laplace

Discussion in 'Math' started by ezekiel, Jun 10, 2007.

  1. ezekiel

    Thread Starter New Member

    Jun 9, 2007
    1
    0
    hi does anyone know how to solve y' + y = e [-t] using laplace?? thanks!
     
  2. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Yes. y' = sY(s), y = Y(s) and e[-t] = 1/(s-a)

    Looking at your basic Laplacian theory should give you the derivations.

    Dave
     
  3. kautilya

    Active Member

    Apr 26, 2007
    39
    0
    y' = sY(s)
    y = Y(s)
    e[-t]

    y' + y = e[-t]

    Y(s){s + 1} = 1/(s+a)

    Y(s) = 1/(s+a)^2
    = te[-t]
     
  4. BlackBox

    Member

    Apr 22, 2007
    20
    0
    Sorry to contraddict, but the Laplace transform of a derivative is not that simple... remember that we are transforming between time and complex frequency, so the "memory" included in the derivative is to be found explicitly in the Laplace domain. Otherwise you are not gonna be able to solve the Cauchy problem deriving form your differential equation:

    = s F(s) + f(t)|_{s=0^+}" alt="L(f'(t)) = s F(s) + f(t)|_{s=0^+}" />

    so the correct solution of the problem (given you don't know any initial states) is:

    sY(s) \ + \ y(0^+) \ + \ Y(s) = \frac{1}{s+1}\\<br />
 Y(s) (s +1) \ = \ \frac{1}{s+1} \ - \ y(0^+) \\<br />
 Y(s) \ = \ \frac{1}{(s+1)^2} \ -\  \frac{y(0^+)}{s+1}

    if the y(t) function is continuous in the origin you could put y(0) instead of y(0^+)
     
Loading...