Laplace transform: function defined by parts

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SUMMARY

The discussion focuses on the Laplace transform of a piecewise function defined as \(f(t) = \left\{ t, t \in [0,2); t + 1, t \in [2,4); 0, t \ge 4 \right\}\). The user initially misapplied the translation theorem, particularly in shifting the Heaviside function \(H(t)\). The correct formulation is \(f(t) = tH(t) + H(t-2) - (t+1)H(t-4)\), which ensures the function vanishes for \(t \ge 4\).

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  • Understanding of Laplace transforms
  • Familiarity with piecewise functions
  • Knowledge of the Heaviside step function \(H(t)\)
  • Basic differential equations concepts
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  • Study the properties of the Laplace transform, particularly the translation theorem
  • Learn about piecewise function transformations in Laplace analysis
  • Explore the application of the Heaviside function in differential equations
  • Review examples of solving differential equations with discontinuous forcing functions
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Students and professionals in mathematics, particularly those studying differential equations and Laplace transforms, as well as engineers applying these concepts in system analysis.

libelec
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I have this DE:

[tex]\[y'' - 4y' + 8y = f(t) = \left\{ \begin{array}{l}<br /> t{\rm{ }},t \in [0,2) \\ <br /> t + 1{\rm{ }},t \in [2,4) \\ <br /> 0{\rm{ }},t \ge 4 \\ <br /> \end{array} \right.\][/tex]

I have problems transforming f(t). I know that when I have a function defined by parts, I have to use the translation theorem. But for t > 4, f(t) = 0 would let me something like [tex]\[f(t) = t.H(t) + 1.H(t + 2) + 0.H(t + 4)\][/tex], the 3rd term being irrelevant.

What am I doing wrong?
 
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First of all, you did your shifting wrong.
You want to shift H(t) to the right, so you substract the shifting amount from the argument.
Second of all, the third term must be relevant, because you need it to cancel out the two first term, so the function will vanish.
So your function looks actually like this:
[tex]f(t)=tH(t)+H(t-2)-(t+1)H(t-4)[/tex]
 
Damn, you're right. I made some stupid mistakes there.

Thanks.
 

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