# Diff. Eq. La place transform

1. Apr 29, 2014

### Jtechguy21

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 = $\frac{s^2-b^2}{(s^2+b^2)^2}$

long way transform of a derivative
(-1)^n $\frac{d^n}{ds^n}$ 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}= $\frac{s}{s^2+16}$

-$\frac{d}{ds}$ $\frac{s}{s^2+16}$

Quotient rule

-$\frac{s^2+16-2s^2}{(s^2+16)^2}$

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:
-$\frac{-s^2+16}{(s^2+16)^2}$

now we distribute the negative

and get
$\frac{s^2-16}{(s^2+16)^2}$

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

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

2. Apr 29, 2014

### Jtechguy21

wow im so stupid. i missed such an easy thing. combing the like terms s^2 and -2s^2. i get it now.