laplacetransform e^{-3t}sin(4t)

Lösung

laplace transformation

e^{- 3 t} sin (4 t)

\frac{4}{( s + 3 ) ^{2} + 16}

Schritte zur Lösung

L {e^{- 3 t} sin (4 t)}

Wende Transformationsregel an: wenn

L {f (t)} = F (s)

dann

L {e^{a t} f (t)} = F (s - a)

Für

e^{- 3 t} sin (4 t) : f (t) = sin (4 t), a = - 3

= L {sin (4 t)} (s + 3)

L {sin (4 t)} : \frac{4}{s ^{2} + 16}

= \frac{4}{( s + 3 ) ^{2} + 16}

d er i v a t i v e f (x) = 6 3 x

\int \frac{25 - 4 x ^{2}}{x} d x

\int cot^{3} (3 x) csc^{5} (3 x) d x

\frac{d}{d x} (4 \frac{e ^{5 x}}{x ^{2} + 1})

n \to \infty lim (sin (\frac{nπ}{2}))