Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Well, can you actually solve the integral and come up with the inverse Laplace transform? –  Amzoti Dec 11 '12 at 6:56
    
Wolfram tells me that the integral does not converge. –  John Stewley Dec 11 '12 at 6:58
    
How about working it from another angle first. What is the inverse Laplace transform of F(s)? (Do it the way you know how.) Can you now figure out how to use that to move forward? –  Amzoti Dec 11 '12 at 7:00
    
Should I use a table to look this up? –  John Stewley Dec 11 '12 at 7:03
    
Do you know the method of partial fractions? –  kigen Dec 11 '12 at 7:09

3 Answers 3

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

share|improve this answer

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.

share|improve this answer

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, $$\mathscr{L}^{-1}\left\{{{6\over (s+3)^2}\over s}\right\}=\int_0^t f(\tau)\,d\tau=\int_0^t \tau e^{-3\tau}\,d\tau=-2 e^{-3 t} t-\frac{2 e^{-3 t}}{3}+\frac{2}{3}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.