Find the Laplace transform of: 2t^5.

Find the Laplace transform of: 2t^5.

The Correct Answer and Explanation is:

To find the Laplace transform of the function:f(t)=2t5f(t) = 2t^5f(t)=2t5

we use the standard Laplace transform formula:L{tn}=n!sn+1,for n≥0\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad \text{for } n \geq 0L{tn}=sn+1n!​,for n≥0

In this case, n=5n = 5n=5. So:L{t5}=5!s6=120s6\mathcal{L}\{t^5\} = \frac{5!}{s^{6}} = \frac{120}{s^6}L{t5}=s65!​=s6120​

Since the original function is 2t52t^52t5, we multiply the Laplace transform of t5t^5t5 by 2:L{2t5}=2⋅120s6=240s6\mathcal{L}\{2t^5\} = 2 \cdot \frac{120}{s^6} = \frac{240}{s^6}L{2t5}=2⋅s6120​=s6240​

Final Answer:

L{2t5}=240s6\boxed{\mathcal{L}\{2t^5\} = \frac{240}{s^6}}L{2t5}=s6240​​


Explanation (300 words):

The Laplace transform is a mathematical tool used to change a function from the time domain to the complex frequency domain. This transformation is useful in solving differential equations, especially in engineering and physics, because it simplifies operations such as differentiation and integration.

For polynomial functions of the form tnt^ntn, where nnn is a non-negative integer, the Laplace transform has a known formula:L{tn}=n!sn+1\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}L{tn}=sn+1n!​

This formula makes it straightforward to compute the Laplace transform of any power function. The “!” symbol denotes factorial, meaning that n!n!n! is the product of all positive integers from 1 to nnn.

In our case, the function is 2t52t^52t5. We treat the constant 2 as a scalar multiplier. The Laplace transform of a constant times a function is equal to the constant multiplied by the Laplace transform of the function. That is:L{af(t)}=a⋅L{f(t)}\mathcal{L}\{af(t)\} = a \cdot \mathcal{L}\{f(t)\}L{af(t)}=a⋅L{f(t)}

So we first find L{t5}\mathcal{L}\{t^5\}L{t5}, which is 5!s6=120s6\frac{5!}{s^6} = \frac{120}{s^6}s65!​=s6120​, and then multiply by 2 to get 240s6\frac{240}{s^6}s6240​.

This result can now be used in solving problems involving differential equations where such terms appear, helping to convert them into simpler algebraic expressions.

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