# Please check this inverse this Laplace transform [closed]

I just want to check if my exercise are right:

Inverse of these Laplace transform

$$F^{-1}\left(\frac{1}{p-2}\right)= e^{2s}$$

$$F^{-1}\left(\frac{e^{-2p}}{p^2}\right)=\frac{2}{s^3(s+2)}$$

$$F^{-1}\left(\frac{1}{p^2+4p+5}\right)= e^{-2s}\cos(s)$$

$$F^{-1}\left(\frac{p}{(p+1)^2}\right)=\frac{e^{-s}}{s^4}$$

$$F^{-1}(e^{-p})=\frac{1}{s+1}$$

Can you just confirm that this results are right and if not please tell me the right solution and how to get there?? Thanks!!!

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Questions portrayed like this, without context, need to be expanded upon. What exactly are $p$ and $s$? I know this is for ILT, but I have seen places where both $p$ and $s$ are the LT independent variable (e.g., a complex frequency). Please provide context, as well as what you have done in obtaining your answers. –  Ron Gordon Apr 28 at 17:48
Do not delete the body of the question again. –  Pedro Tamaroff Apr 30 at 19:43
And do not attempt to make the answer empty. The answer belongs to the answerer. It is not within your rights to attempt to ruin it. –  Jyrki Lahtonen Apr 30 at 19:49
Rolled back to the version with suggested solutions since the posted answer refers to these. –  Did Apr 30 at 21:29
In my opinion, this question should not remain closed as it has already been answered. –  Alexander Gruber May 1 at 10:06
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## closed as not a real question by Steve D, Davide Giraudo, ncmathsadist, Lord_Farin, Ron GordonApr 29 at 7:43

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

## 1 Answer

You can verify your result from the following:

For laplace transform: http://www.wolframalpha.com/input/?i=laplace+transform

For laplace inverse: http://www.wolframalpha.com/input/?i=laplace+inverse

$1)$ $F^{-1}\left(\frac{1}{p-2}\right)= e^{2s}$ is correct

$2)$ $F^{-1}\left(\frac{e^{-2p}}{p^2}\right)=\frac{2}{s^3(s+2)}$ is incorrect

Correction: $F^{-1}\left(\frac{e^{-2p}}{p^2}\right)=\left({s-2}\right)u\left({s-2}\right)$ by time-shifting.

$3)$ $F^{-1}\left(\frac{1}{p^2+4p+5}\right)= e^{-2s}\cos(s)$ is incorrect

Correction: $F^{-1}\left(\frac{1}{p^2+4p+5}\right)= e^{-2s}\sin(s)$

$4)$ $F^{-1}\left(\frac{p}{(p+1)^2}\right)=\frac{e^{-s}}{s^4}$ is incorrect

Correction: $F^{-1}\left(\frac{p}{(p+1)^2}\right)=e^{-s}(1-s)$ by frequency-shifting.

$5)$ $F^{-1}(e^{-p})=\frac{1}{s+1}$ is incorrect

Correction: $F^{-1}(e^{-p})=u\left({s-1}\right)$ by time-shifting.

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Thanks! I just bought the app! Seams good –  Amccds Apr 28 at 17:30
Sorry but 3 is correct. –  Amccds Apr 29 at 9:05
Corrected! Though 3 is incorrect! –  Maazul Apr 29 at 14:15
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