Question

***Differential Equations***
Im trying to do the Laplace of tcos(t). In order to avoid the integral I changed to exponential. .5t[e^(t) + e^(-t)]= .5 Laplace { te^(t) + te^(-t) } which should euqal .5 [1/(s+1)^2 + 1/(s-1)^2 ]. Why is this wrong? correct answer is (s^2)-1/(1+s^2)^2

Answer 1

\[\large \cos(t) = \dfrac{e^{it}+e^{-it}}{2}\]

Answer 2

looks you're using the wrong euler formula

Answer 3

use this if u need to get it done fast
\[L[t^n.f(t)]=(-1)^n.\frac{ d^n }{ ds^n }F(s)\]

Answer 4

I used the same formula except I didnt include the i. How does that change the answer?

Answer 5

\[here ~n=1\\L[t.cost]=(-1)^1.\frac{ d^1 }{ ds^1 }\frac{ s }{ s^2+1 }=-1.\frac{ d }{ ds }(\frac{ s }{ s^2+1 })\]
easy diffrentiation now :)

Answer 6

u cant use this if instructed to do by definition

Answer 7

I see what you mean Sid. Thanks. But do you know where I went wrong in my logic?

Answer 8

yes as ganeshie8 said your identity for cost is wrong

Answer 9

try simplifying this

.5 [1/(s+i)^2 + 1/(s-i)^2 ]

Answer 10

.5* 1/(s^2+1)^2

Answer 11

as you might be knowing the numerator isnt right, try again

Answer 12

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

Answer 13

http://www.wolframalpha.com/input/?i=alternate+forms+%28.5+%281%2F%28s%2Bi%29%5E2+%2B+1%2F%28s-i%29%5E2%29%29

Answer 14

yea my bad ganeshie, and Sid yea I thought the e^at would just cause a shift in the t by one so thats the answer I got

Answer 15

yes same result\[L[t.cost]--1.\frac{ d }{ ds }[\frac{ s }{ s^2+1 }]=-\frac{ (s^2+1)-2s^2 }{ (s^2+1)^2 }=\frac{ s^2-1 }{ (s^2+1)^2 }\]

Answer 16

so te^at would be 1/s^2 shifted by 1

Answer 17

you're right, and "a" could be any number... complex/real

Answer 18

\[\mathcal{L}\left(te^{at}\right) = \dfrac{1}{(s-a)^2}\]

Answer 19

\[\mathcal{L}\left(te^{it}\right) = \dfrac{1}{(s-i)^2}\]

Answer 20

oh I see, k thanks for the help