Diff. Eq. La place transform

1. The problem statement, all variables and given/known data



Evaluate the Laplace of {tcos4t} using the derivative of a transform



Ofcourse i know the shortcut way of doing this, but I need to do it the long way.



2. Relevant equations



shortcut way

t cos bt = [itex]\frac{s^2-b^2}{(s^2+b^2)^2}[/itex]



long way transform of a derivative

(-1)^n [itex]\frac{d^n}{ds^n}[/itex] F(s)



F(s)=L{f(t)}



3. The attempt at a solution



n=1 f(t)=cos4t

f(s)= Laplace of cos4t



L{cos4t}= [itex]\frac{s}{s^2+16}[/itex]



-[itex]\frac{d}{ds}[/itex] [itex]\frac{s}{s^2+16}[/itex]



Quotient rule



-[itex]\frac{s^2+16-2s^2}{(s^2+16)^2}[/itex]



This is as far as I get on my own.



In my notes for class shes goes a step further and I'm not quite sure what happens to the

-2s^2 in the next step



my notes continue as follows:

-[itex]\frac{-s^2+16}{(s^2+16)^2}[/itex]



now we distribute the negative



and get

[itex]\frac{s^2-16}{(s^2+16)^2}[/itex]



I checked the answer using a Laplace transform table(easy way) and did receive [itex]\frac{s^2-16}{(s^2+16)^2}[/itex]



but when i do it the long way i dont know what happens to -2s^2

1. The problem statement, all variables and given/known data







2. Relevant equations







3. The attempt at a solution