Question

what is the laplace transform of y'' - y' -2y = -8cost-2sint, with initial values y(pi/2) = 1 and y'(pi/2) = 0?

Answer 1

did you get the last one?

Answer 2

yea

Answer 3

I dont want to do that again if its not helping lol

Answer 4

did i mess anyhting up?

Answer 5

ooh laplace =P

Answer 6

i didn't get the matrix part, but the rest helped

Answer 7

but i figure if I can get this last problem done, I should be able to figure all the rest out

Answer 8

so are you at the point again where you need to do partial fractions or move some stuff around so your inverse laplaces make since?

Answer 9

sense*

Answer 10

no, I haven't started that yet. honestly, the last few ?'s i posted werent answered withitn ten minutes, so i thought i had some time

Answer 11

you know how to do this outkast?

Answer 12

I just learned it yesterday

Answer 13

hahahaha

Answer 14

any luck? or any hints on how to start out?

Answer 15

aight but give me a sec, it takes way to long for me to type it.

Answer 16

to start out? hold on let me get my book

Answer 17

take the laplace of each one

Answer 18

of the derivatives what do you get

Answer 19

(s^2-s-2)Y - s - 1 = (-8s/(s^2 + B^2)) - (2B/(s^2+B^2))

Answer 20

but what would B be?

Answer 21

\[y''=s^2Y(s)-sy(\pi/2)-y'(\pi/2)=s^2Y(s)-s(1)-0=s^2Y(s)-s\]

Answer 22

\[y'=sY(s)-y(\pi/2)=sY(s)-1\]

Answer 23

y=Y(s)

Answer 24

okay, got the left side correct

Answer 25

yeah man your teacher is a d**k the first term expands to 6 terms and then there are 4 more that I have not expanded so its going to be like 12+ terms. sorry man I dont got that much time. I think its just the partial fractions you have a problem with? Dont worry about the initail conditions just call them to variables for now and deal with them at the end. as far as partial fractions I would just google a good tutorail on them, because when you start adding variables to the numerator it gets odd. But i know there are some slight of hand methods that work just fine...google cover up method for partial fraction decompasition.

Answer 26

the right side is

\[-8(\frac{s}{s^2+k^2})+2\frac{k}{s^2+k^2}\]

Answer 27

it's cool rock, i'm trying to find the easier way here with outkast

Answer 28

\[(s^2-s-2)Y(s)-s-1=-8(\frac{s}{s^2+k^2}+2(\frac{k}{s^2+k^2})\]

Answer 29

but i'm confused on what 'k' is

Answer 30

\[(s^2-s-2)Y(s)=-8(\frac{s}{s^2+k^2}+2(\frac{k}{s^2+k^2})+s+1

Answer 31

\[(s^2-s-2)Y(s)=-8(\frac{s}{s^2+k^2}+2(\frac{k}{s^2+k^2})+s+1\]

Answer 32

\[sin(kt)\]

Answer 33

the form is \[L{sin(kt)}=\frac{k}{s^2+k^2}\]

Answer 34

but k is a number and not a constant

Answer 35

\[(s^2-s-2)Y(s)=-8(\frac{s}{s^2+1})+2(\frac{1}{s^2+1})+s+1\]

Answer 36

divide the whole thing by (s^2-s-2)

\[Y(s)=-8(\frac{s}{(s^2+1)(s^2-s-2)})+2(\frac{1}{(s^2+1)(S^2-s-2)})\]

\[+\frac{s}{s^2-s-2}+\frac{1}{s^2-s-2}\]

Answer 37

can it just be the -8?

Answer 38

one works too

Answer 39

nevermind, i got it

Answer 40

not yet, just what to put there

Answer 41

\[Y(s)=\frac{-8s+2+s(s^2+1)+s^2+1}{(s^2-s-2)(s^2+1}\]

Answer 42

whaaa? what happened to them all being split up?

Answer 43

\[Y(s)=\frac{-8s+2+s^3+s+s^2+1}{(s^2-s-2)(s^2+1)}\]

Answer 44

even if the answer in the back was more like the other formula, with it all split up?

Answer 45

oooooh

Answer 46

\[Y(s)=\frac{s^3+s^2-7s+3}{(s^2-s-2)(s^2+1)}\]

Answer 47

expanding that is going to be nutty

Answer 48

split the denominator up into (s-2)(s+1)(s^2+1)

Answer 49

\[Y(s)=\frac{s^3+s^2-7s+3}{(s-2)(s+1)(s^2+1)}\]

Answer 50

next use partial fractions

\[\s^3+s^2-7s+3=\frac{A}{s-2}+\frac{B}{s+1}+\frac{Cs+D}{s^2+1}\]

Answer 51

using the cover up method woudl be the wisest way or by expanding

Answer 52

which method would you like to do

Answer 53

cover up method?

Answer 54

ill show you what i mean

\[\frac{s^3-s^2-7s+3}{(s-2)(s+1)(s^2+1)}=\frac{A}{s-2}+\frac{B}{s+1}+\frac{Cs+D}{s^2+1}\]

Answer 55

if you multiply both sides by s-2 you get

Answer 56

\[\frac{s^3-s^2-7s+3}{(s+1)(s^2+1}=A+\frac{B(s-2}{s+1}+\frac{Cs+D(s-2)}{s^2+1}\]

Answer 57

letting s=2 wll cancel out the last two terms getting A= something

Answer 58

s^3 - s^2 - 7s+3 = A(s+1)(s^2 + 1) + B(s-2)(s^2 +1) + ???

Answer 59

\[\frac{8-4-14+3}{(3)(5)}=A\]

Answer 60

what is your question?

Answer 61

if i did it my way, what would I have where C and D are supposed to be?

Answer 62

what way ar eyou doing?

Answer 63

are you expanding?

Answer 64

is that what it's called? then yes

Answer 65

it looks like you're expanding.. because you have a polynomial of power n, the numerator is always to the power n-1... so since it's a order 2 polynomial where n=2.. your polynomial on the top must be order 1 or 2-1=1

Answer 66

....sigh, sorry, then ezpanding is the way my teacher is doing it...

Answer 67

the standard form of a order one is As+B ... but A and B are already used therefore it is Cs+D

Answer 68

it doesn't matter . these are the two ways of doing it .. If you want to do expansion i can do that way also

Answer 69

well, if you can just tell me how to expand it so i can then cross multiply everything, I can to the simple math myself

Answer 70

oh crap, what about the initial values?

Answer 71

\[\frac{s^3+s^2-7s+3}{(s-2)(s+1)(s^2+1)}=\frac{A}{s-2}+\frac{B}{s+1}+\frac{Cs+D}{s^2+1}\]

Answer 72

\[s^3+s^2-7s+3=A(s+1)(s^2+1)+B(s-2)(s^2+1)+(Cs+D)(s-2)(s+1)\]

Answer 73

\[s^3+s^2-7s+3=A(s^3+s+s^2+1)+B(s^3+s-2s^2-2)+(Cs+D)(s^2-s-2)\]

Answer 74

\[s^3+s^2-7s+3=As^3+As^2+As+A+Bs^3-2Bs^2+Bs-2B+(Cs+D)(s^2-s-2)\]

Answer 75

wooooh

Answer 76

ah! can't see the rest

Answer 77

\[(Cs+D)(s^2-s-2)=Cs^3-Cs^2-2Cs+Ds^2-Ds-2D\]

Answer 78

and this is the expansion way?

Answer 79

\[As^3+As^2+As+A+Bs^3-2Bs^2+Bs-2B\]
\[+Cs^3-Cs^2-2Cs+Ds^2-Ds-2D=s^3+s^2-7s+3\]

Answer 80

now you have to group

Answer 81

\[s^3(A+B+C)+s^2(A-2B-C+D)+s(A+B-2C-D)\]
\[+1(A+B-2D)=s^3+s^2-7s+3\]

Answer 82

\[A+B+C=1\]
\[A-2B-C+D=1\]
\[A+B-2C-D=-7\]
\[A+B-2D=3\]

Answer 83

now you have to solve all these

Answer 84

that is why i'd say covermethod is the best way to go here

Answer 85

@lgbasallote how about this for DE LOL

Answer 86

so, would you hate me entirely and say no if I asked you to show the cover method?

Answer 87

....yuuuup! lol

Answer 88

i Showed you above to find A @iStutts