Question

inverse Laplace transform of 1/s is
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Answer 1

Answer:

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Step-by-step explanation:

Now the inverse Laplace transform of 2 (s−1) is 2e1 t. Less straightforwardly, the inverse Laplace transform of 1 s2 is t and hence, by the first shift theorem, that of 1 (s−1)2 is te1 t.

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Answer 2

The inverse Laplace transform of 1 is 1/s.

• The Inverse Laplace Transform is the transformation into a function with respect to time.
• Let y(a) be  a unique function that is continuous on [0,  ∞]. If it satisfies $L [y(a)](b) = Y (b)$, then it is called an Inverse Laplace transform of Y(b).

• Proof:

      Let f (t) = 1

      Now,

      F (s) = L { f ( t ) }

    = $\int\limits^\infty_0 {1.e^-^s^t dt$

   = $\lim_{b \to \infty} (\int\limits^b_0 {1.e^-^s^t} \, dt)$

  = $\lim_{b \to \infty} (-\frac{e^-^s^t}{s})^b_0$

  = $\lim_{b \to \infty} \frac{-e^-^s^b}{s} + \frac{1}{s}$

     = $\frac{1}{s}$

Hence, the inverse Laplace transform of 1 will be 1/s.