Laplace Transform (Linear Algebra)

yesiammanu
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Homework Statement


Evaluate the laplace transform of {t2e7tsinh(3t)}


Homework Equations


Laplace transform of {tnf(t)}=(-1)ndn/ds2 * F(s)


The Attempt at a Solution


I've replaced it with (-1)2d2L{e7tsinh(3t)}

I'm not sure how to proceed, though, as I don't really see how to take the laplace without somehow splitting up these functions. Do I use a unit step function? How would I use it if so? I'm not really sure what to do here

Thanks for any help.
 
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The Laplace transform of a function that has been multiplied by an exponential function can be found using the shift theorem for the s-axis. Do you know this theorem?

BiP
 
Is this the theorem you were referring to?

L{e-asF(s)}=f(t-a)*UnitStep(t-a)
 
yesiammanu said:

Homework Statement


Evaluate the laplace transform of {t2e7tsinh(3t)}


Homework Equations


Laplace transform of {tnf(t)}=(-1)ndn/ds2 * F(s)


The Attempt at a Solution


I've replaced it with (-1)2d2L{e7tsinh(3t)}

I'm not sure how to proceed, though, as I don't really see how to take the laplace without somehow splitting up these functions. Do I use a unit step function? How would I use it if so? I'm not really sure what to do here

Thanks for any help.

You can write ##\sinh(3t) = \frac{1}{2} \left( e^{3t} - e^{-3t}\right),## and proceed accordingly.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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