1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving Integral Equation with Laplace Transform

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]y(x) = e^{3x} + \int_0^{x}\hspace{1mm}y(t)\hspace{1mm}e^{x-t}\hspace{1mm}dt[/tex]

    2. Relevant equations

    [tex](f*g)(x) = \int_0^{x}\hspace{1mm}y(t)\hspace{1mm}g(x-t)\hspace{1mm}dt[/tex]

    [tex]L(f*g;s) = L(f;s)L(g;s)[/tex]

    I will use [tex]Y(s)[/tex] to denote [tex]L(y;s)[/tex]

    3. The attempt at a solution

    I tried to solve this like all the other problems I have encountered so far, I took the laplace transform of both sides, giving:

    [tex]Y(s) = s-3 + (s-1)Y(s)[/tex]

    Which gave me:

    [tex]Y(s) = -\frac{s-3}{s-2}[/tex]

    However, this doesn't work out, as far as I'm aware there is no inversion for 1 ( I have only dealt with the standard Laplace transforms and can only invert back to them ).

    The book I have gives:

    [tex]Y(s) = \frac{s-1}{(s-2)(s-3)}[/tex]

    But I'm not sure how this was calculated, I'll appreciate any hints and pointing out where I have gone wrong, thanks.
     
  2. jcsd
  3. Jul 17, 2010 #2

    samalkhaiat

    User Avatar
    Science Advisor


    [tex]Y(s) = \frac{1}{s-3} + \frac{Y(s)}{s-1}[/tex]


    You can also differentiate with respect to x to get

    [tex]y^{'}(x) = 2 e^{3x} + 2y(x)[/tex]

    now take your transform and use

    [tex]L(y^{'}) = sY(s) - y(0)[/tex]

    and (from the integral equation): [itex]y(o) = 1[/itex]

    sam
     
  4. Jul 17, 2010 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You took the Laplace transform of the exponentials incorrectly. You need the reciprocal of what you have. It should be, for example,

    [tex]L[e^{3x}] = \frac{1}{s-3}[/tex]
     
  5. Jul 18, 2010 #4
    Thanks sam and vela, completely missed that mistake.

    Sam, I'm not sure how to use the differentiation approach. Is that starting from:

    [tex]y(x) = e^{3x} + \int_0^{x}\hspace{1mm}y(t)\hspace{1mm}e^{x-t}\hspace{1mm}dt[/tex]

    And then differentiating both with respect to x? If so, how do I go about differentiating the integral?
     
  6. Jul 18, 2010 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

  7. Jul 18, 2010 #6

    Mute

    User Avatar
    Homework Helper

    A note on your notation: [itex](f\ast g)(x)[/itex] is typically used to denote a convolution, which is not what you have in the first line of what I've quoted. The integral limits on a convolution run over the entire domain. You then have the property that the fourier transform of a convolution is the product of the fourier transforms of the convoluted functions.

    Your second line quoted, nevertheless, is correct. I am merely pointing out that most people would assume the first line refers to a convolution, which it really isn't, so be aware of this when you use the (f*g) notation.
     
  8. Jul 18, 2010 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Laplace transforms, unlike Fourier transforms, are traditionally one-sided, so you can take both y(t) and g(t) to vanish for t<0. With that assumption, the convolution integral reduces to the expression MisterMan wrote.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Solving Integral Equation with Laplace Transform
Loading...