laplacetransform 3\delta(t)+2+4e^{-3t+2}

Lösung

laplace transformation

3 δ (t) + 2 + 4 e^{- 3 t + 2}

3 + \frac{2}{s} + \frac{4 e ^{2}}{s + 3}

Schritte zur Lösung

L {3 δ (t) + 2 + 4 e^{- 3 t + 2}}

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)}

= 3 L {δ (t)} + L {2} + 4 e^{2} L {e^{- 3 t}}

L {δ (t)} : 1

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

L {e^{- 3 t}} : \frac{1}{s + 3}

= 3 \cdot 1 + \frac{2}{s} + 4 e^{2} \frac{1}{s + 3}

Fasse

31 + \frac{2}{s} + 4 e^{2} \frac{1}{s + 3}

zusammen:

3 + \frac{2}{s} + \frac{4 e ^{2}}{s + 3}

= 3 + \frac{2}{s} + \frac{4 e ^{2}}{s + 3}

\frac{\partial}{\partial x} (\frac{y}{1 + x ^{2}})

d er i v a t i v e (60 + 2 t)^{\frac{2}{3}}

f (x) = 3^{4 x + 5}

t an g e n t y = - 1 - 8 x^{2}, (- 2, - 33)

f (x) = x^{\frac{2}{5}}