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ranyai12
Can someone please help!!! How do you solve for the inverse laplace of 1/(s(s+4)(s+4))
\[\Large \frac{1}{s(s+4)^2} \]?
Partial fraction decomposition \[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2}\]
so you'd get A(S+4)(S+4)^2 +B(s)(S+4)^2 +C(s)(S+4)^2
If you multiply through you are left with \[\Large 1=A(s+4)^2+B(s+4)+Cs \]
I thought you are supposed to multiply A by the values under B and C?
I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression
\[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]
better now it seems
wouldnt you get A=0, B=0 and C=-1/4
Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.
\[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]
could you please explain how you go that
\[A+B=0 \\ 8A+4B+C=0 \\ 16A=0 \]
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.
\[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2} \] 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.
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.
you will end up with a 3x3 system of linear equations. \[A+B=0 \\ 8A+4B+C=0 \\ 16A=1 \] *corrected the last line*
sorry im still confused on how you got the equation
can you please show me by the work?
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
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i understand that part but after that I get confuesd on how you distributed the variables
\[ \Large 1=(A+B)s^2 +(8A+4B+C)s+16A \] This part?
yea, I'm not sure how you got that
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: \[\Large 1=A(s+4)^2+Bs(s+4)+Cs \]
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.
\[\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]
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: \[0s^2 +0s+\large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]
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.
\[0s^2+0s+1=(A+B)^2+(8A+4B+C)s+16A \]
im sorry but i dont understand the LHS
\[ 0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A\]
where did the numbers come in from
just a correction, was mistyping it.
and why is A added to B
Well the numbers come from because you expand a polynomial, namely (s+4)^2
And \[ As^2+Bs^2 = (A+B)s^2\]
just normal algebra.
ok so (s+4)^2=s^2+8s+16 but how do we incorproate that into the vaues for a,b and c
\[\Large 1=A(s+4)^2+Bs(s+4)+Cs \] 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!
\[\large 0s^2+0s+1=\underbrace{A(s+4)^2+Bs(s+4)+Cs}_{\text{expand and match with LHS}} \]
oh ok! thanks! btw do we have to expand?
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.
For instance if you set \( s=-4\) in the equation above, the first terms will all go to zero and you are left with \[ \Large 1=-4C\]
but I don't recommend doing this, because this isn't an easy partial fraction decomposition.
so then you get As^2+8As+16A+Bs^2+4Bs+Cs
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.
ok now I understand that part! thank you! after that you solve for the variable right
\[0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A \] 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.
yes, match all the coefficients with the the coefficients on the LHS
\[ A+B=0 \\ 8A+4B+C=0 \\ 16A=1\] See how I got that 3x3 System?
Sorry, C is not zero no.
yes, I got it right, confused myself with my silly notes (-:
I get the following (concentrating hard now) \[A= \frac{1}{16}, \ B=- \frac{1}{16}, \ C= -\frac{1}{4} \]
so A=1/16 B=-1/16 C=-1/4
awesome! wat do we do now?
So we have just proved that the left hand side can also be written as \[\Large \frac{1}{16s}- \frac{1}{16(s+1)}-\frac{1}{4(s+1)^2}\]
of course it is, I just typed it wrong. (-:
\[\Large \frac{1}{16s}- \frac{1}{16(s+4)}-\frac{1}{4(s+4)^2} \]
ok awesome! whats the next step
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
ok but the there are two constaNTS
For the first one \[ \Large \frac{1}{16} ( \mathcal{L}\lbrace \frac{1}{s}\rbrace)\]
F(s)=(1/16)[(1/s)-(1/(s+4)]-(1/4)[1/(s+4)^2)]
The inverse of 1/s = 1
what do i do about 1/(s+4)^2
you forgot the last one though.
\[\Large t^ne^{at}= \frac{n!}{(s-a)^{n+1}} \] with \(n=1, 2, 3 ...\) in our case n=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)
for 1/(s+4)-->n!/(s^(n+1))
I use this one, most of the time http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
i understand that we have to use the table bu incorporating it is confusing me a lil what do we plug in for n
\[\Large e^{at}= \frac{1}{s-a} \] In our case \( a=-4\)
non related formulas.
but we have s+4 not s-4
oh nvr mind i confused t^n and e^at
\[ \Large t^1e^{-4t}= \frac{1!}{(s+4)^{2}} \]
this is what I get for my substitution.
ok so I ended up with (1/16)(1)-(1/16)[1/(s+4)]-(1/4)[1/(s+4)^5)]
No, that's not what you are doing.
You will use this for differential equations, so you compute the inverse, the inverse is a function.
\[\Large \frac{1}{16}- \frac{1}{16}e^{-4t}- \frac{1}{4}te^{-4t} \]
That is what I get, if I didn't make any careless mistakes.
so you dont use the transform equation
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.
\[\Large \frac{1}{16}\mathcal{L}\lbrace \frac{1}{s}\rbrace- \frac{1}{16} \mathcal{L}\lbrace\frac{1}{s+4} \rbrace.....\]
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.
The inverse is then the solution to the Laplace Transform, it is commonly used for differential equations.
It basically turns differential equation problems into Algebra problems.
oh i get it thank you sooo much!