What is laplacetransform t*sin(k*t) ?

The answer to laplacetransform t*sin(k*t) is (2ks)/((s^2+k^2)^2)

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laplacetransform tsin(kt)

laplace transform

t \cdot sin (k \cdot t)

\frac{2 k s}{( s ^{2} + k ^{2} ) ^{2}}

Solution steps

L {t sin (k t)}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

t sin (k t) : f (t) = sin (k t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {sin (k t)})

L {sin (k t)} : \frac{k}{s ^{2} + k ^{2}}

\frac{d}{d s} (\frac{k}{s ^{2} + k ^{2}}) = - \frac{2 k s}{( s ^{2} + k ^{2} ) ^{2}}

= (- 1)^{1} (- \frac{2 k s}{( s ^{2} + k ^{2} ) ^{2}})

= \frac{2 k s}{( s ^{2} + k ^{2} ) ^{2}}

x \to - 4 lim (- 5 x - 8)

x y^{^{'}} - 2 y = x^{2} - x - 2

y^{^{'}} = x e^{y - x^{2}}

a re a x = 2 y^{2}, x = 2

y^{^{''}} - 2 y^{^{'}} - 15 y = e^{(3 x)} cos (2 x)

What is laplacetransform t*sin(k*t) ?
The answer to laplacetransform t*sin(k*t) is (2ks)/((s^2+k^2)^2)