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Math Help - [SOLVED] Derivative properites of LaPlace Transforms

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    [SOLVED] Derivative properites of LaPlace Transforms

    Given that g(t) has the Laplace transform of G(s), what is the Laplace transform of H(s):

    h(t)=e^{-t}g'(3t)

    I know that g(t) has to satisfy the Derivatives proptery theorem: f'(t)=s F(s)-f(0), but im not really sure what to do. As always step by step instructions would be amazing
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    Quote Originally Posted by snaes View Post
    Given that g(t) has the Laplace transform of G(s), what is the Laplace transform of H(s):

    h(t)=e^{-t}g'(3t)

    I know that g(t) has to satisfy the Derivatives proptery theorem: f'(t)=s F(s)-f(0), but im not really sure what to do. As always step by step instructions would be amazing
    Note that from the definition it easily follows that LT[f(at)] = \frac{1}{a} F\left(\frac{s}{a}\right) where  F(s) = LT[f(t)].
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    Would these be correct?

    1)
    h(t)=g'(3t)
    H(s)=s*LT[g(3t)]-g(0)
    H(s)=s*\dfrac{1}{3}G(\dfrac{s}{3})-g(0)

    2) Should be the same only only I use e^{at}f(t) = F(s-a)
    h(t)=e^{3t}g'(3t)
    H(s)=(s-3)*\dfrac{1}{3}G(\dfrac{s-3}{3})-g(0)

    3) Now I use t^{n}f(t)=(-1)^{n}\dfrac{d^{n}}{ds^{n}}(F(s))
    h(t)=t^{2}e^{3t}g'(3t)
    H(s)=t^{2}((s-3)*\dfrac{1}{3}G(\dfrac{s-3}{3})-g(0))
    H(s)=(-1)^{2}\dfrac{d^{2}}{ds^{2}}[(s-3)*\dfrac{1}{3}G(\dfrac{s-3}{3})-g(0)]
    H(s)=\dfrac{d^{2}}{ds^{2}}[(s/3-1)*G(\dfrac{s-3}{3})

    I think this is right but im not sure...

    NVM I GOT IT FIGURED OUT
    Last edited by snaes; April 26th 2010 at 12:49 PM.
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