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

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

What is the Laplace of t² cosat

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

Explanation:

Step 1:

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 $\dfrac{d^n}{ds^n}(F(s))$ …(∗)

Step 2:

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

Now,

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

Step 3:

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:

L{t sint}	L{t cost}
L{sin2t}	L{cos2t}
L{sin2t sin3t}	L{t et}
L{cost cos2t}	L{t2}

FAQs