2 Questions on inverse Laplace Transformation

[Reshma]

1] Find the inverse Laplace transform of the given function, i. e.
L^{-1}\left[1\over {s^3 + 1}}\right]

And this is how I proceeded about:
L^{-1}\left[1\over {s^3 + 1}}\right]

= L^{-1}\left[\frac{1}{(s+1)(s^2 - s +1)}\right] (By factorising the denominator)

I am stuck at this step and can't proceed furthur. Someone help.

2]Find the inverse Laplace Transform of the given function, i. e.
L^{-1}\left[9\over {s(s^2 + 9)}}\right]

I did solve this one, just want to know if its correct.
L^{-1}\left[9\over {s(s^2 + 9)}}\right]

= 3L^{-1}\left[3\over {s(s^2 + 9)}}\right]

We know:
L^{-1}\left[3\over {s^2 + 9}}\right] = \sin 3t = F(t)
L^{-1}\left[1\over s}\right] = 1 = G(t)

By the convolution theorem;
L^{-1}{f(s)} = F(t) * G(t) = \int_0^t G(\tau)F(t - \tau)d\tau

So,
L^{-1}\left[9\over {s(s^2 + 9)}}\right]
=3\int_0^t sin3(t - \tau)d\tau = (\cos 3\tau - 1)

Please care to see if this is right. Thanks in advance.

 

[HallsofIvy]

1] Find the inverse Laplace transform of the given function, i. e.
L^{-1}\left[1\over {s^3 + 1}}\right]

And this is how I proceeded about:
L^{-1}\left[1\over {s^3 + 1}}\right]

= L^{-1}\left[\frac{1}{(s+1)(s^2 - s +1)}\right] (By factorising the denominator)

I am stuck at this step and can't proceed furthur. Someone help.

Partial fractions, of course!
\frac{1}{(s+1)(s^2-s+1)}= \frac{A}{s+1}+ \frac{Bs+ C}{s^2- s+ 1}
s²- s+ 1 cannot be factored using real coefficients but you can complete the square: s²- s+ 1= (s- 1/2)²+ 3/4.

2]Find the inverse Laplace Transform of the given function, i. e.
L^{-1}\left[9\over {s(s^2 + 9)}}\right]

I did solve this one, just want to know if its correct.
L^{-1}\left[9\over {s(s^2 + 9)}}\right]

= 3L^{-1}\left[3\over {s(s^2 + 9)}}\right]

We know:
L^{-1}\left[3\over {s^2 + 9}}\right] = \sin 3t = F(t)
L^{-1}\left[1\over s}\right] = 1 = G(t)

By the convolution theorem;
L^{-1}{f(s)} = F(t) * G(t) = \int_0^t G(\tau)F(t - \tau)d\tau

So,
L^{-1}\left[9\over {s(s^2 + 9)}}\right]
=3\int_0^t sin3(t - \tau)d\tau = (\cos 3\tau - 1)

Please care to see if this is right. Thanks in advance.

Well, it is cos (3t)- 1, not cos(3\tau )-1!

Again, I would have done that by partial fractions:
\frac{9}{s(s^2+ 9)}= \frac{1}{s}- \frac{s}{s^2+ 9}[/itex]<br /> which then gives what you have.

 

[Reshma]

Cool, that makes sense. I will follow the technique you've suggested. Thank you so very much for your time.

 

[prakashlava]

hi

hi... reshma thanks a lot for ur updates of iit which is very usefull to me.ho w is ur preparation going.please tell me when is the last date if submitting the application form. which book book ru using for geology.eagerly waiting fror ur reply

from
prakash chennai

 

[Reshma]

hi... reshma thanks a lot for ur updates of iit which is very usefull to me.ho w is ur preparation going.please tell me when is the last date if submitting the application form. which book book ru using for geology.eagerly waiting fror ur reply

from
prakash chennai

Welcome to PF. Please do not hijack other threads to post your questions. Choose a thread relevant to your topic. Please read the Forum Guidelines: https://www.physicsforums.com/showthread.php?t=5374

Good luck!