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- Thread starter gurtaj
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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].

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Ray Vickson

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If '*' means multiplication, then your answer is correct (and Maple confirms this). If '*' means 'convolution' then you answer is wrong.

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Ray Vickson

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I agree with you; if F(s) means the product as given in the 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.

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vela

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I don't get what you did there.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

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?

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I'm confused here... are u saying i should put e^t inside integral and do L{e*t^2}?

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vela

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$$f(t) = \int_0^t \tau^2 e^{t-\tau}\,d\tau.$$ You should recognize the form of that integral.

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oh wow , i get it now. Thanks you so much for helping.

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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.

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