anonymous
  • anonymous
Can someone please help!!! How do you solve for the inverse laplace of 1/(s(s+4)(s+4))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@Spacelimbus
anonymous
  • anonymous
\[\Large \frac{1}{s(s+4)^2} \]?
anonymous
  • anonymous
yes

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anonymous
  • anonymous
Partial fraction decomposition \[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2}\]
anonymous
  • anonymous
so you'd get A(S+4)(S+4)^2 +B(s)(S+4)^2 +C(s)(S+4)^2
anonymous
  • anonymous
If you multiply through you are left with \[\Large 1=A(s+4)^2+B(s+4)+Cs \]
anonymous
  • anonymous
I thought you are supposed to multiply A by the values under B and C?
anonymous
  • anonymous
I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression
anonymous
  • anonymous
\[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]
anonymous
  • anonymous
better now it seems
anonymous
  • anonymous
wouldnt you get A=0, B=0 and C=-1/4
anonymous
  • anonymous
Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.
anonymous
  • anonymous
\[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]
anonymous
  • anonymous
could you please explain how you go that
anonymous
  • anonymous
\[A+B=0 \\ 8A+4B+C=0 \\ 16A=0 \]
anonymous
  • anonymous
Yes, I multiplied the above form by s(s+4)^2 (make sure you cancel on the right hand side right) and then you factor likewise terms. When you did that, you can set both sides of the equation equal, they are equal if and only if their coefficients match.
anonymous
  • anonymous
\[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2} \] General setup of partial fraction decomposition, multiply through by \(s(s+4)^2\), it will disappear entirely on the left hand side of the equation and piecewise on the righthandside of the equation.
anonymous
  • anonymous
Then expand the righthand side and factor likewise terms, set the coefficients equal to the coefficients on the lefthand side to make a valid equation out of it LHS=RHS, in this case a lot of the coefficients have to be zero, because there is only 1 on the LHS.
anonymous
  • anonymous
you will end up with a 3x3 system of linear equations. \[A+B=0 \\ 8A+4B+C=0 \\ 16A=1 \] *corrected the last line*
anonymous
  • anonymous
okay so far?
anonymous
  • anonymous
sorry im still confused on how you got the equation
anonymous
  • anonymous
can you please show me by the work?
anonymous
  • anonymous
The idea of partial fraction decomposition is that you find a different way of writing the quotient, so in the end there will be an easier form for it which is especially useful in integral calculus and for LaPlace Transformations. The idea is that you completely factor the denominator and then you split it up, so each new quotient becomes its own denominator
anonymous
  • anonymous
|dw:1344895643951:dw|
anonymous
  • anonymous
i understand that part but after that I get confuesd on how you distributed the variables
anonymous
  • anonymous
\[ \Large 1=(A+B)s^2 +(8A+4B+C)s+16A \] This part?
anonymous
  • anonymous
yea, I'm not sure how you got that
anonymous
  • anonymous
I multiplied both sides by s(s+4)^2 so the denominator will completely disappear on the LHS, on the right hand side I have: \[\Large 1=A(s+4)^2+Bs(s+4)+Cs \]
anonymous
  • anonymous
you want to solve for A, B, C in the end, but first you distribute all the binomials out and then factor likewise powers together to make a suitable system of equations.
anonymous
  • anonymous
\[\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]
anonymous
  • anonymous
now read the equation like this, on the righthand side you have a quadratic polynomial, but on the left hand side, you can understand it like this: \[0s^2 +0s+\large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]
anonymous
  • anonymous
okay?
anonymous
  • anonymous
so in order that the lefthand side is equal to the righthand side, the coefficients A,B,C on the righthand side have to match the coefficients of the same powers of s on the lefthand side.
anonymous
  • anonymous
\[0s^2+0s+1=(A+B)^2+(8A+4B+C)s+16A \]
anonymous
  • anonymous
im sorry but i dont understand the LHS
anonymous
  • anonymous
I mean RHS
anonymous
  • anonymous
\[ 0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A\]
anonymous
  • anonymous
where did the numbers come in from
anonymous
  • anonymous
just a correction, was mistyping it.
anonymous
  • anonymous
and why is A added to B
anonymous
  • anonymous
Well the numbers come from because you expand a polynomial, namely (s+4)^2
anonymous
  • anonymous
And \[ As^2+Bs^2 = (A+B)s^2\]
anonymous
  • anonymous
just normal algebra.
anonymous
  • anonymous
ok so (s+4)^2=s^2+8s+16 but how do we incorproate that into the vaues for a,b and c
anonymous
  • anonymous
\[\Large 1=A(s+4)^2+Bs(s+4)+Cs \] Could you follow to this step? You see that A, B, and C gets multiplied with all that stuff. A with the first entire term, B with the second entire term, and C with the third. To match the equation we have to write the LHS different and then factor terms together, not only the first expression will have an x^2, also the second one with the Bs in front of it!
anonymous
  • anonymous
\[\large 0s^2+0s+1=\underbrace{A(s+4)^2+Bs(s+4)+Cs}_{\text{expand and match with LHS}} \]
anonymous
  • anonymous
oh ok! thanks! btw do we have to expand?
anonymous
  • anonymous
yes, in this case I suggest it, there are methods called cover up etc which can be applied to get at least some of the solutions easily, but I wouldn't recommend it.
anonymous
  • anonymous
For instance if you set \( s=-4\) in the equation above, the first terms will all go to zero and you are left with \[ \Large 1=-4C\]
anonymous
  • anonymous
but I don't recommend doing this, because this isn't an easy partial fraction decomposition.
anonymous
  • anonymous
so then you get As^2+8As+16A+Bs^2+4Bs+Cs
anonymous
  • anonymous
yes, that's what you get. If you read your equation careful enough you will see that there are multiple s^2 and s in it, you want to collect them and then match them to the left hand side, this is a valid technique.
anonymous
  • anonymous
ok now I understand that part! thank you! after that you solve for the variable right
anonymous
  • anonymous
\[0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A \] See, the right hand side, you multiply A+B times s^2, if A+B are 0, then the right hand side becomes 0s^2, so it is equal to the left hand side and we are happy.
anonymous
  • anonymous
yes, match all the coefficients with the the coefficients on the LHS
anonymous
  • anonymous
would you get C=1
anonymous
  • anonymous
\[ A+B=0 \\ 8A+4B+C=0 \\ 16A=1\] See how I got that 3x3 System?
anonymous
  • anonymous
isnt it 16A=1
anonymous
  • anonymous
Sorry, C is not zero no.
anonymous
  • anonymous
yes, I got it right, confused myself with my silly notes (-:
anonymous
  • anonymous
I get the following (concentrating hard now) \[A= \frac{1}{16}, \ B=- \frac{1}{16}, \ C= -\frac{1}{4} \]
anonymous
  • anonymous
so A=1/16 B=-1/16 C=-1/4
anonymous
  • anonymous
yes
anonymous
  • anonymous
awesome! wat do we do now?
anonymous
  • anonymous
So we have just proved that the left hand side can also be written as \[\Large \frac{1}{16s}- \frac{1}{16(s+1)}-\frac{1}{4(s+1)^2}\]
anonymous
  • anonymous
isnt it s+4
anonymous
  • anonymous
of course it is, I just typed it wrong. (-:
anonymous
  • anonymous
\[\Large \frac{1}{16s}- \frac{1}{16(s+4)}-\frac{1}{4(s+4)^2} \]
anonymous
  • anonymous
ok awesome! whats the next step
anonymous
  • anonymous
well you know that the Laplace is a linear operator, so you can factor out constants and then grab a cheat sheet to see if you can find inverse laplace transforms
anonymous
  • anonymous
ok but the there are two constaNTS
anonymous
  • anonymous
For the first one \[ \Large \frac{1}{16} ( \mathcal{L}\lbrace \frac{1}{s}\rbrace)\]
anonymous
  • anonymous
F(s)=(1/16)[(1/s)-(1/(s+4)]-(1/4)[1/(s+4)^2)]
anonymous
  • anonymous
The inverse of 1/s = 1
anonymous
  • anonymous
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
anonymous
  • anonymous
1/16[e^0-e^-4t]
anonymous
  • anonymous
yes looks good.
anonymous
  • anonymous
what do i do about 1/(s+4)^2
anonymous
  • anonymous
you forgot the last one though.
anonymous
  • anonymous
\[\Large t^ne^{at}= \frac{n!}{(s-a)^{n+1}} \] with \(n=1, 2, 3 ...\) in our case n=1
anonymous
  • anonymous
got that?
anonymous
  • anonymous
kind of
anonymous
  • anonymous
You have to use a cheat sheet for these kind of problems, those formulas have been derived once, (some of them get ridiculously complicated, because of the long integration by parts problems)
anonymous
  • anonymous
for 1/(s+4)-->n!/(s^(n+1))
anonymous
  • anonymous
I use this one, most of the time http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
anonymous
  • anonymous
i understand that we have to use the table bu incorporating it is confusing me a lil what do we plug in for n
anonymous
  • anonymous
\[\Large e^{at}= \frac{1}{s-a} \] In our case \( a=-4\)
anonymous
  • anonymous
non related formulas.
anonymous
  • anonymous
but we have s+4 not s-4
anonymous
  • anonymous
oh nvr mind i confused t^n and e^at
anonymous
  • anonymous
\[ \Large t^1e^{-4t}= \frac{1!}{(s+4)^{2}} \]
anonymous
  • anonymous
this is what I get for my substitution.
anonymous
  • anonymous
ok so I ended up with (1/16)(1)-(1/16)[1/(s+4)]-(1/4)[1/(s+4)^5)]
anonymous
  • anonymous
No, that's not what you are doing.
anonymous
  • anonymous
You will use this for differential equations, so you compute the inverse, the inverse is a function.
anonymous
  • anonymous
what do you do?
anonymous
  • anonymous
\[\Large \frac{1}{16}- \frac{1}{16}e^{-4t}- \frac{1}{4}te^{-4t} \]
anonymous
  • anonymous
That is what I get, if I didn't make any careless mistakes.
anonymous
  • anonymous
so you dont use the transform equation
anonymous
  • anonymous
look at the cheat sheet again, right you have the Laplace Transform, and on the other side you have the inverse, you want to compute the inverse.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
\[\Large \frac{1}{16}\mathcal{L}\lbrace \frac{1}{s}\rbrace- \frac{1}{16} \mathcal{L}\lbrace\frac{1}{s+4} \rbrace.....\]
anonymous
  • anonymous
see that? I didn't carry it out entirely, but that above is a Laplace Transform, whenever you have something of such a form, you want to compute its inverse.
anonymous
  • anonymous
The inverse is then the solution to the Laplace Transform, it is commonly used for differential equations.
anonymous
  • anonymous
It basically turns differential equation problems into Algebra problems.
anonymous
  • anonymous
oh i get it thank you sooo much!
anonymous
  • anonymous
welcome
anonymous
  • anonymous
:)

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