Question

smbdy help me ...laplace transform {t^4 + 2t^3-1}

Answer 1

so laplace transform of \[t^{n}=\frac{ n! }{ s^{n+1} }\]
you can use that formula

Answer 2

how to solve it

Answer 3

i am still early to understand it

Answer 4

ok so using that formula for t^4 we would have :

Answer 5

\[laplace(t^{4})=\frac{ 4! }{ t^{4+1} }\]

Answer 6

when you have laplace of sum of terms like here t^4 + 2t^3+1 because of the property of linearity of the laplace you calculate laplace of each term so you would have :

Answer 7

\[L({t^{4}+2t^{3}+1})=L(t^{4})+2L(t^{3})+L(1)\]
you can pull constants out of the laplace

Answer 8

here the question...i am still need more help to clear it..i

Answer 9

for which one u need help ?

Answer 10

first qstion..n the working step how to calculate it

Answer 11

ok do you have the laplace transforms table in front of you ?

Answer 12

yes

Answer 13

and i already told you the formula you need for laplace of the first one you just need to apply it ..

Answer 14

i am still blur..n my math fail

Answer 15

ok i will do the first one all the way to the result
\[L(t^{4}+2t^{3}+1)=L(t^{4})+2L(t^{3})+L(1)\]
you can pull out constants in front of the laplace like i did there with the 2 in front ot t^3
no using the formula :
\[L(t^{n})=\frac{ n! }{ s^{n+1} }\]
we have :
\[\frac{ 4! }{ s^{4+1} }+2\frac{ 3! }{ s^{3+1} }+\frac{ 0! }{ s^{0+1} }\]
so our result is :
\[\frac{ 24 }{ s^{5} }+\frac{ 12 }{ s^{4} }+\frac{ 1 }{ s }\]
that is our solution .
note that 0!=1

Answer 16

L(t4+2t3-1)

Answer 17

yeah that what i wrote above is the solution of that laplace transform ^^

Answer 18

how u get 24

Answer 19

4!=4*3*2=24

Answer 20

L(3e^2t + 2e^-4t)

Answer 21

\[L(e^{at}) = \frac{ 1 }{ s-a }\]

you should use tables; you'll be allowed one on your exams.

Answer 22

so answer is

\[\frac{ 3 }{ s-2 }+ \frac{ 2 }{ s+4 }\]

Answer 23

the answer only like that?

Answer 24

even if ur not alowed its just few formulas not hard to remember , as i can see most of ur laplace are just table transforms

Answer 25

lol i know :P. i know them all

Answer 26

my prof . dont alow anything except calculator :S

Answer 27

lol funny. we weren't allowed a calculator on our exam for this class

Answer 28

bigget "calculation" is factorials lol

Answer 29

L(cos8t-sin8t
)

Answer 30

for laplace u dont need calculator we had it bec we had numerical math also and some other stuff xD

jeff that is also table transform

Answer 31

try and use this table and let us know if you still need help:

http://mathworld.wolfram.com/LaplaceTransform.html

Answer 32

lol we had an entire class for just this :P

Answer 33

0.0 weird xD

Answer 34

ok

Answer 35

wait 4 me..
i have 10 more qstion to solve.after i did it.i will up n share it.if wrong please correct 4 me

Answer 36

if like this how to solve it (cos8t-sin8t)

Answer 37

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

Answer 38

must use 2 formul?

Answer 39

yeap one for sin(a) and other for cos(at)

Answer 40

sin(at)

Answer 41

after tht

Answer 42

after get the answer euler gve is it any working process to show?

Answer 43

that is the final answer unless you want to replace 8^2 with 64 u can do it but u cant do anything more

Answer 44

just like that?

Answer 45

he just used the formulas from the laplace table for cos(at) and sin(at) ..

Answer 46

(t4 e 3t)

Answer 47

i am sorry buddy i totaly have to go , maybe euler can help you with the rest ^^ , good luck ;D

Answer 48

ok tq

Answer 49

let's denote the L[f(t)] as F(s)

when you have e^(at)*f(t)

the L[e^(at)*f(t)] = F(s-a)

so you do the laplapce xform of f(t) then plug (s-a) for every s in F(s)

\[L[t^4] = \frac{ 24 }{ s^5 }\]
\[L[e^{3t}t^4] = \frac{ 24 }{ (s-3)^5 }\]