# Laplace transform of te^t

The Laplace transform of tet is equal to 1/(s-1)2. In this article, we will learn how to find the Laplace transform of the product function tet.

## What is the Laplace Transform of tet?

Answer: The Laplace transform of tet is 1/(s-1)2 when s>1.

Proof:

We will find the Laplace transform of tet by definition. Recall the definition of the Laplace transform of f(t) which is given below:

L{f(t)} = $\int_0^\infty$ f(t) e-st dt.

Put f(t) = tet.

∴ L{tet} = $\int_0^\infty$ tet e-st dt

= $\lim\limits_{T\to \infty} \int_0^T$ t e-(1-s)t dt

Using integrating by parts formula, the above is

= $\lim\limits_{T\to \infty} \Big( \left[ t \cdot \dfrac{e^{(1-s)t}}{1-s} \right]_{t=0}^T$ $- \int_0^T \dfrac{e^{(1-s)t}}{1-s} \, dt\Big)$

= $\dfrac{1}{1-s} \lim\limits_{T \to \infty} \Big(T e^{(1-s)T}$ $- \int_0^T e^{(1-s)t} \, dt\Big)$

= $\dfrac{1}{s-1} \lim\limits_{T \to \infty} \int_0^T e^{(1-s)t} \, dt$ [as s>1, we have e(1-s)T 0 when T∞]

= $\dfrac{1}{s-1} \lim\limits_{T \to \infty} \Big[ \dfrac{e^{(1-s)t}}{1-s} \Big]_{t=0}^T$

= $\dfrac{1}{s-1} \lim\limits_{T \to \infty} \Big[ \dfrac{e^{(1-s)T}}{1-s} – \dfrac{1}{1-s} \Big]$

= $\dfrac{1}{s-1} \Big[ 0 – \dfrac{1}{1-s} \Big]$

= $\dfrac{1}{(s-1)^2}$

So the Laplace transform of tet by definition is 1/(s-1)2.

Prove that Laplace of tet is 1/(s-1)2.

Proof:

Recall the multiplication by tn Laplace transform formula, we have that:

$L\{t f(t)\} = – \dfrac{d}{ds}(F(s))$, where L{f(t)}=F(s)

Put f(t) = et. Note that F(s) = L{et}=1/(s-1).

Then L{tet} = $- \dfrac{d}{ds}\left( \dfrac{1}{s-1} \right)$

= $- \left(- \dfrac{1}{(s-1)^2} \right)$

= $\dfrac{1}{(s-1)^2}$.

Thus, the Laplace of tet is 1/(s-1)2.

## Laplace of tet by Differentiation

We prove that the Laplace of tet is 1/(s-1)2.

Proof:

We know that the Laplace transform of first derivative of f(t) is given as follows:

$L\{f'(t)\}=f(0)+s L\{f(t)\}$ …(I)

Let f(t) = tet

Then we have from (I) that

L{et+tet} = 0+sL{tet}

By the linearity of Laplace transform, we get that

L{et} + L{tet} = s L{tet}

⇒ L{et} = (s-1) L{tet}

⇒ 1/(s-1) = (s-1) L{tet} as the Laplace of et is 1/(s-1).

⇒ L{tet} = 1/(s-1)2.

Thus, we prove that the Laplace transform of tet by first derivative formula is equal to 1/(s-1)2.

## FAQs

Q1: What is Laplace of tet?

Answer: The Laplace of tet is 1/(s-1)2.

Q2: What is Laplace of et?

Answer: The Laplace of et is 1/(s-1).

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