Need help compute the laplace transform

[slug]

need help
compute the laplace transform of
h(t)= -5t^2
I took the integral((-5t^2)e^-(st)) i forgot integration by parts so i did it the hard way in my head and finally got
lim t -> oo 5[(-1/s)(t^2)(e^-st) - (2te^-st) - (2se^-st)] 0--oo
i got 10s but the book says -10/s^3
did i get the inetgration wrong or is it the second part that's wrong

i also have to verify that L[t^n]= n!/s^(n+1) basicaly the same problem but general
muchos gracias and stuff

 

[saltydog]

$$\mathcal{L}\{-5t^2\}=-5\int_0^\infty e^{-st}t^2 dt$$

Let:

$$u=t^2$$

$$dv=e^{-st}dt$$

Really need to use parts for this. Also, this is a standard one that most ODE books will illustrate.

 

[M98Ranger]

It seems like you may have made a mistake in your integration. The correct way to compute the Laplace transform of h(t) = -5t^2 would be to use the definition of the Laplace transform, which is the integral of e^(-st) multiplied by the function being transformed. In this case, it would be:

L[h(t)] = ∫e^(-st)(-5t^2)dt from 0 to infinity

To solve this integral, you can use integration by parts. Let u = -5t^2 and dv = e^(-st)dt. Then, du = -10t dt and v = (-1/s)e^(-st). Plugging this into the integration by parts formula, we get:

∫e^(-st)(-5t^2)dt = (-1/s)e^(-st)(-5t^2) from 0 to infinity - ∫(-1/s)e^(-st)(-10t)dt from 0 to infinity

= (-1/s)e^(-st)(-5t^2) + (10/s^2)e^(-st) from 0 to infinity

= (-5/s)e^(-st)t^2 + (10/s^2)e^(-st) from 0 to infinity

Now, taking the limit as t approaches infinity, we get:

lim t -> infinity (-5/s)e^(-st)t^2 + (10/s^2)e^(-st) = (10/s^2) - (5/s)(0) = 10/s^2

So the correct Laplace transform of h(t) = -5t^2 is 10/s^2, not 10s as you had calculated.

For the general case of L[t^n], we can use induction to prove that L[t^n] = n!/s^(n+1). First, we can easily verify that this formula holds for n = 0. L[t^0] = L[1] = 1/s = 0!/s^(0+1).

Now, let's assume that the formula holds for some arbitrary value of n. That is, L[t^n] = n!/s^(n+1). We want to show that this formula also holds for n+1. Using the definition of the Laplace transform, we have:

L[t^(n+1)] =