What is laplacetransform te^{2t}sin(6t) ?

The answer to laplacetransform te^{2t}sin(6t) is (6(2s-4))/((s^2-4s+40)^2)

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laplacetransform te^{2t}sin(6t)

laplace transform

t e^{2 t} sin (6 t)

\frac{6 ( 2 s - 4 )}{( s ^{2} - 4 s + 40 ) ^{2}}

Solution steps

L {e^{2 t} t sin (6 t)}

Use Laplace Transform table:

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

For

t e^{2 t} sin (6 t) : f (t) = e^{2 t} sin (6 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{2 t} sin (6 t)})

L {e^{2 t} sin (6 t)} : \frac{6}{( s - 2 ) ^{2} + 36}

\frac{d}{d s} (\frac{6}{( s - 2 ) ^{2} + 36}) = - \frac{6 ( 2 s - 4 )}{( s ^{2} - 4 s + 40 ) ^{2}}

= (- 1)^{1} (- \frac{6 ( 2 s - 4 )}{( s ^{2} - 4 s + 40 ) ^{2}})

= \frac{6 ( 2 s - 4 )}{( s ^{2} - 4 s + 40 ) ^{2}}

\frac{d}{d x} (e^{x} cos (x) - x^{3} sin (x) + π)

y^{^{'}} = \frac{y}{50} 5, y (0) = 30

(7, 6), (- 7, - 7)

\frac{d y}{d x} = \frac{y}{x} + (\frac{y}{x})^{2} - 1

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{1 + e ^{t}}

What is laplacetransform te^{2t}sin(6t) ?
The answer to laplacetransform te^{2t}sin(6t) is (6(2s-4))/((s^2-4s+40)^2)