Re: Laplace transformation

Quote:

Originally Posted by

**hampos** Hi,

I'm trying to Laplace transform this function:

$\displaystyle f(t)=2\int_{0}^{t}f'(u)sin8(t-u)du\,+\,9cos(8t)$

But I don't manage to get it right, the f'(u) within the integral makes me confused. Would really appreciate your help![FONT='Lucida Grande', 'Trebuchet MS', Verdana, Helvetica, Arial, sans-serif]

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Split it into two laplace transforms (one for the integral, one for the cosine function). The first is a convolution, and you make use of the fact that $\displaystyle \displaystyle \begin{align*} \mathcal{L}\{f'(t)\} = s\,F(s) - f(0) \end{align*} $.