Question

laplace transform

Answer 1

kind of messy because I've made a mistake. note to self....no matrices. @UnkleRhaukus is convolution required?

Answer 2

there is an error on your second line ):

Answer 3

\[\mathcal L\big\{y''(t)\big\}=s^2Y(s)-sy(0)-y'(0)\]

Answer 4

ffffffffffffffffffffffffffffffffffffffff great...I have to start over...wait a sec...yeh *sticks finger* probably because I did it while I was waiting for dinner.

Answer 5

also the laplace transform of \(x(t)\) is not 1/s
it is \(X(s)\)

Answer 6

ugh this is shi........

Answer 7

you know what my book is saying that

Answer 8

oh wait , y is y(x) , so

laplace of {x} = 1/s^2

Answer 9

( i thought maybe y(t), x(t) )

Answer 10

what do you have for second line now?

Answer 11

shoots hold up need to redo

Answer 12

that is good , keep going

Answer 13

now divide by (s^2+1), keep four terms, (dont merge the fractions)

Answer 14

O_O oh narfklsdjlafdks;j I've done long divison doh!

Answer 15

did you get something nice?

Answer 16

no... but I've divided with the s^2+1 and got the bottom exponential to be greater than the top

Answer 17

i'm not sure why you have done that

Answer 18

y?

Answer 19

\[(s^2+1)Y(s)=\frac1{s^2}+\frac1{s-1}+1+2s\\
Y(s)=\frac1{s^2(s^2+1)}+\frac1{(s-1)(s^2+1)}+\frac1{s^2+1}+2\frac s{s^2+1 }\]

Answer 20

now take the laplace transform of of both sides, term by term

\[y(x)=\mathcal L^{-1}\Big\{\frac1{s^2(s^2+1)}\Big\}+\mathcal L^{-1}\Big\{\frac1{(s-1)(s^2+1)}\Big\}+\mathcal L^{-1}\Big\{\frac1{s^2+1}\Big\}+2\mathcal L^{-1}\Big\{\frac s{s^2+1 }\Big\}\]

Answer 21

the last two terms are on the table,

Answer 22

ugh so I'm going through partial fractions for jacks

Answer 23

practice is always good ,

you might need partial fraction or convolution for the first two terms

Answer 24

foo foo dude my prof didn't even look at the book. He just assigned random jizz and he didn't teach convolution

Answer 25

I already told him what happened to that previous problem with the sin3x on the right hand side. he removed it

Answer 26

careful the second blue term is 1/[(s-1)(s^2+1)] right?

Answer 27

yes

Answer 28

oh ( '-1' looks a bit like a '+' , carry on

Answer 29

Do you have to use Laplace transforms for this? I think it's much easier way to solve this.

Answer 30

yeah I have to use stupid Laplace...if it were up to me, I would use variation of parameters to end my suffering

Answer 31

\[y(x)=\mathcal L^{-1}\Big\{\frac1{s^2(s^2+1)}\Big\}+\mathcal L^{-1}\Big\{\frac1{(s-1)(s^2+1)}\Big\}+\sin x+2\cos x\]

Answer 32

Haha yep, that's what I just got through doing to make sure. At least you have a way to check yourself... lol

Answer 33

variation of parameters is awesome, but Laplace is f**********ing B**********hit. YOu know what's b*****hit? You have to use a specific method just to get the answer, Why couldn't I just used something that I already learned in class? By the time you realized that there's an alternative, you'll already handed in your homework. Solving with Laplace is B***********hit

Answer 34

There are some fun linear algebra ways of solving differential equations too, I don't know if we'd be able to do that here or not. I think you might need at least a 3x3 matrix to solve this one...

Answer 35

Wait, maybe if you consider taking the second derivative as all 1 operation, you could make this a 2x2... Ok I'm gonna play around with this, not that it helps you even slightly haha.

Answer 36

hate this method...

Answer 37

Actually, I guessed a really simple answer to the differential equation,
\[y=x+\frac{ 1 }{ 2 }e^x\]

Unfortunately I can't seem to change it to get it to meet the initial conditions, anyone have ideas on how I might change this?

Answer 38

I'm getting close. I got a sign error somewhere....otherwise I got the whole thing

Answer 39

\[\mathcal L^{-1}\Big\{\frac1{s^2(s^2+1)}\Big\}=\mathcal L^{-1}\Big\{\frac{1}{s^2}-\frac{1}{s^2+1}\Big\}=\mathcal L^{-1}\Big\{\frac{1}{s^2}\Big\}-\mathcal L^{-1}\Big\{\frac{1}{s^2+1}\Big\}\]

Answer 40

I'm supposed to have -3/2cosx

Answer 41

oops sorry I meant 3/2cosx

Answer 42

supposed to have that which means a sign error is lurking. otherwise it's done

Answer 43

wait found it ...

Answer 44

yay boy that took forever

Answer 45

pffft 4 down 7 to go T_T

Answer 46

(:

Answer 47

you know that problem with the sin3x at the end? I convinced my professor to remove it XD

Answer 48

so instead of 12 problems, there's 11 XD

Answer 49

sounds like your professor isn't that great ,
how can one understand laplace without the convolution theorem?

Answer 50

he's from another country D: and using a ghetto book D:

Answer 51

kahn academy is alright for laplace

https://www.khanacademy.org/math/differential-equations/laplace-transform

Answer 52

what's the first derivative for laplace?

Answer 53

y(s)s-y(0)?

Answer 54

yep

Answer 55

k doing #7 now... I want to get this out of the way so It doesn't ruin my holiday

Answer 56

L{y'(t)} = sY(s)-y(0),

L{y''(t)} = s^2Y(s)-sy(0)-y'(0),

Answer 57

it is much easier to take the inverse laplace of a sum of fractions, (rather than the inverse laplace of one crazy fraction)

Answer 58

damn! I got the exponentials, but what happened here? sign issues?

Answer 59

2 is supposed to be positive, not negative ahhhhh T_T, and a sinx should be there after reverse laplace x.x

Answer 60

sigh some of these laplace problems are easy...others not so much. I got through 8 and 9 with ease

Answer 61

L^-1 {-2/s}= ?

Answer 62

that's -2 unless I'm missing a property which makes it positive

Answer 63

1/s is 1

Answer 64

oh yeah

Answer 65

I got the 2 but it's supposed to be positive

Answer 66

sign issues -__- too many terms on that one

Answer 67

your answer is correct

Answer 68

i bet the book is wrong again

Answer 69

O_O how? really:? ?!?!?!

Answer 70

http://www.wolframalpha.com/input/?i=y%27%27%27-y%27%3D2sinx%2C+y%280%29%3D0%2C+y%27%280%29%3D0%2C+y%27%27%280%29%3D0

Answer 71

the book has it at 2-sinx+2e^x-e^-x. I don't see how that's possible

Answer 72

woowwwwwww!!!!

Answer 73

your ancient text book is wrong wrong wrong

Answer 74

the book it is le coo coo ...

Answer 75

tell that to my prof XD. He swears on the thing

Answer 76

man I should e-mail what happened...maybe he'll remove that problem too XD

Answer 77

he did confess that he just assigned the problems and didn't even read what it was about

Answer 78

haha, i'm so glad i'm not in that class,

Answer 79

I'm gonna try conquer 11 and 12 and see what's happening XD. Maybe it will end up like 10.

Answer 80

uh oh #11 has a complex root xxx

Answer 81

unfortunately i can not find an errata for your text

(when i buy a text book i like to get the errata too, and ill go through the text and fix all the typographic mistakes, before trying to read it)