Find the Laplace transform of t^2 u(t-1)

The Laplace transform of t^2 u(t-1) is equal to e^-s[2/s³ + 2/s² +1/s]. Here we find the Laplace of t²u(t-1) using the second shifting property of Laplace transforms.

The formula of the Laplace of t²u(t-1) is given as

L{t²u(t-1)} = $e^{-s}\Big[ \dfrac{2}{s^3} + \dfrac{2}{s^2} +\dfrac{1}{s}\Big]$.

What is the Laplace of t²u(t-1)?

Answer: The Laplace of t2u(t-1) is e-s[2/s3 + 2/s2 +1/s].

Explanation:

To find the Laplace of t²u(t-1) we will use the second shifting theorem of Laplace transforms which says that if L{f(t)} = F(s), then

L{f(t-a) u(t-a)} = e^-as F(s) for a>0 …(∗)

Comparing f(t-a) u(t-a) with the given function t²u(t-1), we get that

a=1 and f(t-1)=t².

So f(t) = (t+1)² = t²+2t+1.

Hence, F(s) = L{t²+2t+1} = 2/s³ + 2/s² +1/s, by the formula L{tⁿ} = n!/sⁿ⁺¹.

Now, using the formula (∗), we obtain that

L{t²u(t-1)} = $e^{-s}\Big[ \dfrac{2}{s^3} + \dfrac{2}{s^2} +\dfrac{1}{s}\Big]$.

So the Laplace transform of t²u(t-1) is equal to e^-s[2/s³ + 2/s² +1/s], and this is proved by using the second shifting property of Laplace transforms.

Related Topics:

Laplace transform of unit step function, L{u(t)}

Find the Laplace transform of u(t-1)

Find the Laplace transform of u(t-2)

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