Inverse laplace 2/((s-1)^2+1)^2

Question

Find inverse laplace transform of $$\frac{2}{((s-1)^2+1)^2}$$

I tried to decompose the fraction using

$$\because (s-1)^2+1=s^2-2s+2$$

$$\rightarrow \frac{2}{((s-1)^2+1)^2}=\frac{As+B}{s^2-2s+2}+\frac{Cs+D}{(s^2-2s+2)^2}$$

yet I get D=2, which leads me back to the same exact equation, any help?

(I have also tried the $\frac{-d}{ds}(L{\{sint\}})$ propertie, but failed since there's no variable s in the numerator)

Answer 1

While there is a defined inverse Laplace transform, the integral is often difficult. Usually, it is easier to take the transforms we know and noodle with them until we find a fit.

$L\{e^t\cos t\} = \frac {s-1}{(s-1)^2+1}\\ L\{e^t\sin t\} = \frac 1{(s-1)^2+1} = \frac {s^2-2s +2}{((s-1)^2+1)^2}\\ L\{te^t\cos t\} = -\frac {d}{ds} \frac {(s-1)}{(s-1)^2+1} = \frac {s^2 - 2s}{((s-1)^2+1)^2}\\ L\{te^t\sin t\} = -\frac {d}{ds} \frac 1{(s-1)^2+1} = \frac {2s-2}{((s-1)^2+1)^2}$

Some combination of $(s^2 - 2s + 2), (s^2-2s),(2s-2) = 2$

$e^t \sin t - te^t \cos t$

Answer 2

HINT:

Use the following properties

$$\mathscr{L}\{tf(t)\}(s)=-\frac{d}{ds}\mathscr{L}\{f(t)\}(s)$$

$$\mathscr{L}\{e^{s_0t}f(t)\}(s)=\mathscr{L}\{f(t)\}(s-s_0)\,\,\,\,\,\,$$

along with these Laplace Transform pairs

$$\mathscr{L}\{\cos(t)\}(s)=\frac{s}{s^2+1}$$

$$\mathscr{L}\{\sin(t)\}(s)=\frac{1}{s^2+1}$$