(s^2 - 4s + 4) I need help to solve partial fractions in
First factor it.
It's complex roots are. s=2-6i s=2+6i
Where is the fraction?
@marielen123 you have to have a fraction to partial it. yours is not a fraction, what should we do ?
I would like to solve partial fraction, using the formula
but s^2 -4s + 4 = (s^2 -2) ^2 no get to result
yes, one moment
y'' - 4y' + 4y = 3 δ ( t - 1) + δ ( t-2) it is the question
laplace, because solving the question
y(s) =[ 3e^-s + e^-2s + s -3] / s^2 -4s + 4
in this site, I know many many mathematicians who are experts this problem, let me tag them @Psymon @wio @UsukiDoll
ok, thank you
@marielen123 when using Laplace, we need initial value, what are they?
I tag others but I still work by myself. If I got it, I will show you
the initial values it is y(0) = 1 and y'(0) = 1 @Loser66
First factor the denominator.
@marielen123 got you @wio I need work from the beginning by myself, what if she makes mistake at somewhere?
@marielen123 I don't think you have to use partial fraction there, from the laplace table, you have formula #13 which talks about \(e^{-cs}F(s)\) your F(s) is \(\dfrac{1}{s^2-4s+4}\) just by that way, you can figure out the inverse of that Laplace
for the last 2 parts, s/ wholething, , you have to apply formula #16 . only the last one, which is 3/ wholething you can manipulate to get the form of formula #11(this is the easiest one) I don't see any thing else can apply to your case. I am sorry.
I think apply formula #10
he is not here, but when he came online, you can tag him. I am pretty sure that he is willing to help and his major is Math. he is @SithandGiggle I apply #10 and got stuck, since yours is 4s when it needs 6s for the complete square.
ok, apply: [(s-2) -1] / (s-2) ^2
I understand, thank you
hehe ok
I am very good at poking others. heehe.. good luck, I have a bunch of homework to do.
ahaaha... you are right at the post above, you are sooooo smart
hehe, thank you
flattering you first, poking you later. wait there. girl
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