Laplace of t*x''(t) Pulling the derivative through the integral

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SUMMARY

The discussion focuses on solving the differential equation \( t x''(t) + x'(t) + t x(t) \) using the Laplace transform. The key point is the ability to pull the derivative through the integral in the Laplace transform, specifically for the term \( L[tx''(t)] \). This is valid under the condition that the function \( x(t) \) is "well-behaved," meaning it must be sufficiently smooth and ensure the integral \( \int_0^\infty (x'' e^{-st})\,dt \) exists and is differentiable with respect to \( s \).

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  • Familiarity with the concept of "well-behaved" functions in analysis
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Saladsamurai
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Homework Statement



I am working on solving the DE: t*x" + x' +t*x
by use of the Laplace transform. Now if we just look at the first term and its transform, we have

[tex]L[tx''(t)] = \int_0^\infty tx''e^{-st}\,dt = -\int_0^\infty \frac{d}{ds}\left(x''e^{-st}\right)\,dt[/tex]


Now in the next step of my text, they say that we can pull the derivative through the integral if we assume that the unknown x(t) is "well-behaved enough." Can someone please explain to my feeble mind what that means and how that allows us to reverse the limit processes? I would like to be able to understand it well enough that I can apply it to future problems.

Thanks! :smile:
 
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Someone that knows more analysis might be able to give a better answer, but I think that it's usually enough that [tex]x(t)[/tex] is sufficiently well-behaved that the integral

[tex]\int_0^\infty \left(x''e^{-st}\right)\,dt[/tex]

exists and is a differentiable function of s.
 

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