laplacetransform f(t)=(e^t-e^{-t})^2

Lösung

laplace transformation

f (t) = (e^{t} - e^{- t})^{2}

\frac{8}{s ( s - 2 ) ( s + 2 )}

Schritte zur Lösung

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

Schreibe

(e^{t} - e^{- t})^{2}

um:

e^{2 t} - 2 + e^{- 2 t}

= L {e^{2 t} - 2 + e^{- 2 t}}

Verwende die lineare Eigenschaft der Laplace-Transformation:

Für Funktionen

f (t), g (t)

und Konstanten

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {e^{2 t}} - L {2} + L {e^{- 2 t}}

L {e^{2 t}} : \frac{1}{s - 2}

L {2} : \frac{2}{s}

L {e^{- 2 t}} : \frac{1}{s + 2}

= \frac{1}{s - 2} - \frac{2}{s} + \frac{1}{s + 2}

Vereinfache

\frac{1}{s - 2} - \frac{2}{s} + \frac{1}{s + 2} : \frac{8}{s ( s - 2 ) ( s + 2 )}

= \frac{8}{s ( s - 2 ) ( s + 2 )}

y^{^{'}} + 8 y = 3 x

\frac{\partial}{\partial u} (2 uv)

d er i v a t i v e (x - 2) (3 x + 1)^{2}

\frac{d y}{d x} = e^{6 x} + 9 y

a re a 5 x - x^{2}, x, 4, 4