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:
[tex]\frac{9}{s(s^2+ 9)}= \frac{1}{s}- \frac{s}{s^2+ 9}[/itex]
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

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