The Laplace transform of t sin2t is equal to L{t sin2t}= 4s/(s2+4)2. In this post, we learn how to find the Laplace of t times sin2t.
The Laplace formula of tsin2t is given by
L{t sin2t}= $\dfrac{4s}{(s^2+4)^2}$.
Table of Contents
Find Laplace of t sin2t
Question: What is the Laplace of tsin2t?
Solution:
If L{f(t)}=F(s), then the multiplication by t Laplace formula states that
L{tn f(t)} = (-1)n $\dfrac{d^n}{ds^n} \big(F(s) \big)$ …(I)
Put n=1 and f(t) = sin2t.
So F(s) = L{sin2t} = $\dfrac{2}{s^2+4}$ as we know L{sin at} = a/(s2+a2).
By the above formula (I),
L{t sin2t} = (-1)1 $\dfrac{d}{ds} \Big( \dfrac{2}{s^2+4} \Big)$
= $-2\dfrac{d}{ds} (s^2+4)^{-1}$
= -2 × $[-1(s^2+4)^{-2} \frac{d}{ds}(s^2+4)]$ by the power rule of derivatives.
= -2 × $-2s(s^2+4)^{-2}$
= $\dfrac{4s}{(s^2+4)^2}$
So the Laplace transform of tsin2t is 4s/(s2+4)2.
More Laplace Transforms:
Laplace transform of sin2t sin3t
Laplace transform of sint sin2t sin3t
FAQs
Answer: The Laplace transform of tsin2t is equal to 4s/(s2+4)2.
Answer: L{t sin2t}= 4s/(s2+4)2.
