Question

x'' + 4x = 0 ; x(0) = 5, x'(0) = 0 . use laplace to solve ivp

Answer 1

did u try this ?
what u got ?

Answer 2

i got something with -5s and 4/s^2 and its 4cos2t or something

Answer 3

answer is 5cos2t haha but im having a hard time undertstanding the book

Answer 4

L[x''] = ... ?

Answer 5

so L(x'' + 4x) = L(0) ?

Answer 6

if i can just figure out how to do this one then i can figure out the rest

Answer 7

ok, L[x''] = s^2 F(x)-s x(0)-x'(0)

Answer 8

how do we know this? and is it s^2*F(x) - s *x() - x'(0) ?

Answer 9

and what is F(x) haha

Answer 10

thats general formula
and F(x) .... i had to write X(s) is the L.T of x.

Answer 11

okay so its L[x''] = s^2*X(s) - s*x(0) - x'(0) ?

Answer 12

yup, and you are given x'(0)

Answer 13

and x(0)

Answer 14

im so lost still

Answer 15

dont you have to find the L of 4x too?

Answer 16

yes, ofcourse.

Answer 17

it would be 4 X(s)

Answer 18

x'' + 4x = 0
taking L.T
s^2*X(s) - s*x(0) - x'(0) + 4 X(s) =0

Answer 19

okay so L = s^2 *X(s of x) - sx(0) - x'(0) + 4X(s of x) ?

Answer 20

yeah, now put the given initial conditions.

Answer 21

what is X(s) for the 4X(s) ? is it still x?

Answer 22

how do they get 5cos2t haha

Answer 23

the Lt of x is 1/s^2, right?

Answer 24

isolate X(s) from that equation

Answer 25

then take inverse L.T

Answer 26

okay right now i have s^2X(s) - sx(0) - x'(0) + 4X(s) = 0

Answer 27

put the initial values.

Answer 28

okay so s^2X(s) - s(5) - 0 + 4X(s) = 0

Answer 29

yes, now isolate X(s)

Answer 30

s^2X(s) + 4X(s) = 5s

Answer 31

= 5cos2t omg i got it! well i dont understand it but i got an answer by struggling my way through it , thank you for all your help :)

Answer 32

what did u not understand ?

Answer 33

the very first equation, where you got that

Answer 34

and how to apply it

Answer 35

thats a standard equation...like a formula.

Answer 36

oh okay that makes sense

Answer 37

L[x'] = sX(s) - x(0)

Answer 38

okay and what is X(s) again?

Answer 39

oh its the laplace, right?

Answer 40

X(s) is the L.T of x.

Answer 41

yes.

Answer 42

L[x'] = sX(s) - x(0)
L[x''] = s^2X(s) - s x(0)- x'(0)
L[x'''] = s^3X(s) -s^2 x(0) - sx'(0) -x''(0)
and so on...

Answer 43

OH thank you so much!!!!!

Answer 44

welcome ^_^