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Laplace Transform help please

  1. Dec 15, 2011 #1
    So the question is: A curve rise from the origin of the xy plane into the 1st quadrant. The area under the curve from (0,0) to (x,y) is 1/5 the area of the rectangle with these points as opposite vertices.

    So i'm solving for f(x):
    So far what i have is:

    Area(D)=1/5 xy=integral 0 to x y(t)dt
    and then

    rewrite as: 1/5xy = integral 0 to x g(x-t)y(t) dt, where g(t)=1

    then the next step i get stuck, because when i take the Laplace of both sides i get: 1/5L[xy]=L[y]

    Thanks in advance for any help.
     
  2. jcsd
  3. Dec 15, 2011 #2

    hunt_mat

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    So in LaTeX you have:
    [tex]
    \int_{0}^{x}f(u)du=\frac{xf(x)}{5}
    [/tex]
    You know that you can differentiate to get and ODE don't you? Or do you have to use Laplace transforms?
     
  4. Dec 15, 2011 #3
    unfortunately i must use Laplace transforms to solve. :(
     
  5. Dec 15, 2011 #4

    hunt_mat

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    According to tables:
    [tex]
    \mathscr{L}(xf(x))=-\mathscr{L}(f)'(s)
    [/tex] and
    [tex]
    \mathscr{L}\left(\int_{0}^{x}f(t)dt\right) =\frac{\mathscr{L}(f)}{s}
    [/tex]
     
  6. Dec 15, 2011 #5
    thank you, but i still have no idea how to proceed from here.?
     
  7. Dec 15, 2011 #6

    hunt_mat

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    You obtain a differential equation for the Laplace transform, solve this ODE and then invert the answer.
     
  8. Dec 15, 2011 #7
    Can you please check if this is correct:

    after applying the process of Laplace i get this, -1/5 y'-y=0. and now i just solve this ODE?
     
  9. Dec 16, 2011 #8

    hunt_mat

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    No, there is an s floating around (s being the laplace transform variable) which comes from taking the laplace transform of the integral.
     
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