Laplace Transform

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  • #1
gurtaj
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I'm having trouble with this question. Can anyone please guide me.

My Attempt :

= 1/(s-1) * L{t^2*e^-t}
= 1/(s-1) * (2/(s+1)^3)
= 2/((s-1)(s+1)^3))

but that's not the answer , its 2/((s-1) s^3) somehow.
The question is attached below.
 

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Answers and Replies

  • #2
Ray Vickson
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I'm having trouble with this question. Can anyone please guide me.

My Attempt :

= 1/(s-1) * L{t^2*e^-t}
= 1/(s-1) * (2/(s+1)^3)
= 2/((s-1)(s+1)^3))

but that's not the answer , its 2/((s-1) s^3) somehow

You are correct: it really is 2/[(s-1)(s+1)^3].
 
  • #3
gurtaj
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But on the test it says its 2/(s^3 *(s-1)), and wolfram alpha confirms it too... The question asked to find F(2).. according to my solution i will get 2/(3^3) which is 2/27 but the answer is 1/4
 
  • #4
Ray Vickson
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But on the test it says its 2/(s^3 *(s-1)), and wolfram alpha confirms it too... The question asked to find F(2).. according to my solution i will get 2/(3^3) which is 2/27 but the answer is 1/4

If '*' means multiplication, then your answer is correct (and Maple confirms this). If '*' means 'convolution' then you answer is wrong.
 
  • #5
gurtaj
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In my answer which is 2/((s-1)*(s+1)^3)) , '*' does mean multiplication, so F(2) should be 2/27 , is that correct? ... test got 1/4 however.
 
  • #6
Ray Vickson
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In my answer which is 2/((s-1)*(s+1)^3)) , '*' does mean multiplication, so F(2) should be 2/27 , is that correct? ... test got 1/4 however.

I agree with you; if F(s) means the product as given in the question you wrote, then F(2) = 2/27, NOT 1/4. Maybe you wrote the wrong question?

I can see no way to get the answer 2/[(s-1)s^3] using any familiar kinds of operations. A product does not give this, and a convolution also does not give this. You say that Wolfram Alpha gives this; well, I don't believe it. Maple does not give this, and it is essentially equivalent to Mathematica in strength and applicability, so is at least as powerful as Wolfram Alpha (which uses a subset of Mathematica routines). If you say Wolfram Alpha gets this you will need to present exactly what commands you used; perhaps you were solving a different problem without realizing it.
 
  • #7
gurtaj
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This is the exact question... no option of 2/27 :(
 

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  • #8
vela
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I'm having trouble with this question. Can anyone please guide me.

My Attempt :

= 1/(s-1) * L{t^2*e^-t}
= 1/(s-1) * (2/(s+1)^3)
= 2/((s-1)(s+1)^3))

but that's not the answer , its 2/((s-1) s^3) somehow
I don't get what you did there.

Pull the exponential in front back inside the integral and then identify the integral as a convolution of two functions. What two functions are they?
 
  • #9
gurtaj
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I'm confused here... are u saying i should put e^t inside integral and do L{e*t^2}?
 
  • #10
vela
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No, how did you get that? You do realize the integral isn't a Laplace transform, right? When you pull the exponential back in, you get
$$f(t) = \int_0^t \tau^2 e^{t-\tau}\,d\tau.$$ You should recognize the form of that integral.
 
  • #11
gurtaj
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oh wow , i get it now. Thanks you so much for helping.
 
  • #12
Ray Vickson
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oh wow , i get it now. Thanks you so much for helping.

You still did not answer my question: you said "Wolfram Alpha gets this too...", and I essentially said I did not believe that claim; I asked you to back it up, by supplying the actual commands you gave to Wolfram Alpha. Please do this; we have tried to help you, now you can try to help us.
 

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