Question

Hi everyone! I'm doing an inverse Laplace transform s/(s^2 + 4s + 5)...then I use completing the square to get to s/ [(s+2)^2 + 1]...now this problem asks you to use the first translation theorem L{e^at * f(t)} = F(s-a) so the next step is to add and subtract 2 on the numerator like this >>>
(s + 2 - 2)/ [(s+2)^2 + 1]...my question is why do they use 2? Thanks! :o)

Answer 1

why not +9-9 for example?

Answer 2

@rational

Answer 3

@ParthKohli can you help me? :o)

Answer 4

Ah, I'm not too familiar with DEs yet, but your question seems to be mostly about algebraic manipulation. I don't know how the theorem is applied, but I can tell that they did \((s+2) -2\) keeping in mind the \((s+2)^2 + 1\) in the denominator.

Answer 5

i'm trying to figure out how to word my question..one sec lol

Answer 6

when they did +2-2, it looks like that order is somehow significant when they eventualy split up the problem because the s+2 on top and bottom get separated from the -2 like this>>>
[ (s+2)/(s+2)^2+1 ] - [ 2/ )s+2)^2 +1 ]...see what I mean?

I wonder if they would have gotten the wrong answer if they put the s-2 in the numerator on the left side ?

Answer 7

Yeah, I believe you just answered your own question! Expressions of the form \(\frac{t}{t^2 + 1}\) and \(\frac{1}{t^2 + 1}\) are simple to work with, at least, in calc I. (Here, \(t := s+2\).)

I do not know the Laplace Transform part of it, but what I can tell from context is that you wouldn't have been able to do the same with any other kind of splitting.

Answer 8

@rational knows this stuff, and he may be able to add to my argument.

Answer 9

I tagged him but he must be busy

Answer 10

@Callisto?

Answer 11

test today and I need to figure this out

Answer 12

Do you know the rest of the solution? Can you see in the later steps how this manipulation was used?

Answer 13

i wonder...if they had instead (s-9) in the denominator, would they have chosen -9+9 on top and both (s-9) would have traveled off together to the left hand side?

Answer 14

Yes, that's correct!

Answer 15

yes I see what they did, but don't know why

Answer 16

If it led to the solution, you do.

Answer 17

I guess you need to keep the like quantities together because ultimately, that s+2 is actually s-(-2) and a=-2 then

Answer 18

you need the actual value of "a" in order to get the final solution

Answer 19

I'm pretty sure that is the reasoning, at least I can't think of anything else

Answer 20

I will leave this question open for a bit in case someone can confirm what we have written...thanks Parth! :o)

Answer 21

\[\mathcal{L}\{\sin t\}=\frac{s}{s^2+1};\quad\mathcal{L}\{\cos t\}=\frac1{s^2+1}\] so \[\mathcal{L}\{e^{-at}\sin t\}=\frac{s+a}{(s+a)^2+1};\quad\mathcal{L}\{e^{-at}\cos t\}=\frac1{(s+a)^2+1}\]
in our case, we have: $$\begin{align*}\mathcal{L}^{-1}\left\{\frac{s}{(s+2)^2+1}\right\}&=\mathcal{L}^{-1}\left\{\frac{(s+2)-2}{(s+2)^2+1}\right\}\\&=\mathcal{L}^{-1}\left\{\frac{s+2}{(s+2)^2+1}\right\}-2\cdot \mathcal{L}^{-1}\left\{\frac1{(s+2)^2+1}\right\}\\&=e^{-2t}\sin t-2e^{-2t}\cos t\end{align*}$$

Answer 22

so you MUST keep the (s+2) with the other (s+2) or it will not work correct?

Answer 23

we split the numerator \(s=(s+2)-2\) in such a manner because we notice the similarity of our expression to that of the Laplace transform of \(e^{-at}\sin t\) with \(a=2\). the purpose of splitting $(s+2)-2$ rather than $(s-2)+2$ is for the sole reason that the numerator and squared expression in the Laplace transform of \(e^{-at}\sin t\) are both identical (i.e. \(s+a\) in numerator, \((s+a)^2\) in denominator). we're trying to make our expression look like Laplace transform templates that we recognize in order to invert the transform by eye

Answer 24

awesome explanation!!! wow
i want to give you a medal but I already gave it away!
I will find you later and give you one on another thread! :o)
that is such an awesome explanation! really...wow!

Answer 25

if you instead split \(s=(s-2)+2\) you would still not be able to recognize any specific Laplace transform after breaking into two rational expressions, and inverting the Laplace transform in this manner is ultimately in recognizing the forms of common Laplace transforms

Answer 26

I totally get it! :o)

Answer 27

hey @ParthKohli ...give this gentleman a medal will you? :o)

Answer 28

that being said, it's not a matter of *having* to do it in this way--there are ways of using Laplace properties to tackle this in a slightly different manner, which is generally the case. for example, as you noted: \[\mathcal{L}\left\{e^{-at}f(t)\right\}=F(s+a)\] where \(F(s)=\mathcal{L}\{f(t)\}\)

in our case, we notice that we have a pesky \(s+2\) in our denominator, suggesting that we could apply the above Laplace transform property to rewrite our expression as follows: \[\mathcal{L}^{-1}\left\{\frac{(s+2)-2}{(s+2)^2+1}\right\}=e^{-2t}\cdot\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+1}\right\}\]i.e. drawing out \(e^{-2t}\) in order to replace \(s+2\mapsto s\). now, you can split into two rational expressions naturally with the same result of \(e^{-2t}(\sin t-2\cos t)\)

Answer 29

wow...do you know how to do L^-1{ 2s+5/s^2+6s+34} ?

Answer 30

2S-5/(S+3)^2 +25 SO FAR

Answer 31

May I try? :P\[\dfrac{2s+5}{(s+3)^2 + 25}=\dfrac{2(s+3) - 1}{(s+3)^2 + 25}\]Not sure if that works, but I just did what I saw you do in the previous question.

Answer 32

\[L^-1(2S-5+3-3)/(S+3)^2+25\]

Answer 33

how did you factor the 2?

Answer 34

\[2s + 5 = 2s + 6 - 1 = 2(s+3) - 1\]

Answer 35

did you do something other than +-3 ?

Answer 36

nevermind...i see it...tricky

Answer 37

that should work!...lemme finish :o)

Answer 38

just to make a connection here, @ParthKohli is 'regrouping' in a way much like mental division; if you use polynomial division to evaluate \((2s+5)\div (s+3)\) you will find: $$\frac{2s+5}{s+3}=2\text{ with a remainder of }{-1}=2-\frac1{s+3}$$multiplying both sides by \(s+3\) gives \(2s+5=2(s+3)-1\)

Answer 39

i get as far as e^-3tsin5t-e^-3t(something)

Answer 40

so you're stuck at $$\mathcal{L}^{-1}\left\{-\frac{1}{(s+3)^2+25}\right\}$$?

Answer 41

dont know

Answer 42

well, recognize that we know: $$\mathcal{L}\{e^{-at}\cos(bt)\}=\frac{b}{(s+a)^2+b^2}$$; in this case we can see \(a=3\) and \(b^2=25\) suggesting we want \(b=5\). but our numerator looks like \(-1\) while we want it to look like \(5\)

Answer 43

so \[-\frac1{(s+3)^2+5^2}\cdot\frac55=-\frac15\cdot\frac5{(s+3)^2+5^2}\] and then \[\mathcal{L}^{-1}\left\{-\frac15\cdot\frac5{(s+3)^2+5^2}\right\}=-\frac15\cdot\mathcal{L}^{-1}\left\{\frac5{(s+3)^2+5^2}\right\}=-\frac15e^{-3t}\cos(5t)\]

Answer 44

actually I don't have that memorized...the teacher only made us remember 7 basic ones

Answer 45

omg! that is seriously tricky algebra!

Answer 46

well, do you have the following 'memorized'? $$\mathcal{L}\left\{\cos(bt)\right\}=\frac{b}{s^2+b^2}$$ what about this? $$\mathcal{L}\left\{e^{-at}f(t)\right\}=F(s+a)\quad\text{where }F(s)=\mathcal{L}\{f(t)\}$$

Answer 47

so should I just keep in mind to make k^2 on the bottom and then think about fancy forms of 1?

Answer 48

1st one yes...second one no, and my book has the signs reversed also

Answer 49

in this case, we've just applied both at the same time :-) you can also apply the exponential property first as I demonstrated earlier. and yes, you should recognize the \(b^2\) in the denominator suggests our numerator needs to look like \(b\) and we can accomplish that using a 'fancy one', right

Answer 50

the second identity can also be written by substituting \(a\mapsto-a\) which gives $$\mathcal{L}\left\{e^{at}f(t)\right\}=F(s-a)\quad\text{where }F(s)=\mathcal{L}\{f(t)\}$$

Answer 51

so then how do i do 5/(s+3)^2 + 5^2 part then?

Answer 52

to be completely honest with you, I'm rather aversive to memorizing properties and instead I typically derive both of those results on the fly from the integrals :)

Answer 53

for that part, we'll notice we can map \(s+3\mapsto s\) if we pull out an \(e^{-3t}\) using the exponential property: $$\mathcal{L}\left\{\frac5{(s+3)^2+5^2}\right\}=e^{-3t}\cdot\mathcal{L}^{-1}\left\{\frac{5}{s^2+5^2}\right\}=e^{-3t}\cos(5t)$$