Question

Pls help:)

Answer 1

with what?

Answer 2

need help in laplace transformations

Answer 3

which one?

Answer 4

all if possible.

Answer 5

for the first one
use this property\[\Large L[e^{at}]=\frac{ 1 }{ s-a }\]
so firstly considerinf only e^(2t)\[\Large L[e^{2t}]=\frac{ 1 }{ s-2 }\]
then use this
\[L[t^nf(t)]=(-1)^n \frac{ d^n }{ ds^n }[F(s)]\]

Answer 6

so
multipliocation with t means derivative only

Answer 7

then
\[\Large L[t^2*e^{2t}]=(-1)^2\frac{ d^2 }{ ds^2 }[\frac{ 1 }{s-2}]\]

Answer 8

if there is a t^3
then u have to differentiate it 3 times
getting this?

Answer 9

and if there is a (1/t) then integration

Answer 10

ya. so my final answer is gonna be 2(s-2)^(-3). am i right?

Answer 11

yes :)

Answer 12

thnx:) but isnt it this for L(t^n)=n!/(s^(n+1)),s>0.

Answer 13

yes this is true for only t like\[L[t]=1/s^2\\L[t^2]=2/s^3\]
but when t is multiplied with any other functions then u have to use the above property
take another examplt like
\[L[t*\sin(at)] \\so ~for ~this ~one~firstly\\L[\sin(at)]=\frac{ a }{ a^2+s^2}\\then\\for ~the ~t\\L[t*\sin(at)]=(-1)^1\frac{ d }{ ds}[\frac{ a }{ a^2+s^2}]\]

Answer 14

oh k. thnx.:)

Answer 15

another thing u should know that u can have a problem like this also--\[L[\frac{ \sin(2t) }{ t }]=?\]then u should use this
\[L[f(t)/t]=\int\limits_{s}^{\infty}F(s)ds\]so it will become\[L[\sin2t/t]=\int\limits_{s}^{\infty}\frac{ 2 }{ s^2+4 }ds\]

Answer 16

oh k

Answer 17

^_^

Answer 18

for ur second problem you can use-\[\sin3t=3sint-4\sin^3t\\\\sin^3t=\frac{ 3sint-\sin3t }{ 4 }\]
now u can do it ;)

Answer 19

but it is sin^3 3t right?

Answer 20

sin^3 3t=(3sin3t-sin9t)/4 isnt it

Answer 21

yup :)

Answer 22

ok:)

Answer 23

again basic trigs for the 3rd one

Answer 24

\[\sin^2tcost=\frac{ 1-\cos2t }{ 2 }*cost\\\\=cost/2-\frac{ 2\cos(2t)*cost }{ 4 }\]

Answer 25

is the final answer for 2nd 1 this
\[\frac{ 1 }{4 }[\frac{ 9 }{ s^2+9 }-\frac{ 9 }{s^2+81}]\]

Answer 26

now use
\[2cosa*cosb=\cos(a+b)+\cos(a-b)\]

Answer 27

yeah good thats correct

Answer 28

final answer
\[\frac{ 1 }{ 4 }[\frac{ s }{ s^2+1 }-\frac{ s }{ s^2+9}]\]
am i right?

Answer 29

so it will be
\[=\frac{ cost }{2 }-\frac{ 1 }{ 4 }*[\cos(3t)+\cos(t)]\]
now its easy

Answer 30

ok??

Answer 31

ya

Answer 32

is my final answer for 3rd crct?

Answer 33

u mean (2)-(ii) ??

Answer 34

ya

Answer 35

yeah the first part is correct but dont forget the\[t^{5/2} ~term\]

Answer 36

use
\[\Large L[t^n]=\frac{ n! }{ s^{n+1}}\]

Answer 37

ya. totally forgot it. so it will be
\[\frac{ 1 }{ 4 }[\frac{ s }{ s^2+1 }-\frac{ s }{s^2+9 }]+\frac{ \frac{ 5 }{ 2} !}{ s^\frac{ 7 }{ 2}}\]

Answer 38

lol yeah :)

Answer 39

thnx:)

Answer 40

for next one use
\[\sinh(at)=\frac{ e^{at}-e^{-at} }{ 2} \\sinh^2(at)= \frac{ [e^{at}-e^{-at}]^2 }{ 4 }=\frac{ e^{2at}-2+e^{-2at} }{ 4 }\]
ok!!

Answer 41

ya k

Answer 42

and this time dont forget the 5t^2
lol!! ;)

Answer 43

lol k:)

Answer 44

what is L(a)=?

Answer 45

in this case a=3 because it is sinh^2(3t) k?

Answer 46

ya. what i was trying to ask is L(2)=?

Answer 47

L(2)=2*L(1)=2/s

Answer 48

oh k. thnx:)

Answer 49

so my final answer is
\[\frac{ 1 }{ 4}[\frac{ 1 }{ s-6 }-\frac{ 2 }{ s }+\frac{ 1 }{ s+6}]+\frac{ 10 }{ s^3 }\]. right?

Answer 50

yup awesome ;)

Answer 51

thnx:)

Answer 52

can u do the next one ?

Answer 53

let me try

Answer 54

ok :)

Answer 55

tried it. but finding it a bit hard.

Answer 56

ok
\[L[\frac{ e^{-at}*t^{n-1} }{ (n-1)! }]\\first ~taking~(n-1)! ~out\\ \frac{ 1 }{ (n-1)! }L[e^{-at}*t^{n-1}]\]
now we know that
\[L[t^n]=\frac{ n! }{ s^{n+1} }\\now ~just ~replace~n ~by~(n-1)\\L[t^{n-1}]=\frac{ (n-1)! }{ s^n }\]
ok so far?

Answer 57

K? having any prob :P

Answer 58

ya k

Answer 59

now shifting property means if u multiply e^at then s will become (s-a) and if its e^(-at) then (s+a)
so
\[L[e^{-at}t^{n-1}]=\frac{ (n-1)! }{( s+a)^n }\\\]

Answer 60

look here it is e^(-at) so s^n became (s+a)^n

Answer 61

so finally it is
\[\frac{ 1 }{ (n-1)! }*\frac{ (n-1)! }{ (s+a)^n }=\frac{ 1 }{ (s+a)^n }\]
got this ur highness Aj lol :)

Answer 62

lol ya i get it

Answer 63

i missed most of the lectures. nw only all of these makes sense. thnk u so much:)

Answer 64

happy to help :)

Answer 65

do u mind helpng me with the last two.

Answer 66

yeah can u plot those functions

Answer 67

if u r busy i am sry for taking ur time

Answer 68

nope i'm okay lets try the next one

Answer 69

oh k thnx.:)

Answer 70

all right consider the first part
\[f(t)=t;~~0<t<\pi\]
this will look like

okay?