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}= $\dfrac{4(3s^2-4)}{(s^2+4)^3}$.
Table of Contents
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:
FAQs
Answer: The Laplace transform of t2 sin2t is 4(3s2-4)/(s2+4)3.
Answer: L{t2 sin 2t}= (12s2-16)/(s2+4)3.
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