# 2 Questions on inverse Laplace Transformation

1. Nov 26, 2006

### 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.

Last edited: Nov 26, 2006
2. Nov 26, 2006

### HallsofIvy

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.

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

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

3. Nov 26, 2006

### Reshma

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

4. Dec 1, 2006

### prakashlava

hi

hi.... reshma thanks a lot for ur updates of iit which is very usefull to me.ho w is ur preparation going.plz 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

5. Dec 2, 2006