Question

y'' + 16y = {(2t, 0<=t<=8) , (0, t => 8)}

and: y(0) = 0 , y'(0) = 0
find the Laplace transform.

the left side is equal to: Y(s^2 + 6) how can i construct the appropriate laplace transform for the right side?

Answer 1

You can construct the right side out of a modified unit-step (heaviside) functions

Answer 2

right but how? like u(t)-u(t-8) or something?

Answer 3

2t*h(t-8) I believe

Answer 4

does the second inequality not matter?

Answer 5

2t-2t*h(t-8)

Answer 6

how did you get that?

Answer 7

You want the function to be 2t until the heaviside "turns on" at 8. At that point the heaviside takes the value that is multiplied to it (2t) and it is subtracted from 2t giving you 0 after 8

Answer 8

so from 0 to 8, it goes to 2t. from 8 onwards it goes to 0. the heaviside equation jumps from 0 to 1 at 8?

Answer 9

Yes heaviside would normally take on the value 1 at t=8 onwards. But since were multiplying it by 2t it'll take on that value instead.

Answer 10

oh and because anything less than 0 is not defined it starts AT 2t

Answer 11

The heaviside starts at 0 but we have 2t until then by having the heaviside subtract it

Answer 12

Oh sorry yeah it "starts" at 8, before then is zero

Answer 13

what was the issue here?

Answer 14

your transform on the left is wrong as well

Answer 15

oh yea its Y(s^2 + 16) but my problem is I dont know how he arrived to that answer. Its a little confusing

Answer 16

to be clearer, i don't know the method to set up the unit step function.

Answer 17

@TuringTest

Answer 18

@jamej

Answer 19

you don't understand the left side either?

Answer 20

http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx
let's use this table
according to #36 we get\[s^2Y(s)-sy'(0)-y(0)-Y(s)=s^2Y-Y=Y(s^2-1)\]

Answer 21

now on the right side what do we know we need to start out with in the equation?

Answer 22

@babacusazn ^

Answer 23

the fact that the unit step function is 0 at t<c
and 1 at t>c

Answer 24

well the left side is simple, reduced it is Y(s^2 + 16) = ?

Answer 25

because the conditions make the s(Y) + y'(0) = 0

Answer 26

ok, but the right has an inconsistent definition already
you have f defined as both 0 and 2t at x=8
is it\[f=2t|t<8\]\[f=0|t\ge8\]?

Answer 27

exactly oh i made a type 2t, \[0 \le t <8\]

Answer 28

so that is what I meant above
so would you agree then that just looking at \(t<8\) we have \(2t\) on the right?

Answer 29

\[y''+16y=2t\]covers us up until \(x=8\), agreed?

Answer 30

correct because it t is not greater than 8, it is not zero

Answer 31

right y'' + 16' = 2t

Answer 32

but when it hits \(x=8\) we need an "off" switch
we do that by turning "on" the \(inverse\) of \(2t\) at \(x=8\)

i.e. we turn on \(-2t\) at \(x=8\)

Answer 33

x's and t's are all mixed up above :S
imagine all t's

Answer 34

yea no problem

Answer 35

how can we turn on the inverse of \(2t\) at \(t=8\) using a step function?

Answer 36

so -2t to turn it off at t = 8 or above that

Answer 37

right, and how do we write that then?

Answer 38

u(t-c)?

Answer 39

c being the point of "inflection" /jump

Answer 40

c=? here?

Answer 41

8 :)

Answer 42

hold on a second, doesn't that mean at c=8, u(t-8) = ehh 1? or

Answer 43

yes 8

Answer 44

and so the final expression is...?

Answer 45

because the unit step function is defined that at 0 t<c and 1 at t>c (or probably \[\ge\])

Answer 46

befor the laplace I mean

Answer 47

2t - 2t(u(t-8))

Answer 48

and from there its a simple look at the table and invert it

Answer 49

yup :)

Answer 50

I see, could I post a nother question after I have attempted it? you cleared up the unit step function so clearly thanks!

Answer 51

I'm goin' to bed bro, it's late for me
but ping me when you need help and I'll do my best :)
g'night

Answer 52

thanks alot! :)

Answer 53

anytime =)