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transform of elementary functions

Posted November 24th, 2010 by sya
in

evaluate :

$$ L( t^3-t^2+4t)=\frac{6}{s^4}-\frac{2}{s^3}+\frac{4}{s^2}$$
$$ L ( 3e^ {4t} - e^{-2t} )=\frac{3}{s-4}-\frac{1}{s+2}$$
$$ L ( \sinh kt )=\frac{k}{s^2-k^2}$$
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can you make a proof

On April 23rd, 2011 kolokoyx says:

can you make a proof solution that will make the Laplace transform of $ e^{at} t^n h(t) $ to be $ \frac{n!}{(s-a)^{n+1}} $

$$\mathcal{L}(g(t))(s)=\mathcal{L}(e^{\lambda t}f(t))(s)=\int_{0}^{+\infty}e^{\lambda t}f(t)e^{-st}dt=\int_{0}^{+\infty}f(t)e^{-(s-\lambda )t}dt=F(s-\lambda)$$
$$\mathcal{L}(h(t))(s)=\frac{1}{s}\:\: \; \mathcal{L}(th(t))(s)=\frac{1!}{s^2}\:\:\ ;<br /> \mathcal{L}(t^2h(t))(s)=\frac{2!}{s^3}\:\: \; \mathcal{L}(t^3h(t))(s)=\frac{3!}{s^4}\:\:\ ;\mathcal{L}(t^nh(t))(s)=\frac{n!}{s^{n+1}}$$
$$\mathcal{L}(e^{at}t^nh(t))(s)=\frac{n!}{(s-a)^{n+1}}$$

laplace transform

On November 25th, 2010 sya says:

evaluate ;

L{ phi(t) } , where phi (t) = 1, 0< t< 2
t , t >2

answer

On November 24th, 2010 Structure says:
$$ L( t^3-t^2+4t)=\frac{6}{s^4}-\frac{2}{s^3}+\frac{4}{s^2}$$
$$ L ( 3e^ {4t} - e^{-2t} )=\frac{3}{s-4}-\frac{1}{s+2}$$
$$ L ( \sinh kt )=\frac{k}{s^2-k^2}$$

see laplace-transform-elementary-functions

laplace transform

On November 25th, 2010 sya says:

thanks a lot ... but what is the different between

L { sin kt } and L {| sin kt | } ???

answer

On November 25th, 2010 Structure says:

there is big difference between the two functions.
See Laplace transform of periodic functions

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