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

Lösung

laplace transformation

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

\frac{5 ( 2 s + 10 )}{( s ^{2} + 10 s + 50 ) ^{2}}

Schritte zur Lösung

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

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

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

\frac{d}{d s} (\frac{5}{( s + 5 ) ^{2} + 25}) = - \frac{5 ( 2 s + 10 )}{( s ^{2} + 10 s + 50 ) ^{2}}

= (- 1)^{1} (- \frac{5 ( 2 s + 10 )}{( s ^{2} + 10 s + 50 ) ^{2}})

= \frac{5 ( 2 s + 10 )}{( s ^{2} + 10 s + 50 ) ^{2}}

\int 6 e^{7} d x

\frac{d}{d x} (e^{2 x} + 3)

y^{^{''}} + 5 y^{^{'}} + 6 y = e - 3 x

\int \frac{e ^{2 t}}{e ^{t} + 3} d t

\frac{d y}{d x} + 6 y = 4