Find Laplace transform of t sin2t

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

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/t

Laplace transform of t sint

Laplace transform of sin²t

Laplace transform of sin2t sin3t

Laplace transform of sint sin2t sin3t

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