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

Lösung

laplace transformation

t e^{(- 3 t)} 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}}

\frac{\partial}{\partial y} (y^{2} + x y^{3})

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

\int (2 sin (4 x) - 8 cos (3 x)) d x

\int_{0}^{1} x e^{- 5 x} d x

\frac{d}{d x} ((x + 4)^{3} (x - 3)^{6})