inverse Laplace transform of the function 1(s−4)3
is

a.
e4tt36

b.
e−4tt22

c.
None

d.
e4tt22

To find the inverse Laplace transform of the function \(\frac{1}{(s-4)^3}\), we can use the following formula:

\[
\mathcal{L}^{-1} \left\{ \frac{1}{(s-a)^n} \right\} = \frac{t^{n-1}}{(n-1)!} e^{at}
\]

Here, \(a = 4\) and \(n = 3\). Plugging in these values:

\[
\mathcal{L}^{-1} \left\{ \frac{1}{(s-4)^3} \right\} = \frac{t^{3-1}}{(3-1)!} e^{4t} = \frac{t^2}{2} e^{4t}
\]

This corresponds to option **d** \( \frac{t^2}{2} e^{4t} \).

So the correct answer is:

**d. \( \frac{t^2}{2} e^{4t} \)**

Would you like further details or have any questions? Here are some related questions:

1. What is the Laplace transform of \(t^2 e^{4t}\)?
2. How do you find the inverse Laplace transform of \(\frac{1}{(s-a)^n}\) in general?
3. What is the inverse Laplace transform of \(\frac{1}{(s-3)^4}\)?
4. How can we apply the Laplace transform to solve differential equations?
5. What are the conditions for the existence of a Laplace transform?

**Tip:** Always remember that the inverse Laplace transform of a shifted function \(\frac{1}{(s-a)^n}\) involves an exponential term \(e^{at}\) multiplied by a polynomial in \(t\).

Learn how to find the inverse Laplace transform of the function 1/(s-4)^3. Detailed solution using the inverse Laplace transform formula. Suitable for advanced mathematics students.

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