Use the transform rule $\mathcal{L}^{-1}\{\frac{F(s)}{s}\} = \int_0^t{f(\tau)d\tau}$ to find the inverse transform of $F(s) = \frac{6}{s(s+3)(s+3)}$

Question

How does the transform rule help us solve this problem? Does this just mean I can rewrite the problem as:

$$\mathcal{L}^{-1}\left\{\frac{6}{(s+3)(s+3)}\right\} = \int_0^t \frac{6}{\tau(\tau+3)(\tau+3)}d\tau$$

Which I can then do what with?

Answer 1

Hint:

If $A,B,C$ are constants, then given the rational function

$\displaystyle f(x) = \frac{A}{(x-B)(x-C)}$

we can write

$\displaystyle f(x)=\frac{\alpha}{x-B}+\frac{\beta}{x-C}$ for some constants $\alpha, \beta$. This in fact generalizes to when

$\displaystyle f(x) = \frac{A}{(x-B)(x-C)(x-D)}$, and when

$\displaystyle f(x) = \frac{A}{(x-B)(x-C)(x-D)(x-E)}$, and so on, and when the terms in the denominator repeat.

If you know how to get $\alpha$ and $\beta$, then the rest of the problem will be easily solved.

Answer 2

$$ F(s) = \frac{1}{s} \frac{6}{(s+3)^2} $$ $$ \frac{1}{s} \triangleq \int_0^t $$ $$ f(t)=\int_0^t \mathcal{L^{-1}}[\frac{6}{(s+3)^2}]d\tau $$ We know : $$ \mathcal{L}[t^n e^{-at}]=\frac{n!}{(s+a)^{n+1}} $$ $$ \mathcal{L^{-1}}[\frac{6}{(s+3)^2}]=6te^{-3t} $$ So : $$ f(t)= \int_0^t 6\tau e^{-3 \tau } d \tau =6[\frac{-t}{3}e^{-3t}-\frac{1}{9}e^{-3t}]|_0^t $$ $$ f(t)=-2te^{-3t}-\frac{2}{3}e^{-3t}+\frac{2}{3} $$

Answer 3

Amir Alizadeh has all the pieces, I'm just tying it together/explaining it slightly differently.

You want $\displaystyle{\mathscr{L}^{-1}\left\{{{6\over (s+3)^2}\over s}\right\}}$ and this matches the form of your hint with $F(s)={6\over (s+3)^2}$ which implies $f(t)=6te^{-3t}$. Thus, $$\mathscr{L}^{-1}\left\{{{6\over (s+3)^2}\over s}\right\}=\int_0^t f(\tau)\,d\tau=\int_0^t \tau e^{-3\tau}\,d\tau=-2 e^{-3 t} t-\frac{2 e^{-3 t}}{3}+\frac{2}{3}.$$