laplace transform

Question

find L[t^3e^3t]

Answer 1

ℒ{f(t)}(s)=₀∫^∞f(t)e^-stdt; f(t)=t³e^3t.

ℒ{f(t)}(s)=₀∫^∞t³e^3te^-stdt=₀∫^∞t³e^(3-s)tdt.

Integrate by parts:

Let u=t³, du=3t²dt; dv=e^(3-s)tdt, v=(1/(3-s))e^(3-s)t. We'll consider the integration limits later.

F(s)=ℒ{f(t)}(s)=∫udv=∫t³e^(3-s)tdt=(t³/(3-s))e^(3-s)t-(3/(3-s))∫t²e^(3-s)tdt. (This is simply a rearrangement of d(uv)=vdu+udv or uv=∫vdu+∫udv, ∫udv=uv-∫vdu.)

Let J=∫t²e^(3-s)tdt, u=t², du=2tdt; v=(1/(3-s))e^(3-s)t as before.

J=t²e^(3-s)t/(3-s)-(2/(3-s))∫te^(3-s)tdt.

Let K=∫te^(3-s)tdt, u=t, du=dt; J=t²e^(3-s)t/(3-s)-2K/(3-s).

K=te^(3-s)t/(3-s)-(1/(3-s))∫e^(3-s)tdt=te^(3-s)t/(3-s)-e^(3-s)t/(3-s)².

Assume 3-s<0, i.e., s>3. All terms containing e^(3-s)t are zero when t→∞ and are 1 when t=0, which zeroises (cancels) all t terms.

Now apply the integration limits (indicated by |₀^∞):

K|₀^∞=1/(3-s)²; J|₀^∞=2K|₀^∞/(3-s)=2/(3-s)³;

F(s)=[t³e^(3-s)t/(3-s)-3J|₀^∞/(3-s)]_t=0^t=∞=6/(3-s)⁴, which can also be written 3!/(s-3)⁴.

So ℒ{t³e^3t}=3!/(s-3)⁴.

ℒ{tⁿe^at}=n!/(s-a)ⁿ⁺¹ in general.