laplacetransform te^{-t}sin(5t)

Lösung

laplace transformation

t e^{- t} sin (5 t)

\frac{5 ( 2 s + 2 )}{( s ^{2} + 2 s + 26 ) ^{2}}

Schritte zur Lösung

L {e^{- t} t sin (5 t)}

Wende Laplace-Transformationstabelle an:

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

Für

t e^{- t} sin (5 t) : f (t) = e^{- t} sin (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- t} sin (5 t)})

L {e^{- t} sin (5 t)} : \frac{5}{( s + 1 ) ^{2} + 25}

\frac{d}{d s} (\frac{5}{( s + 1 ) ^{2} + 25}) = - \frac{5 ( 2 s + 2 )}{( s ^{2} + 2 s + 26 ) ^{2}}

= (- 1)^{1} (- \frac{5 ( 2 s + 2 )}{( s ^{2} + 2 s + 26 ) ^{2}})

= \frac{5 ( 2 s + 2 )}{( s ^{2} + 2 s + 26 ) ^{2}}

\frac{d}{d x} (\frac{1}{2 + 3 x})

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