# Laplace Transform of t^2 cosat | L{t^2 cost}

The Laplace transform of t2cosat is equal to L{t2 cosat}= 2s(s2-3a2)/(s2+a2)3. In this post, we will learn how to find the Laplace of t square cosat.

The Laplace formula of t2cosat is given by

$\boxed{ L\{t^2 \cos at \}= \dfrac{2s(s^2-3a^2)}{(s^2+a^2)^3} }$.

Table of Contents

## What is the Laplace of t2 cosat

Explanation:

Using the multiplication by t Laplace formula we can compute the Laplace transform of t2 cosat. If L{f(t)}=F(s) the the formula states as follows:

L{tn f(t)} = (-1)n $\dfrac{d^n}{ds^n}(F(s))$ …(∗)

Put n=2 and f(t) = cosat.

Now,

F(s) = L{cosat} = $\dfrac{s}{s^2+a^2}$ using the Laplace formula of cosat.

So from the above formula (), we have that

L{t2 cosat} = (-1)2 $\dfrac{d^2}{ds^2} \Big( \dfrac{s}{s^2+a^2} \Big)$

∴ L{t2 cosat} = $\dfrac{d}{ds} \dfrac{d}{ds} \Big( \dfrac{s}{s^2+a^2} \Big)$

= $\dfrac{d}{ds}$ $\Big( \dfrac{(s^2+a^2) \frac{d}{ds}(s) – s \frac{d}{ds}(s^2+a^2)}{(s^2+a^2)^2} \Big)$ using the quotient rule of derivatives.

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

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

= $\dfrac{(s^2+a^2)^2 \frac{d}{ds}(a^2-s^2) – (a^2-s^2) \frac{d}{ds}(s^2+a^2)^2}{(s^2+a^2)^4}$

= $\dfrac{-2s(s^2+a^2)^2 +(s^2-a^2) \cdot 2(s^2+a^2)\cdot 2s}{(s^2+a^2)^4}$

= $\dfrac{(s^2+a^2)(-2s^3-2a^2s+4s^3-4a^2s)}{(s^2+a^2)^4}$

= $\dfrac{(s^2+a^2)(2s^3-6a^2s)}{(s^2+a^2)^4}$

= $\dfrac{2s(s^2-3a^2)}{(s^2+a^2)^3}$

So the Laplace transform of t2cosat is 2s(s2-3a2)/(s2+a2)3.

## Laplace of t2 cost

From above we know that the Laplace of t2cosat is given by

L{t2cosat} = 2s(s2-3a2)/(s2+a2)3.

Here we put a=1.

So the Laplace of t2cost is equal to L{t2cost} = 2s(s2-3)/(s2+1)3.

More Laplace Transforms:

## FAQs

Q1: What is the Laplace transform of t2cosat?

Answer: The Laplace transform of t2 cosat is 2s(s2-3a2)/(s2+a2)3.

Q2: Find L{t2 cos at}.

Answer: L{t2 cos at}= 2s(s2-3a2)/(s2+a2)3.

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