laplacetransform te^{-3t}sin(5t)

Lösung

laplace transformation

t \cdot e^{- 3 t} \cdot sin (5 t)

\frac{5 ( 2 s + 6 )}{( s ^{2} + 6 s + 34 ) ^{2}}

Schritte zur Lösung

L {e^{- 3 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^{- 3 t} sin (5 t) : f (t) = e^{- 3 t} sin (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 3 t} sin (5 t)})

L {e^{- 3 t} sin (5 t)} : \frac{5}{( s + 3 ) ^{2} + 25}

\frac{d}{d s} (\frac{5}{( s + 3 ) ^{2} + 25}) = - \frac{5 ( 2 s + 6 )}{( s ^{2} + 6 s + 34 ) ^{2}}

= (- 1)^{1} (- \frac{5 ( 2 s + 6 )}{( s ^{2} + 6 s + 34 ) ^{2}})

= \frac{5 ( 2 s + 6 )}{( s ^{2} + 6 s + 34 ) ^{2}}

d er i v a t i v e \frac{x ^{2}}{5 + x}

d er i v a t i v e - \frac{1}{( x - 1 ) ^{2}}

\int 2 sin (x) + cos (x) - 3 d x

\frac{d}{d y} (e^{- y})

in v erse l a pl a ce x^{\frac{1}{2}}