Question

Use the Laplace transform to solve y^(4)-4y'''+6y"-4y'+y=0; y(0)=0, y'(0)=1, y"(0)=0, y'''(0)=1. ( I got L(y^(4))-4L(y''')+6L(y'')-4L(y')+L(y)=L(0) but I don't know what's L(y^(4)) and L(y'''), please help.)

Answer 1

\(\huge \color {green }{L(y^n)=s^nL(y)-\sum \limits_{k=1}^ns^{k-1}y^{n-k}(0)}\)

Answer 2

L (y') = sL(y) -y(0)
L(y'')= s^2 L(y) -s y(0) -y'(0)
L(y''') = s^3 L(y) -s^2 y(0) -s y'(0) - y''(0)
L(y^4) = s^4 L(y) -s^3 y(0) -s^2 y'(0)-s y''(0)-y'''(0)

Answer 3

phew....

Answer 4

Wait a minute, please don't leave.

Answer 5

Thanks for being patient, please wait a little more. I'm almost done.

Answer 6

ok, do you need me to verify your answer ?
because i wasn't solving in the mean time :P

Answer 7

It's alright, let me finish the problem.

Answer 8

Okay, so I got (s^2+4s-5)/(s^4-4s^3+6s^2-4s+1), what to do next?

Answer 9

try to factorize
s^4-4s^3+6s^2-4s+1
if you remember pascal triangle, you will be able to do it in a blink of an eye

Answer 10

But how to do the pascal triangle for this one?

Answer 11

the co-efficients are 1,-4,6,-4,1
which would make that \( \large (s+(-1))^4=(s-1)^4 \)

Answer 12

Could you do synthetic division?

Answer 13

after guessing one root as +1, yes.
finally you'll get (x-1)^4

Answer 14

Wait like 3 minutes, I'll be back.

Answer 15

So you get (s+5)/(s-1)^3, right?

Answer 16

yup, correct :)
know what to do next ?

Answer 17

To take the laplace transform? I need to look at the table, right?

Answer 18

yeah, but you need to apply one property first...

Answer 19

What is it?

Answer 20

ever seen something like this?
\(L^{-1}(F(s+a))=e^{-at}L^{-1}F(s)\)
?

Answer 21

No. But how does that apply to the problem?

Answer 22

you have s-1 in the denominator, so
e^x will come out
\(\large e^x L^{-1}[\dfrac{(s+1)+5}{s^3}]\)

Answer 23

Sorry, I don't understand what you wrote.

Answer 24

then how were you thinking og getting inverse laplace transform of
(s+5)/(s-1)^3
?

Answer 25

I have no idea.

Answer 26

thats why i gave you the property, which needs to be used here....
\(L^{-1}(F(s+a))=e^{-at}L^{-1}F(s)\)

Answer 27

to make , s+a as just s, we have to multiply e^-at
or, s-a would require e^at

here, to have just s^3 instead of (s-1)^3 , we will multiply e^x (e^(+1x)) to inverse laplace after replacing s by s+1

Answer 28

I got it. Thanks for your time.

Answer 29

oh, good!
welcome ^_^

Answer 30

You're right. Thanks, my friend. Can you help me with other question? http://openstudy.com/study#/updates/52056454e4b01dab33705ac7

Answer 31

@hartnn Please , explain more about them. I would like to know.
\[L^{-1}(F(s+a))=e^{-at}L^{-1}F(s)\]\(\rightarrow\)\[\large e^x L^{-1}[\dfrac{(s+1)+5}{s^3}]\]