What is laplacetransform 20te^{-2t}*sin(4t) ?

The answer to laplacetransform 20te^{-2t}*sin(4t) is (80(2s+4))/((s^2+4s+20)^2)

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laplacetransform 20te^{-2t}*sin(4t)

laplace transform

20 t e^{- 2 t} \cdot sin (4 t)

\frac{80 ( 2 s + 4 )}{( s ^{2} + 4 s + 20 ) ^{2}}

Solution steps

L {20 e^{- 2 t} t sin (4 t)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 20 L {t e^{- 2 t} sin (4 t)}

L {t e^{- 2 t} sin (4 t)} : \frac{4 ( 2 s + 4 )}{( s ^{2} + 4 s + 20 ) ^{2}}

= 20 \cdot \frac{4 ( 2 s + 4 )}{( s ^{2} + 4 s + 20 ) ^{2}}

Refine

20 \frac{4 ( 2 s + 4 )}{( s ^{2} + 4 s + 20 ) ^{2}} : \frac{80 ( 2 s + 4 )}{( s ^{2} + 4 s + 20 ) ^{2}}

= \frac{80 ( 2 s + 4 )}{( s ^{2} + 4 s + 20 ) ^{2}}

ln (ln (x))

x \to 0 + lim (\frac{x}{- 3 x ^{3} - 10 x})

y = \frac{1}{x ^{5}}, (- 2, - \frac{1}{32})

\int \frac{2}{( x - 4 ) ( x ^{2} + 2 x + 6 )} d x

y = e^{4 x^{2} + 1}

What is laplacetransform 20te^{-2t}*sin(4t) ?
The answer to laplacetransform 20te^{-2t}*sin(4t) is (80(2s+4))/((s^2+4s+20)^2)