Laplace Transform of t sint | L{t sin at}

The Laplace transform of t sint is 2s/(s²+1)², that is, L{t sint}=2s/(s²+1)². The function t sint is the product of t and the sine of t. In this article, we will find the Laplace transform of both t sint and t sin(at).

What is the Laplace Transform of t sin(t)?

Answer: The Laplace transform of t sin t is 2s/(s²+1)².

Proof:

To find the Laplace transform of t sin(t), we will use the formula for the Laplace transform of a function f(t) multiplied by t, denoted by L{t f(t)}, which is given by the following formula: (multiplication by t Laplace transform formula)

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

Step 1: Put f(t) = sin t in the above formula.

∴ F(s) = L{f(t)} = L{sin t} = 1/(s²+1).

Step 2: So the Laplace transform of tsin(t) by (∗) is equal to

$L\{t\sin t\} = – \dfrac{d}{ds}\left(\dfrac{1}{s^2+1}\right)$

Step 3: By quotient rule of derivatives, we obtain that

$L\{t\sin t \}$ $= – \dfrac{(s^2+1)\frac{d}{ds}(1)-1 \frac{d}{ds}(s^2+1)}{(s^2+1)^2}$

$= – \dfrac{(s^2+1)\cdot 0- 1\cdot 2s}{(s^2+1)^2}$

$= – \dfrac{-2s}{(s^2+1)^2}$

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

So the Laplace transform of tsin t is 2s/(s²+1)².

Also Read:

What is the Laplace Transform of t sin(at)?

Answer: The Laplace transform of t sin at is 2as/(s²+a²)².

Proof:

Put f(t) = t sin(at) in the above formula (∗).

Note that L{sin at} = a/(s²+a²). So the Laplace transform of t sin(at) by the above formula (∗) is equal to

$L\{t\sin(at)\} = – \dfrac{d}{ds}\left(\dfrac{a}{s^2+a^2}\right)$

$= – \dfrac{(s^2+a^2)\frac{d}{ds}(a)-a \frac{d}{ds}(s^2+a^2)}{(s^2+a^2)^2}$

$= – \dfrac{(s^2+a^2)\cdot 0-a \cdot 2s}{(s^2+a^2)^2}$

$= – \dfrac{-2as}{(s^2+a^2)^2}$

$= \dfrac{2as}{(s^2+a^2)^2}$.

So the Laplace transform of tsin at is 2as/(s²+a²)².

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Dr. T

This article is written by Dr. T, an expert in Mathematics (PhD). On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.