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

The answer to laplacetransform te^{2t}sin(t) is (2s-4)/((s^2-4s+5)^2)

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laplacetransform te^{2t}sin(t)

laplace transform

t e^{2 t} sin (t)

\frac{2 s - 4}{( s ^{2} - 4 s + 5 ) ^{2}}

Solution steps

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

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

L {e^{2 t} sin (t)} : \frac{1}{( s - 2 ) ^{2} + 1}

\frac{d}{d s} (\frac{1}{( s - 2 ) ^{2} + 1}) = - \frac{2 s - 4}{( s ^{2} - 4 s + 5 ) ^{2}}

= (- 1)^{1} (- \frac{2 s - 4}{( s ^{2} - 4 s + 5 ) ^{2}})

= \frac{2 s - 4}{( s ^{2} - 4 s + 5 ) ^{2}}

\frac{d y}{d x} + 2 y = e^{- x}

X^{3} - X^{2} + 4 X + 5

f (x) = \frac{125}{x ^{2} + 24}, at x = - 1

\frac{d}{d x} (cos (9 x^{2}))

What is laplacetransform te^{2t}sin(t) ?
The answer to laplacetransform te^{2t}sin(t) is (2s-4)/((s^2-4s+5)^2)