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\[\Large \frac{1}{s(s+4)^2} \]?

yes

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?

\[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]

better now it seems

wouldnt you get A=0, B=0 and C=-1/4

\[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]

could you please explain how you go that

\[A+B=0 \\ 8A+4B+C=0
\\
16A=0 \]

okay so far?

sorry im still confused on how you got the equation

can you please show me by the work?

|dw:1344895643951:dw|

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

\[\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]

okay?

\[0s^2+0s+1=(A+B)^2+(8A+4B+C)s+16A \]

im sorry but i dont understand the LHS

I mean RHS

\[ 0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A\]

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

\[\large 0s^2+0s+1=\underbrace{A(s+4)^2+Bs(s+4)+Cs}_{\text{expand and match with LHS}} \]

oh ok! thanks! btw do we have to expand?

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

ok now I understand that part! thank you! after that you solve for the variable right

yes, match all the coefficients with the the coefficients on the LHS

would you get C=1

\[ A+B=0 \\ 8A+4B+C=0
\\
16A=1\]
See how I got that 3x3 System?

isnt it 16A=1

Sorry, C is not zero no.

yes, I got it right, confused myself with my silly notes (-:

so A=1/16
B=-1/16
C=-1/4

yes

awesome! wat do we do now?

isnt it s+4

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

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

http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

1/16[e^0-e^-4t]

yes looks good.

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

got that?

kind of

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

\[\Large e^{at}= \frac{1}{s-a} \]
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.

what do you do?

\[\Large \frac{1}{16}- \frac{1}{16}e^{-4t}- \frac{1}{4}te^{-4t} \]

That is what I get, if I didn't make any careless mistakes.

so you dont use the transform equation

oh ok

It basically turns differential equation problems into Algebra problems.

oh i get it thank you sooo much!

welcome

:)