Laplace DE Problem: Proving Laplace Delta(t-2)

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SUMMARY

The discussion centers on proving that the Laplace transform of the Dirac delta function, specifically δ(t - 2), is e^{-2s}. Participants clarify that the Laplace transform of δ(t - a) is e^{-as}, and they emphasize the importance of understanding the defining property of the Dirac delta function. The proof involves applying the integral property of the delta function within the context of Laplace transforms. This foundational knowledge is crucial for students preparing for exams in differential equations and transforms.

PREREQUISITES
  • Understanding of Laplace transforms
  • Familiarity with the Dirac delta function
  • Knowledge of integral calculus
  • Basic differential equations
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  • Study the properties of the Dirac delta function in detail
  • Learn how to apply Laplace transforms to solve differential equations
  • Explore the proof of the Laplace transform of the delta function
  • Review examples of Laplace transforms involving unit step functions
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Students in mathematics or engineering disciplines, particularly those studying differential equations and Laplace transforms, will benefit from this discussion.

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Homework Statement



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Homework Equations



Laplace's transforms.

The Attempt at a Solution



Okay, so applying the Laplace on both sides yields:

s^2Y(s) - sy(0) - y'(0) + 2sY(s) - 2y(0) + 10Y(s) = ? + 13 / s - 1

Is e^{-2s} / s the Laplace \delta(t - 2)? Our instructor gave us the result of the Lapace but he did not prove it. The only thing he gave us was that the Laplace of f(t - a)\delta(t - a) = e^{-as}F(s). Can someone point me on how to prove this? It seems our instructor told us that we need to know the proof without him giving it to us. I know that this unit step function is defined to be 0 when 0 <= t < 2 and 1 when t >= 2.
 
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Are you sure ##\delta(t)## is the unit step function? It typically denotes the Dirac delta function.
 
Yes yes. I mixed up the function. Sorry about that.

So is the Laplace of \delta(t - 2) typically e^{-2s} / s?
 
Which function are you talking about? If it's the delta function, then no, that's not correct. If it's the unit step, then that's right.
 
The delta function. Then is it simply e^{-2s}?

If so, how to prove it starting with \int^\infty_0 e^{-st}\delta(t - a)dt = e^{as}? The proof is apparently required for our exam yet our instructor failed to provide the solution.
 
What's the defining property of the Dirac delta function?
 
..that the \int^{a+e}_{a-e} f(t)\delta(t - a)dt = f(a) for e &gt; 0?

I appreciate your patience but I'm relatively 'new' to this concept.
 
Yup. Just apply that to the Laplace transform integral you have.
 

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