# Thread: laplace transforms

1. ## laplace transforms

I need to find the laplace transforms of these functions:
a) (t-1)^4 <---dont know which property to use
b) t*exp(-t)*sin(2t)
c) cos(kt)^2 k = constant

Any help would be great.
My attempts are:
b) 1/s^2 * F(s+1)*4/(s^2+4)
Answer should be: 4(s+1)/[(s+1)^2+4]^2

c) Answer should be: 4(3s^2-4)/(s^2+4)^3

2. Originally Posted by ben.mahoney@tesco.net
I need to find the laplace transforms of these functions:
a) (t-1)^4 <---dont know which property to use
b) t*exp(-t)*sin(2t)
c) cos(kt)^2 k = constant

Any help would be great.
My attempts are:
b) 1/s^2 * F(s+1)*4/(s^2+4)
Answer should be: 4(s+1)/[(s+1)^2+4]^2

c) Answer should be: 4(3s^2-4)/(s^2+4)^3
a) Expand $\displaystyle (t - 1)^4$ and take the Laplace transform of each term.

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b) Here is the chain of logic you need to use:

1. $\displaystyle LT \left[e^{at} F(t)\right] = f(s - a)$ where $\displaystyle f(s) = LT[F(t)]$.

So $\displaystyle LT \left[e^{-t} t \sin (2t) \right] = f(s + 1)$ where $\displaystyle f(s) = LT[t \sin (2t)]$.

2. $\displaystyle LT[t G(t)] = - g'(s)$ where $\displaystyle g(s) = LT[G(t)]$.

So $\displaystyle LT[t \sin (2t)] = - g'(s)$ where $\displaystyle g(s) = LT[\sin (2t)]$.

And I assume you can look up the relevant tables and get $\displaystyle LT[\sin (2t)]$.

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c) Note that $\displaystyle \cos^2 (kt) = \frac{\cos (2kt) + 1}{2}$ using the usual double angle formula.

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# what is the laplace transform of sin^2kt

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