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anonymous
 3 years ago
Can someone please help!!! How do you solve for the inverse laplace of 1/(s(s+4)(s+4))
anonymous
 3 years ago
Can someone please help!!! How do you solve for the inverse laplace of 1/(s(s+4)(s+4))

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \frac{1}{s(s+4)^2} \]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Partial fraction decomposition \[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you'd get A(S+4)(S+4)^2 +B(s)(S+4)^2 +C(s)(S+4)^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you multiply through you are left with \[\Large 1=A(s+4)^2+B(s+4)+Cs \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I thought you are supposed to multiply A by the values under B and C?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wouldnt you get A=0, B=0 and C=1/4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0could you please explain how you go that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[A+B=0 \\ 8A+4B+C=0 \\ 16A=0 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2} \] 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
 3 years ago
Best ResponseYou've already chosen the best response.0Then 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
 3 years ago
Best ResponseYou've already chosen the best response.0you will end up with a 3x3 system of linear equations. \[A+B=0 \\ 8A+4B+C=0 \\ 16A=1 \] *corrected the last line*

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry im still confused on how you got the equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you please show me by the work?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1344895643951:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i understand that part but after that I get confuesd on how you distributed the variables

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \Large 1=(A+B)s^2 +(8A+4B+C)s+16A \] This part?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea, I'm not sure how you got that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0you 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now 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
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[0s^2+0s+1=(A+B)^2+(8A+4B+C)s+16A \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im sorry but i dont understand the LHS

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ 0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0where did the numbers come in from

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just a correction, was mistyping it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and why is A added to B

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well the numbers come from because you expand a polynomial, namely (s+4)^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And \[ As^2+Bs^2 = (A+B)s^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so (s+4)^2=s^2+8s+16 but how do we incorproate that into the vaues for a,b and c

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large 1=A(s+4)^2+Bs(s+4)+Cs \] 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large 0s^2+0s+1=\underbrace{A(s+4)^2+Bs(s+4)+Cs}_{\text{expand and match with LHS}} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok! thanks! btw do we have to expand?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0For 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
 3 years ago
Best ResponseYou've already chosen the best response.0but I don't recommend doing this, because this isn't an easy partial fraction decomposition.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so then you get As^2+8As+16A+Bs^2+4Bs+Cs

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok now I understand that part! thank you! after that you solve for the variable right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A \] 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
 3 years ago
Best ResponseYou've already chosen the best response.0yes, match all the coefficients with the the coefficients on the LHS

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ A+B=0 \\ 8A+4B+C=0 \\ 16A=1\] See how I got that 3x3 System?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, C is not zero no.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, I got it right, confused myself with my silly notes (:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I get the following (concentrating hard now) \[A= \frac{1}{16}, \ B= \frac{1}{16}, \ C= \frac{1}{4} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so A=1/16 B=1/16 C=1/4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0awesome! wat do we do now?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0of course it is, I just typed it wrong. (:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \frac{1}{16s} \frac{1}{16(s+4)}\frac{1}{4(s+4)^2} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok awesome! whats the next step

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok but the there are two constaNTS

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For the first one \[ \Large \frac{1}{16} ( \mathcal{L}\lbrace \frac{1}{s}\rbrace)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0F(s)=(1/16)[(1/s)(1/(s+4)](1/4)[1/(s+4)^2)]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The inverse of 1/s = 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do i do about 1/(s+4)^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you forgot the last one though.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large t^ne^{at}= \frac{n!}{(sa)^{n+1}} \] with \(n=1, 2, 3 ...\) in our case n=1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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
 3 years ago
Best ResponseYou've already chosen the best response.0for 1/(s+4)>n!/(s^(n+1))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I use this one, most of the time http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i understand that we have to use the table bu incorporating it is confusing me a lil what do we plug in for n

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large e^{at}= \frac{1}{sa} \] In our case \( a=4\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0non related formulas.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but we have s+4 not s4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh nvr mind i confused t^n and e^at

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \Large t^1e^{4t}= \frac{1!}{(s+4)^{2}} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is what I get for my substitution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so I ended up with (1/16)(1)(1/16)[1/(s+4)](1/4)[1/(s+4)^5)]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, that's not what you are doing.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You will use this for differential equations, so you compute the inverse, the inverse is a function.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \frac{1}{16} \frac{1}{16}e^{4t} \frac{1}{4}te^{4t} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is what I get, if I didn't make any careless mistakes.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you dont use the transform equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0look 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \frac{1}{16}\mathcal{L}\lbrace \frac{1}{s}\rbrace \frac{1}{16} \mathcal{L}\lbrace\frac{1}{s+4} \rbrace.....\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0see 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
 3 years ago
Best ResponseYou've already chosen the best response.0The inverse is then the solution to the Laplace Transform, it is commonly used for differential equations.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It basically turns differential equation problems into Algebra problems.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i get it thank you sooo much!
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