace Transform Involving Initial Value Problem

  1. #16
    Member
    Joined
    Nov 2009
    Posts
    92
    So, is the answer i posted correct?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Prove It View Post
    Are Laplace Transforms really necessary here?
    Not to solve the ODE. But to do the question yes, it asks the student to use Laplace transforms.

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Belowzero78 View Post
    So, is the answer i posted correct?
    You can check this yourself. There are two requirements that the solution must full fill, it must satisfy the differential equation tofether with the initial condition and be continuous at the knot (the point where the forcing function changes its' form).

    You solution appears to meet these requirements.

    CB
    Last edited by CaptainBlack; November 27th 2010 at 09:15 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Member
    Joined
    Nov 2009
    Posts
    92
    Quote Originally Posted by CaptainBlack View Post
    You can check this yourself. There are two requirements that the solution must full fill, it must satisfy the differential equation tofether with the initial condition and be continuous at the knot (the point where the forcing function changes its' form).

    You solution appears to meet these requirements except it does not satisfy the initial condition.

    CB
    f(t)=\left\{\begin{array}{cc}3 - 8e^{-t}&\mbox{ for }<br />
0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.

    Why doesn't this satisfy the initial condition? I'll show my work below.

     \displaystyle{Y(s) = \frac{-5}{s+1} -\frac{6e^{-7s}}{s(s+1)} + \frac{3}{s(s+1)}}

     \displaystyle{Y(s) = \frac{-5}{s+1} +(\frac{-6}{s} + \frac{6}{s+1})e^{-7s} + \frac{3}{s} - \frac{3}{s+1}}

     \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t)  = \mathfrak{L^{-1}}\{\frac{-5}{s+1}\}(t)  +\mathfrak{L^{-1}}\{(\frac{-6}{s} + \frac{6}{s+1})e^{-7s}\}(t)  + \mathfrak{L^{-1}}\{\frac{3}{s}\}(t) - \mathfrak{L^{-1}}\{\frac{3}{s+1})\}(t)}

     \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = -5e^{-t} + \mathfrak{L^{-1}}\{\frac{-6}{s} + \frac{6}{s+1}\}(t - 7)\cdot U_7(t) + 3 - 3e^{-t}}

     \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{-t})|_{t = t - 7} \cdot U_7(t)

     \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{7 - t}) \cdot U_7(t)

    So this is what gives me the answer. I still am not sure why it doesn't satisfy the initial conditions given the fact I used it right at the start when  Y(0) = -5.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Belowzero78 View Post
    f(t)=\left\{\begin{array}{cc}3 - 8e^{-t}&\mbox{ for }<br />
0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.

    Why doesn't this satisfy the initial condition? I'll show my work below.

     \displaystyle{Y(s) = \frac{-5}{s+1} -\frac{6e^{-7s}}{s(s+1)} + \frac{3}{s(s+1)}}

     \displaystyle{Y(s) = \frac{-5}{s+1} +(\frac{-6}{s} + \frac{6}{s+1})e^{-7s} + \frac{3}{s} - \frac{3}{s+1}}

     \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t)  = \mathfrak{L^{-1}}\{\frac{-5}{s+1}\}(t)  +\mathfrak{L^{-1}}\{(\frac{-6}{s} + \frac{6}{s+1})e^{-7s}\}(t)  + \mathfrak{L^{-1}}\{\frac{3}{s}\}(t) - \mathfrak{L^{-1}}\{\frac{3}{s+1})\}(t)}

     \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = -5e^{-t} + \mathfrak{L^{-1}}\{\frac{-6}{s} + \frac{6}{s+1}\}(t - 7)\cdot U_7(t) + 3 - 3e^{-t}}

     \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{-t})|_{t = t - 7} \cdot U_7(t)

     \displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{7 - t}) \cdot U_7(t)

    So this is what gives me the answer. I still am not sure why it doesn't satisfy the initial conditions given the fact I used it right at the start when  Y(0) = -5.
    Arithmetic error on my part.

    To check your solution does not require that we look at your work, though it is nice to see it finally.



    CB
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Member
    Joined
    Nov 2009
    Posts
    92
    So, the function f(t) i posted is correct given the intervals of t?
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Belowzero78 View Post
    So, the function f(t) i posted is correct given the intervals of t?
    It has already been confirmed.

    CB
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 15th 2011, 11:28 PM
  2. Laplace transform involving Heaviside functions
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 14th 2011, 07:50 AM
  3. Laplace Transforms to solve Initial-Value problem
    Posted in the Differential Equations Forum
    Replies: 19
    Last Post: December 10th 2010, 02:40 AM
  4. Laplace transform to solve the initial value
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: September 4th 2009, 12:57 AM
  5. Initial Value Problem 3rd order DE - using laplace
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 30th 2009, 04:10 AM

/mathhelpforum @mathhelpforum