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

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SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large \frac{1}{s(s+4)^2} \]?
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
so you'd get A(S+4)(S+4)^2 +B(s)(S+4)^2 +C(s)(S+4)^2
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
I thought you are supposed to multiply A by the values under B and C?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
better now it seems
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
wouldnt you get A=0, B=0 and C=1/4
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
could you please explain how you go that
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[A+B=0 \\ 8A+4B+C=0 \\ 16A=0 \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2} \] 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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
you will end up with a 3x3 system of linear equations. \[A+B=0 \\ 8A+4B+C=0 \\ 16A=1 \] *corrected the last line*
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
sorry im still confused on how you got the equation
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
can you please show me by the work?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
dw:1344895643951:dw
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
i understand that part but after that I get confuesd on how you distributed the variables
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[ \Large 1=(A+B)s^2 +(8A+4B+C)s+16A \] This part?
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
yea, I'm not sure how you got that
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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: \[\Large 1=A(s+4)^2+Bs(s+4)+Cs \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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: \[0s^2 +0s+\large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
im sorry but i dont understand the LHS
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
where did the numbers come in from
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
just a correction, was mistyping it.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
and why is A added to B
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
Well the numbers come from because you expand a polynomial, namely (s+4)^2
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
And \[ As^2+Bs^2 = (A+B)s^2\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
just normal algebra.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
ok so (s+4)^2=s^2+8s+16 but how do we incorproate that into the vaues for a,b and c
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large 1=A(s+4)^2+Bs(s+4)+Cs \] 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!
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
oh ok! thanks! btw do we have to expand?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
For instance if you set \( s=4\) in the equation above, the first terms will all go to zero and you are left with \[ \Large 1=4C\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
but I don't recommend doing this, because this isn't an easy partial fraction decomposition.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
so then you get As^2+8As+16A+Bs^2+4Bs+Cs
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
ok now I understand that part! thank you! after that you solve for the variable right
 one year ago

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

SpacelimbusBest ResponseYou've already chosen the best response.1
yes, match all the coefficients with the the coefficients on the LHS
 one year ago

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

SpacelimbusBest ResponseYou've already chosen the best response.1
Sorry, C is not zero no.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
yes, I got it right, confused myself with my silly notes (:
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
so A=1/16 B=1/16 C=1/4
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
awesome! wat do we do now?
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
So we have just proved that the left hand side can also be written as \[\Large \frac{1}{16s} \frac{1}{16(s+1)}\frac{1}{4(s+1)^2}\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
of course it is, I just typed it wrong. (:
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large \frac{1}{16s} \frac{1}{16(s+4)}\frac{1}{4(s+4)^2} \]
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
ok awesome! whats the next step
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
ok but the there are two constaNTS
 one year ago

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

ranyai12Best ResponseYou've already chosen the best response.0
F(s)=(1/16)[(1/s)(1/(s+4)](1/4)[1/(s+4)^2)]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
The inverse of 1/s = 1
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
what do i do about 1/(s+4)^2
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
you forgot the last one though.
 one year ago

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

SpacelimbusBest ResponseYou've already chosen the best response.1
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)
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
for 1/(s+4)>n!/(s^(n+1))
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
I use this one, most of the time http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
i understand that we have to use the table bu incorporating it is confusing me a lil what do we plug in for n
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large e^{at}= \frac{1}{sa} \] In our case \( a=4\)
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
non related formulas.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
but we have s+4 not s4
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
oh nvr mind i confused t^n and e^at
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[ \Large t^1e^{4t}= \frac{1!}{(s+4)^{2}} \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
this is what I get for my substitution.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
ok so I ended up with (1/16)(1)(1/16)[1/(s+4)](1/4)[1/(s+4)^5)]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
No, that's not what you are doing.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
You will use this for differential equations, so you compute the inverse, the inverse is a function.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large \frac{1}{16} \frac{1}{16}e^{4t} \frac{1}{4}te^{4t} \]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
That is what I get, if I didn't make any careless mistakes.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
so you dont use the transform equation
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
\[\Large \frac{1}{16}\mathcal{L}\lbrace \frac{1}{s}\rbrace \frac{1}{16} \mathcal{L}\lbrace\frac{1}{s+4} \rbrace.....\]
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
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.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
The inverse is then the solution to the Laplace Transform, it is commonly used for differential equations.
 one year ago

SpacelimbusBest ResponseYou've already chosen the best response.1
It basically turns differential equation problems into Algebra problems.
 one year ago

ranyai12Best ResponseYou've already chosen the best response.0
oh i get it thank you sooo much!
 one year ago
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