Question

Find the Laplace Transform of the given function; a and b are real numbers and n is a positive integer.

Answer 1

\[\huge f(t)=t e^{at} \sin(bt)\]

Here's what I have..
\[\large \text{Let}~~g(t)=e^{at} \sin(bt)\]\[\large G(s)=\mathcal{L}[g(t)]=\frac{b}{(s-a)^2+b^2}\]\[\text{Then}~~F(s)=G'(s)\](which I haven't done yet.. just wanted to check if I'm on the right track here..)

Answer 2

@ganeshie8 X)

Answer 3

are you trying to use convolution ?
@dan815 is online

Answer 4

convolution may not be needed here, i think your method works nicely

Answer 5

I'm not sure.. I found the solution to a similar problem for \(t^2 \sin(bt)\) and they followed the same process using this property:
\[\mathcal{L}[t^n f(t)]=(-1)^n \cdot \frac{d}{ds^n}F(s)\]

Answer 6

which.. i forgot the -1

Answer 7

\[\mathcal{L}\{t e^{at} \sin(bt)\} = -\partial_s\mathcal{L}\{e^{at}\sin (bt)\}\]

Answer 8

Is that partial derivative with respect to s?

Answer 9

next you want to use
\[\mathcal{L}\{e^{at}f(t)\} = F(s-a)\]

right ?

Answer 10

yes it is w.r.t s

Answer 11

Just checkin! and yeah

Answer 12

Looks good to me!

Answer 13

@dan815 may show you another way to work this using convolution

Answer 14

@ganeshie8 Okay, cool! And just for future reference.. will there be a time when this property can't be used for problems that may *look* similar?

Answer 15

Also, my final answer is:

\[\huge F(s)=\frac{ 2b(s-a) }{ ((s-a)^2+b^2)^2 }\]

Answer 16

hey sorry there is some issue with notifications,

below property works always
\[\mathcal{L}\{e^{at}f(t)\} = F(s-a)\]


ofcourse laplace transform of f(t) need to exist for above property to make any sense

Answer 17

that looks good
http://www.wolframalpha.com/input/?i=laplace+transform+te%5E%28at%29sin%28bt%29

Answer 18

@ganeshie8 It's all good and sweet! X) Thanks