laplacetransform e^tsin(2t)

Lösung

laplace transformation

e^{t} sin (2 t)

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

Schritte zur Lösung

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

Wende Transformationsregel an: wenn

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

dann

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

Für

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

= L {sin (2 t)} (s - 1)

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

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

\int - \frac{1}{n ^{2}} cos (n x) d x

t \to \infty + lim (\frac{1}{4 t})

\frac{d}{d x} (a + b cos (x) + cs in (x))

\frac{d}{d x} (\frac{9}{1 - x})

x \to 5 + lim (\frac{1}{x - 5})