laplacetransform e^{-t}tsin(2t)

Lösung

laplace transformation

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

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

Schritte zur Lösung

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

Wende Laplace-Transformationstabelle an:

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

Für

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

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

L {e^{- t} sin (2 t)} : \frac{2}{( s + 1 ) ^{2} + 4}

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

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

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

\frac{d}{d t} (2 t^{- \frac{3}{4}})

x \to \infty lim ((e)^{x})

\frac{\partial}{\partial x} (x^{2} (\frac{y}{x}))

(f (x) (x)^{g (x)})^{'}

x \to 2 lim (\frac{4 x ^{2} - 2 x + 5}{x})