Laplace transform of t^2 sin2t | Find L{t^2 sin2t}

The Laplace transform of t^2 sin2t is equal to L{t^2 sin 2t}= 4(3s²-4)/(s²+4)³. Here we learn how to find the Laplace of t square sin2t.

The Laplace formula of t²sin2t is given by

L{t² sin 2t}= $\dfrac{4(3s^2-4)}{(s^2+4)^3}$.

Find Laplace of t² sin2t

Answer: The Laplace of t2 sin2t is 4(3s2-4)/(s2+4)3.

Explanation:

The Laplace transform of t² sin2t can be computed using the multiplication by t Laplace formula, which we state below:

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

where L{f(t)}=F(s). Put n=2 and f(t) = sin2t.

∴ F(s) = L{sin2t} = $\dfrac{2}{s^2+4}$ by the formula L{sin at} = a/(s²+a²). Now, by the above formula (I), we get

L{t² sin2t} = (-1)² $\dfrac{d^2}{ds^2} \Big( \dfrac{2}{s^2+4} \Big)$

= $\dfrac{d}{ds} \dfrac{d}{ds} \Big( \dfrac{2}{s^2+4} \Big)$

= $\dfrac{d}{ds}$ $\Big( \dfrac{(s^2+4) \frac{d}{ds}(2) – 2 \frac{d}{ds}(s^2+4)}{(s^2+4)^2} \Big)$ using the quotient rule of derivatives.

= $\dfrac{d}{ds} \Big( \dfrac{-4s}{(s^2+4)^2} \Big)$ as the derivative of a constant is zero, that is, d/ds(2)=0.

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

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

= $\dfrac{(s^2+4)(-4s^2-16+16s^2)}{(s^2+4)^4}$

= $\dfrac{12s^2-16}{(s^2+4)^3}$

So the Laplace transform of t²sin2t is (12s²-16)/(s²+4)³.

More Laplace Transforms:

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

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