Results 1 to 5 of 5

Math Help - Inverse Laplace problems....

  1. #1
    Senior Member
    Joined
    Aug 2009
    Posts
    349

    Inverse Laplace problems....

    ok can someone tell me why my work is wrong, it seems the book has what i have up to this point, except the have negative signs where i have positive ones...



    thanks in advance...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,832
    Thanks
    1602
    Why don't you just use Partial Fractions?

    .

    A + D = 0, B - A = 0, C - B = 0, -C = 1.

    Therefore C = -1, B = -1, A = -1, D = 1.

    So .


    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Aug 2009
    Posts
    349
    i didnt even think to use partial fractions....thx!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by slapmaxwell1 View Post
    ok can someone tell me why my work is wrong, it seems the book has what i have up to this point, except the have negative signs where i have positive ones...



    thanks in advance...
    Since it seems the book wanted you to use the convolution theorem (A very powerful tool) lets give it a go. This can be inverted by taking three easy integrals.

    \mathcal{L}^{-1}\left\{ F(s)G(s)\right\}=\int_{0}^{t}f(t-\tau)g(\tau)d\tau

    So start with

    \mathcal{L}^{-1}\left\{\frac{1}{s}\cdot \frac{1}{s-1} \right\}=\int_{0}^{t}1\cdot e^{\tau}d\tau=e^{t}-1

    Now use the convolution theorem again to get

    \mathcal{L}^{-1}\left\{\frac{1}{s}\left(\frac{1}{s}\cdot \frac{1}{s-1}\right) \right\}=\int_{0}^{t}1\cdot (e^{\tau}-1)d\tau=e^{t}-t-1

    And one more time to get

    \mathcal{L}^{-1}\left\{\frac{1}{s}\left(\frac{1}{s^2}\cdot \frac{1}{s-1}\right) \right\}=\int_{0}^{t}1\cdot (e^{\tau}-t-1)d\tau=e^{t}-\frac{t}{2}-t-1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Aug 2009
    Posts
    349
    Quote Originally Posted by TheEmptySet View Post
    Since it seems the book wanted you to use the convolution theorem (A very powerful tool) lets give it a go. This can be inverted by taking three easy integrals.

    \mathcal{L}^{-1}\left\{ F(s)G(s)\right\}=\int_{0}^{t}f(t-\tau)g(\tau)d\tau

    So start with

    \mathcal{L}^{-1}\left\{\frac{1}{s}\cdot \frac{1}{s-1} \right\}=\int_{0}^{t}1\cdot e^{\tau}d\tau=e^{t}-1

    Now use the convolution theorem again to get

    \mathcal{L}^{-1}\left\{\frac{1}{s}\left(\frac{1}{s}\cdot \frac{1}{s-1}\right) \right\}=\int_{0}^{t}1\cdot (e^{\tau}-1)d\tau=e^{t}-t-1

    And one more time to get

    \mathcal{L}^{-1}\left\{\frac{1}{s}\left(\frac{1}{s^2}\cdot \frac{1}{s-1}\right) \right\}=\int_{0}^{t}1\cdot (e^{\tau}-t-1)d\tau=e^{t}-\frac{t}{2}-t-1
    that was awesome thx! and just in time, i have my exam today!..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laplace Inverse
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 24th 2011, 05:50 PM
  2. laplace inverse
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: October 9th 2010, 10:15 PM
  3. Laplace/Inverse Laplace Questions
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: August 14th 2010, 12:29 PM
  4. Inverse Laplace
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: June 1st 2009, 09:54 PM
  5. Inverse Laplace
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 1st 2008, 04:41 AM

Search Tags


/mathhelpforum @mathhelpforum