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Laplace initial value problem HELP! PLEASE!

  1. Dec 28, 2007 #1
    Laplace initial value problem.... HELP! PLEASE!

    Hello all!
    I'm stuck on this question:

    y' + y = t sin t

    y(0) = 0

    solve it using laplace transform,... my final is tomorrow, and its 2 am, i would appreciate a quick respone
    thanks in advance!
    Last edited: Dec 28, 2007
  2. jcsd
  3. Dec 28, 2007 #2

    [tex]\mathcal{L}\{y(t)\}=F(s)=\int_0^{+\infty} e^{-s\,t}\,y(t)\,d\,t[/tex]

    be the Laplace transform of [tex] y(t)[/tex], then the Laplace transform for [tex] \mathcal{L}\{y'(t)\} [/tex] would be

    [tex]\mathcal{L}\{y'(t)\}=\int_0^{+\infty} e^{-s\,t}\,y'(t)\,d\,t\Rightarrow \mathcal{L}\{y'(t)\}=s\,F(s)-y(0) \Rightarrow \mathcal{L}\{y'(t)\}=s\,F(s)[/tex]

    The Laplace transform of the right hand side of your eq is

    [tex] \mathcal{L}\{t\,\sin t\}=-\frac{2\,s}{(1+s^2)^2} [/tex]

    Plugging the above values into your equation, you can evaluate [tex]F(s)[/tex] and applying the inverse Laplace transformation


    you will arrive at [tex] y(t)[/tex]
  4. Dec 28, 2007 #3
    I'm arriving at these results:

    2s / {(s^2+1)^2(2+1)}, but i cant continue from there.... I have an appendix of Laplace transforms in my text but none of them seem to fit this one...
  5. Dec 28, 2007 #4
    Split the fraction into simpler ones

  6. Dec 28, 2007 #5
    i tried partial fractions, but how do you assign the contants in the numerator fro such a complex polynomial in the denominator...?
  7. Dec 28, 2007 #6
    I don't understand the question (my English are pretty poor!)

    " ...complex polynomial in the denominator..."

    If you mean the term


    it's inverse Laplace transform can be evaluated by

    [tex]\mathcal{L}^{-1}\{\frac{s+\gamma}{(s+\alpha)^2+\beta^2}\}=e^{-\alpha\,t}\left(\cos(\beta t)+\frac{\gamma-\alpha}{\beta}\,\sin(\beta\,t)\right)[/tex]
  8. Dec 28, 2007 #7


    A good place for a table of these transormations could be

  9. Dec 29, 2007 #8


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    Science Advisor

    The standard technique for finding inverse Laplace transformations of complicated fractions (I wouldn't say "complex"; in mathematics that is too closely connected with complex numbers) is to use partial fractions. Surely you've seen that before?
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