Question

How do you find the Laplace transform of
`9t e^(-t) sin(3t)`?

Answer 1

If
`F(s)` is the Laplace transform of
`f(t)`, then first recall the phase shift property of the transform:


`\mathcal L_s\{e^(ct)f(t)\}=F(s-c)`

In this case,
`c=-1`, and
`F(s)` is the transform of
`f(t)=9t\sin3t`.

We can easily (if tediously) derive the following result:


`\mathcal L_s\{t\sin(at)\}=(2as)/((s^2+a^2)^2)`

so that


`F(s)=(9(6s))/((s^2+9)^2)=(54s)/((s^2+9)^2)`

and so


`F(s+1)=\mathcal L_s\left\{9te^(-t)\sin3t\right\}=(54(s+1))/(((s+1)^2+9)^2)`

`F(s+1)=(54s+54)/(s^4+4s^3+24s^2+40s+100)`