What is the inverse Laplace of (5-s)/(s^2+6s+9)?

Question

abb0t · Answer

try and fctor your s^2+6s+9

anonymous · Answer

(5-s)/(s+3)^2

zepdrix · Answer

If you split up your fractions, you can do these pieces separately.$$\large \frac{5}{(s+3)^2}-\frac{s}{(s+3)^2}$$

Hmm I can explain the first piece. I'm trying to remember how to do the second one though. heh.

abb0t · Answer

note your denom
$$(\frac{ 1 }{ s+3 })(\frac{ 1 }{ s+3 })$$ which is familiar

abb0t · Answer

But, the bottom is 3, so you should miltiply by a number to get it into the form to get an exponential

abb0t · Answer

Im also trying to remember how to work this.

zepdrix · Answer

$$\huge \mathcal L\left[t\cdot f(t)\right]=(-1)\frac{d}{ds}F(s)$$Does this formula look familiar yury? :o

anonymous · Answer

maybe partial fraction

anonymous · Answer

yes

zepdrix · Answer

We can use that on the first part, but in reverse.

zepdrix · Answer

We're basically integrating the term, and attaching a d/ds to it. :D

zepdrix · Answer

$$\large \frac{5}{(s+3)^2} \qquad \rightarrow \qquad (-1)\frac{d}{ds}\frac{-5}{(s+3)}$$

zepdrix · Answer

Understand what I did there? :o I integrated, but attached a derivative operator to it at the same time, so it hasn't affected the value of the fraction.

anonymous · Answer

the question starts with
 y"+6y'+9y=0 y(0)=-1  y'(0)=6

anonymous · Answer

the answer in the book is -e^(-3t)+3te^(-3t)

zepdrix · Answer

Yes, see the second term? That's where I was headed with the steps I was showing you.

zepdrix · Answer

err wait, we would end up with a 5 instead of a 3 coefficient, so yah you musta run into a tiny mistake somewhere earlier on.

zepdrix · Answer

|dw:1355526190110:dw|Hmm I don't think that 5 should be there, check to see if you forgot to distribute the 6 in the second term.