Question

HI everyone! Can someone help me solve the inverse laplace transform of -(1/4)L^-1{1/(s+2)^3) please? I know the answer is -(1/8)t^2e^-2t but I need help getting there. Thanks! :o)

Answer 1

I know 1/(s+2)^3 resembles n! / s^(n+1)
so n would be equal to 2...
then you have to multiply the top and bottom by a fancy form of 1, in the case, 2factorial/2factorial (2!/2!)...
so then the problem would look like this...
-(1/4)(1/2!) L^-1 { 2/(s+2)^(2+1)}...
THEN...

Answer 2

i also know that L^-1{n!/s^(n+1)} = t^n
this lead me to want to make the answer...
-(1/4)(1/2!) times t^2 which would make the answer -(1/8)t^2 but that's not the answer...so I need help with this step
how do I get from -(1/4)(1/2!) L^-1 { 2/(s+2)^(2+1)} to -(1/8)t^2e^-2t ???

Answer 3

@rational @Michele_Laino if anyone can help, please do

Answer 4

do you want to know how to get the inverse transform of this function:
\[\Large \frac{1}{{{{\left( {s + 2} \right)}^3}}}\]

Answer 5

yes basically...not forgetting the -1/4 in front...but yes

Answer 6

i typed everything above pretty neatly...you can see the step I get stuck on if you take a second and review my work above :o)

Answer 7

ok!

Answer 8

I also tried to explain my thought process as I was typing as well

Answer 9

In general the subsequent identity holds:
\[\Large L\left( {{t^n}{e^{ct}}} \right) = \frac{{n!}}{{{{\left( {s - c} \right)}^{n + 1}}}}\]

Answer 10

one mistake I made...
this L^-1{n!/s^(n+1)} = t^n should have been L^-1{t^n}=n!/s^(n+1)

Answer 11

michele...where did you get that transform? from some table?

Answer 12

no, we can demonstrate it, using the concept of differetiation of the L-transform

Answer 13

uhm

Answer 14

for example we have to start from the definition of L-transform, like this:
\[L\left( {f\left( t \right)} \right) = \int_0^\infty {f\left( t \right){e^{ - st}}dt} \]

Answer 15

\[\Large L\left( {f\left( t \right)} \right) = \int_0^\infty {f\left( t \right){e^{ - st}}dt} \]

Answer 16

okay...you don't have to go through that integral...i get it...it will equal t^2e^at

Answer 17

but my problem is this...

Answer 18

when you are working on an inverse laplace transform, you most likely will not have that integral and its answer sitting around...I need to know how to work backwards towards that without actually knowing it

Answer 19

please it is my starting point.
That function is a function of the variable s, so we can rewrite it as below:
\[\Large F\left( s \right) = \int_0^\infty {f\left( t \right){e^{ - st}}dt} \]
Now we have to consider, the subsequent incremental ration:
\[\Large \frac{{F\left( {s + \sigma } \right) - F\left( s \right)}}{\sigma } = \int_0^\infty {\frac{{{e^{ - \sigma t}} - 1}}{\sigma }{e^{ - st}}f\left( t \right)dt} \]

Answer 20

how do I get from -(1/4)(1/2!) L^-1 { 2/(s+2)^(2+1)} to -(1/8)t^2e^-2t ???
is it possible or do you ABSOLUTELY HAVE TO KNOW that integral?

Answer 21

ratio*

Answer 22

i don't even know what a subsequent incremental ratio is...i never studied that

Answer 23

ok!
the general formulas are these:
the inverse transform of this function:
\[\Large \frac{1}{{{{\left( {s - a} \right)}^k}}}\]
is:
\[\Large \frac{1}{{\left( {k - 1} \right)!}}{t^{k - 1}}{e^{at}}\]

Answer 24

my teacher only required us to remember 7 basic transforms...i have never seen this before

Answer 25

we can demonstrate that formula recursively

Answer 26

i don't even know what that means lol

Answer 27

what is recursively?

Answer 28

it is simple, since those formulas depend on natural parameter k, then we can apply the mathematical induction principle to k

Answer 29

i don't even know what the mathematical induction principal is michele
i don't think you are going to be able to help me...too much "mathese" to understand

Answer 30

here is what i know...the top 8 transforms plus the hyperbolics on this page...that's it
http://www.egr.msu.edu/~priezjev/ME391/Laplace.pdf

Answer 31

you can memorize those formulas, then

Answer 32

i have them memorized...i just need to know how to get from
-(1/4)(1/2!) L^-1 { 2/(s+2)^(2+1)} to -(1/8)t^2e^-2t

Answer 33

all those formulas, in your table can be demonstrated usin the concept of derivative of L-transform, and the mathematical induction principle

Answer 34

using*

Answer 35

okay but why are you telling me that? i don't understand...does that help me solve my problem somehow?

Answer 36

because in order to solve your problem I have to explain you how I'm able to get some formulas

Answer 37

i don't even know what concept of derivative of L-transform is...is that having a laplace trransform sitting on your paper, then taking its derivative or something?

Answer 38

Notice that \(\large \mathcal{L}^{-1}(\frac{1}{s^3})= \frac{t^2}{2!}\)

Answer 39

familiar with exponential shift formula ?

Answer 40

exponential shift formula : \[\mathcal{L}(e^{at}f(t))=F(s-a)\]

Answer 41

wait...exponential shift...that is the same thing as the first translation theorem right?

Answer 42

Yep!

Answer 43

okay great but wait...

Answer 44

k

Answer 45

that 1st translation theorem is in chapter 7.3
the problem I am trying to solve is in 7.2
I shouldn't have to use that

Answer 46

Ohk

Answer 47

so frustrated with this problem...how about I take a pic of the book...it works out the problem...then maybe you can just tell me how they did it eh?

Answer 48

exponential shift formula is right after the definition of laplace transform in my textbook

Answer 49

sure post it

Answer 50

well my book is dumb then lol

Answer 51

one sec

Answer 52

the top left says L^-1

Answer 53

also attach example6 of section 7.1

Answer 54

it's just a long old integral

Answer 55

then they're just using the previous result
thats all there is to it.

Answer 56

so they just magically expect me to recall an obscure integral from a previous section and "know" to apply it?

Answer 57

For now yes
once you know "exponential shift formula", you don't need to memorize these
it seems they're covering it in next section itself, so have patience

Answer 58

omg! gurrrrrr

Answer 59

so are you telling me without knowing the answer to that integral, there is no "mechanical" way to solve this problem?

Answer 60

But wait a second, you're actually using "exponential shift formula" for calculating the inverse of 1/(s+2) right ?

Answer 61

i don;'t know...i would have to think about that for a sec

Answer 62

how do you know \(\mathcal{L}^{-1}(\frac{1}{s+2})=e^{-2t}\) ?

Answer 63

because L{e^at} = 1/s-a

Answer 64

so a=-2

Answer 65

good, then there is a nice trick to work \(\mathcal{L}^{-1}(\frac{1}{(s+2)^3})\)

Answer 66

lets see

Answer 67

forget about inverses for sometime

Answer 68

from that I know n=2
and I also know you have to multiply by a fancy form of 1, in this case 2factorial/2factorial

Answer 69

so I would have -(1/4)(1/2!) L^-1 { 2/(s+2)^3}

Answer 70

we know that \[\mathcal{L}(1)=\frac{1}{s}\]

then consider \(\mathcal{L}(e^{at}f(t))\)

Answer 71

\(f(t)\) is just some arbitrary function

Answer 72

yes

Answer 73

\[\mathcal{L}(e^{at}f(t))=\int\limits_0^\infty e^{at}f(t)e^{-st}\,dt\]

yes ?

Answer 74

yes

Answer 75

\[\begin{align}\mathcal{L}(e^{at}f(t))&=\int\limits_0^\infty e^{at}f(t)e^{-st}\,dt\\~\\
&=\int\limits_0^\infty f(t)e^{-(s-a)t}\,dt\\~\\
&=F(s-a)
\end{align}\]

yes ?

Answer 76

yes...I've actually worked that integral 3 times in the past...so yes

Answer 77

Sweet :) we're almost done then

Answer 78

we know that \(\large \mathcal{L}(t^2)=\frac{2!}{s^3}\)

can you guess the value of \(\large \mathcal{L}(e^{at}t^2)\) ?

Answer 79

do you mean the inverse laplace?

Answer 80

no, laplace transform only

Answer 81

use the previous result
http://gyazo.com/8ebe92d7699329f175da0ef7da0f1014

Answer 82

nevermind what I said...getting inverses mixed up

Answer 83

ok

Answer 84

uhmmm

Answer 85

we know that \(\large \mathcal{L}(t^2)=\frac{2!}{s^3}\)

and we also know that \(\large \mathcal{L}(e^{at}f(t)) = F(s-a)\)

therefore \(\large \mathcal{L}(e^{at}t^2) = \frac{2!}{(s-a)^3}\)

Answer 86

basically multiplying any function by\(e^{at}\) shifts its transform by \(a\)

Answer 87

okay, but without knowing the 1st translation theorem, I still can't mechanically solve this problem

Answer 88

when you don't know anything, it is good because you don't have a choice... u go and use the definition of laplace transform

Answer 89

okay

Answer 90

okay my turn...

Answer 91

look at the right hand side of that problem I posted...its an inverse correct?

Answer 92

without knowing the 1st translation theorem and without mysteriously remembering that crazy integral...can you solve the problem I posted?

Answer 93

can you mechanically solve it just knowing the basic transforms?

Answer 94

Ahh nope, it wont reduce further in partial fraction decomposition

if you're not allowed to use 1st translation rule/convolution stuff, you just need to evaluate the laplace transform integral of \(e^{at}t^2\) and remember it somehow

Answer 95

but you're thinking too much on this small problem
memorizing these is unnecessary once you get to know the 1st translation rule

Answer 96

yeah but what I am saying is that if you only have the problem to look at, how on earth would you know to apply the e^att^2 thing...that seems almost impossible of a leap